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## Introduction

In recent years, two-dimensional (2D) nanomaterials have been described as strongholds across the fields of nanotechnology with extensive applications in electronics, catalysis, drug delivery, photonics or magnetics.1 Layered double hydroxides (LDHs) are a particular class of 2D materials. Described as a member of the anionic clay family, LDHs are composed of ‘brucite-like’ cationic layers where an inclusion of trivalent cations introduces an overall positive charge to the nanosheets. Charge compensating anions are located in the interlayer galleries leading to the generalised LDH formula, [M2+1−xM3+ x (OH)2]x+[An]x/n·mH2O, where M2+/M3+ = divalent/trivalent metal cation and An = interlayer.2 A representation of a general LDH structure is displayed in Fig. 1.

LDHs are versatile materials with high industrial and academic interest, having applications in drug delivery,3 water oxidation,4 catalysis,5 supercapacitors,6 gas absorbents,7 nanocomposites,8 and transistors,9. Two exciting members of the LDH family are magnesium-aluminium (Mg-Al) LDH and nickel-iron (Ni-Fe) LDH. The former has previously been applied in flame retardant studies,10,11 polymer composites12 and catalyst supports.13 The latter also exhibits broad applicability in flame retardancy,14 sensors,15 electrocatalysis,16 water oxidation,17 oxygen evolution18 and energy storage.19 The application of Ni-Fe LDH in these fields is largely owed to their inherent material properties such as simplistic methods of fabrication, low cost, large specific surface areas, greater active sites and ability to interact with various catalytic supports. In fact, the calcination of LDH materials in general has received an abundance of attention from both an applications and scientific perspective.20,21,22 The associated mechanisms of this thermal degradation have also been well established and described via a complex procedure of dehydration, dehydroxylisation and decarbonisation of the lamellar material.23,24 While factors such as composition,25 morphology,26 structure and atmospheric conditions27 can affect these calcination behaviours, many studies have utilised the calcination procedures for applications in areas such as oxygen evolution16 and catalytic supports.13 In particular, it is the generation of metal oxides and spinel structure composites derived from LDH calcinations that contribute to the enhancement of catalytic properties across several catalysis-related fields including nanocatalyst design,28,29 hydrogenation30 and photocatalysis.31 Moreover, the application of LDH materials in flame retardant composites has attracted interest in recent times. The incorporation of the lamellar structures into polymer materials has shown beneficial results of thermal stability, melting temperatures and smoke suppression as well as an attenuation of peak heat release rate in a number of cases.32 Evidently, the thermal decomposition of the LDH structures plays an important role in both catalytic performance and flame retardant properties of LDH nanocomposites. Despite this extensive work throughout such areas of research, the majority of these works have largely relied on macroscopic characterisation techniques such as X-ray diffraction (XRD), Fourier transform infrared spectroscopy (FTIR) and thermogravimetric analysis (TGA).8,27,33,34,35 In addition to these characterisation techniques, pre-mortem and post-mortem electron microscopy can provide important information. However, none of these pre-mortem and post-mortem methods can be used to elucidate on how to optimise the processes that occur under real-time conditions at single nanoparticle level. A detailed in situ TEM characterisation, observing thermal transformations at individual nanoparticle level, has yet to be established. Determining the correct structure–property–function relationships requires a detailed description of the material in its working state. Recent technological advances in EM allow us to image materials with a range of in situ techniques that have been developed to follow the evolution of materials in the presence of such external stimuli. In addition to observing materials in their working state, in situ techniques can also capture important intermediate transitional forms that may be involved in phase transformations. As such, previous research acknowledges that these exact transformations in which the LDH decomposes are not fully understood.10 Utilising such in situ techniques help to answer the underlying questions regarding the thermal transformations of the LDH material from a nanoscale perspective. Ultimately, these behaviours can be monitored and analysed in real time, providing a further understanding and optimisation of the related processes. Here, we report on the application of state-of-the-art in situ transmission electron microscopy to reveal the processes by which LDHs thermally decompose and to characterise the stages at which these morphological and crystallographic alterations occur. Due to their broad application and well-established calcination processes in the previous literature, we elected to focus our attention on Mg-Al LDH and Ni-Fe LDH for in situ TEM analysis. Undoubtedly, there is an urgent need to fully characterise and understand these materials from a single particle perspective, given their prospective nanoscale applications in flame retardancy, energy storage and catalysis.

## Results and discussion

In both Mg-Al LDH and Ni-Fe LDH cases, a molar ratio of the cationic sites of 3 was selected as a representative of these LDH compositions due to its well-established behaviours and previous successful applications.36,37,38,39,40 The as-synthesised materials were first characterised by XRD, FTIR, TGA and atomic absorption spectroscopy (AAS). The FTIR spectrum of the precursor Ni-Fe and Mg-Al LDH samples is shown in Figure S1a. The broad bands at approximately 3,400 and 1,645 cm−1 are due to the various types of O-H bonding in both materials such as the hydroxyl groups on the brucite-like layers and the adsorbed/inter-gallery water molecules. The sharp absorption band at approximately 1350 cm−1 is ascribed to the stretching modes of CO32−, originating from the intercalated carbonates.39 There is also evidence of C-N stretching modes coming from tertiary amines at approximately 1,155 and 1,040 cm−1 absorption bands in the FTIR spectrum. Moreover, absorption bands below 1,000 cm−1 are attributed to lattice stretching modes of metal-O sites. TGA analysis was also conducted on both LDH compositions (Figure S1b). Firstly, there is a significant mass loss observed (10%) in the 75–150 °C range and is attributed to the dehydration of surface adsorbed and interstitial water. Further mass losses (approximately 20%) in the 200–800 °C temperature range is also observed for both compositions, and is related to the dehydroxylation and decomposition of the counter ions and brucite-like layers themselves.41 AAS analysis demonstrated a Ni:Fe molar ratio of 3.54:1.

In situ and ex situ heating TEM experiments were comparatively carried out in parallel to fully understand the development of LDH thermally induced calcination, as well as the various crystallographic transitions occurring before that. For this purpose, the as-synthesised LDH samples were subjected to identical heating ramp conditions both in ex situ and in situ experimental set-ups (Figure S2). Further details can be found in the Methods section.

This parallel set-up determined if the behaviours of LDH transformations were independent of the different in situ and ex situ experimental conditions. For example, the significant pressure variations as the samples were heated may have impacted the thermal behaviours of the LDH nanomaterials. More importantly, there can be an appreciable contribution of radiation beam damage (in the form of radiolysis or knock-on) or Joule sample heating when the electron beam interacts with the sample.42 In view of these possible issues, full TEM characterisation of the various LDH samples were performed pre and post ex situ heating experiments, as well as during in situ analysis; this comparison was of crucial importance in order to rule out any eventual electron beam-induced transition.

We investigated if the LDH degradation processes are dependent on both the composition and platelet size of the material. As such, the structural transformation of a Ni-Fe LDH material was studied via the application of in situ heating TEM. Fig. 2 displays the BFTEM and associated SAED patterns at various stages of the in situ heating experiments for the Ni-Fe LDH material. Initially, the material was synthesised to have a well-defined hexagonal morphology with a lateral dimension on the order of microns. This starting material was found to have an LDH crystallographic structure, with the associated SAED pattern revealing the (101), $$\left( {01\bar 1} \right)$$, $$\left( {11\bar 2} \right)$$ and (110) crystallographic planes (Fig. 2e). Upon heating, both a morphological and crystallographic change was evidenced. As the sample reached 250 °C, the loss of the (101), $$\left( {01\bar 1} \right)$$ and $$\left( {11\bar 2} \right)$$ reflections indicates a collapse in the crystal structure and a reduction in the non-basal crystallographic phases while still retaining its hexagonal morphology. The loss of such planes is believed to be due to the dehydration of the interstitial galleries. Moreover, the contrast variations of BFTEM images (Fig. 2a, b) indicates a surface alteration at this temperature, possibly as a consequence of the additional loss of water both from the surface of the material and the interlayer galleries. The irradiation of the electron beam perhaps also contributed to the dehydration of the LDH materials; however, this is difficult to avoid.43 As the Ni-Fe LDH was heated to 450 °C, the well-defined hexagonal morphology is retained (Fig. 2c). However, the evolution of spherical particles, <50 nm in size, were seen to generate and randomly distribute themselves onto the hexagonal parent material (Fig. 2c). There appears to be minimal crystallographic tranformations from 250 °C to 450 °C (Fig. 2f, g). This is indicated by the conservation of the (110) LDH plane. As the sample was calcined to 850 °C, there is clear evidence of a further generation of these particles (Fig. 2d). These nucleations become more numerous, increase in size and randomly distribute themselves across the hexagonal LDH platelet (Fig. 2d). The SAED pattern of the calcined sample at 850 °C shows a crystal structure alteration (Fig. 2h). In contrast to the patterns recorded at 250 °C and 450 °C, there is a generation of new reflections in the diffraction pattern, corresponding to newly arranged crystallographic planes. The evolution of such particles and a variance in the diffraction patterns during the calcination procedures confirms a rearrangement of the LDH structures into a new crystallographic phase. It is believed that this is a topotactic transformation caused by the in situ heating effects. Further analysis of the SAED patterns recorded at 850 °C from in situ TEM experiments (Fig. 2h) portrays the emergence of the {200} and {220} family of planes of the NiO and trevorite structures, respectively. The reflections occurring at 13.1 nm−1 may also derive from the parent LDH structure. This suggests that certain LDH planes may be retained even after thermal degradation procedures. In this regard, the SAED findings revealed an evolution to a mixed phase of LDH, metal oxides and trevorite structures (Fig. 2e–h).

To further understand the characteristics of the calcined Ni-Fe structures, we post-analysed the Ni-Fe LDH samples using energy-filtered TEM (EFTEM) and high-resolution TEM (HRTEM) (Figs. 3 and 4, respectively). Further details of the EFTEM technique can be found in the Methods section. EFTEM was favoured over scanning TEM (STEM) for acquiring spectroscopic data in situ such as energy-dispersive x-ray (EDX) and electron energy loss spectroscopy (EELS) to reduce the risk of inducing further transformations due to the beam-sample interaction and not as a consequence of applied thermal environments. Also, the application of EFTEM allowed for the direct analysis of the Ni and Fe from the whole platelet via the acquisition of two EFTEM maps (Fig. 3a). This application was preferred over STEM-EELS methods as a longer timescale for spectroscopic acquistions is required for such LDH platelet regions. The STEM approach may have also introduced issues of drift and potential alterations of crystallographic environments and, in turn, would have impacted the true reflection of the Ni-Fe LDH calcination behaviours.

EFTEM revealed the distribution of Ni and Fe throughout the calcined products (Fig. 3a). In contrast to the Ni evolution, Fe was found to have a more regular distribution across the platelet, with a localisation in the centre of the calcined material. Larger NiO particles tended to form on the central sections of the platelets, with smaller NiO crystallites arranged on the platelets edge regions. This may have been due to the non-uniform local crystallographic environments of the cationic sites, whereby the different environments at the edge regions of the platelets affected the growth of both NiFe2O4 trevorite and NiO oxide nanoparticles. Moreover, it is perceived that the catalytic activity at these edge regions were not only influenced by Fe sites44 but were also influenced by smaller sized NiO particles. We also used the EFTEM technique to directly visualise the Ni-Fe thermal transformations as they took place (Movie V1 Supplementary Information). Our experimental set-up utilised signals from the excitations of the electrons in the Ni 2p3/2 orbital (855 eV), that is, the Ni L3 EELS edge, to form the intensities in such video representations.45 As the platelets were subject to 850 °C, the nucleation of Ni containing particles was clearly evidenced. Particles initially evolved and migrated in a random fashion on the platelet surfaces. Further exposure at this temperature led to a sudden transformation where the remaining material transformed to an array of smaller similar Ni containing particles. This somewhat restricted size is perhaps due to the limited local availability of Ni cations after the initial transformations occurred.

Complimentary to the EFTEM data, an elemental distribution of the calcined Ni-Fe LDH material was further analysed in STEM. An associated EDX elemental map confirmed that the spherical nanoparticles were largely composed of nickel (green signal) (Figure S6B). The effect of the background was corrected for by the subtraction of the average intensity value in the related Fe map (red signal) (Figure S6B). It is also noted that small amounts of Fe may have been present either within or in the vicinity of the particle, suggested by the feint red intensities across the elemental map. Moreover, the low amounts of Fe was illustrated by the recorded EDX sum spectra from areas 1 to 3 across the particle (Figure S6C and D). Although this may have suggested that there is an existence of Fe migration into the NiO particles, the Fe-K EDX line peak at 6.4 keV was not statistically significant across the particle and was deemed to be undetectable above background counts. Hence, the Fe signals are speculated to be due to EDX background counts.

Furthermore, we employed electron diffraction in STEM mode to investigate the variation in the orientation of the metal oxides (Supporting Information Figure S3). This overcame the drawback of SAED techniques in TEM, where the dimensions of the selected area apertures limit the recording of diffraction patterns from regions of minimum 100 nm in size. Whereas the application of electron diffraction using a STEM electron probe permitted the characterisation of features of approximately 10 nm, that is, the size range of the products of calcination procedures. In addition, the application of electron diffraction studies in STEM allowed us to investigate the crystallographic orientations across the calcined Ni-Fe structures. The variations in the recorded patterns show that the NiO particles assembled with random orientations relative to the basal plane of the LDH platelet (Figure S3d and f). This is also viewed in the NiFe2O4 trevorite regions of the calcined structure (Figure S3c and e). It is deduced that the products of mixed metal oxides derived from the Ni-Fe LDH calcinations arrange themselves randomly in a topotactic fashion throughout the parent material, but confine themselves to within the platelet domains.

Energy-dispersive x-ray spectroscopy in STEM quantitatively assessed the calcined products of the Ni-Fe degradations. While the Fe content appeared to be ubiquitously distributed, STEM-EDX revealed that there is a notable higher Ni content on the thermally evolved spherical particles (Fig. 3b and Table 1 (Supplementary Information)). This yielded additional evidence that the evolved particles are that of NiO, and also correlated well with the data acquired by EFTEM methods.

An investigation by HRTEM showed the presence of the NiFe2O4 trevorite lattice structure surrounding the spherical particles resulting in a heterogeneous structure (Fig. 4e). FFT analysis of the associated regions (Fig. 4g) confirmed the existence of the {220} and {242} family of planes (indicated by red indices in Fig. 4g), where the crystallite region had a $$[\bar 11\bar 1]$$ orientation. This can also be corroborated to the Fe signal as found by EFTEM methods (Fig. 3a). Similar analysis of the spherical nanoparticles directly resolved the lattice spacings attributed to the {200} planes of the NiO structure with an orientation along the [023] axis, hence confirming the evolution of metal oxide particles as a result of the Ni-Fe LDH calcination. However, there was a discrepancy in the correlation of the STEM nanoprobe diffraction and HRTEM of similar particle regions, whereby the $$\left( {11\bar 1} \right)$$ family of planes found in the STEM nanoprobe acquisitions did not fully coincide with the FFT analysis of our HRTEM data (Fig. 4c) and Supporting Information S3f. We perceive this to be due to the evolution of a non-stoichiometric NiO compound as opposed to a pure NiO phase. The existence of non-stochiometric NiOs has been previously investigated by Da Rocha and Rougier.46

As aforementioned, the parallel TEM analysis of samples heated ex situ investigated the applicability of our in situ EM methodologies as well as assessed the effect of the electron beam on the LDH thermal evolution. Figure 5 presents the TEM and SAED studies applied to Ni-Fe LDH samples heated with ex situ experimental conditions.

The samples heated to 250 °C ex situ presented an irregular platelet morphology (Fig. 5a). This evidently contrasts the precursor LDH platelets as well as the LDH materials heated to 250 °C using in situ methods (Fig. 2a, b, respectively). The corresponding SAED pattern showed that these irregular sheet-like materials display an LDH crystallographic structure, demonstrated by the {101} and {110} planes (Fig. 5d). Similarly, these crystallographic planes were evidenced as the sample was heated to 450 °C using ex situ methods (Fig. 5e). The BFTEM findings (Fig. 5b) are also consistent with in situ studies at this temperature, in particular the irregular morphology and formation of particles on the platelets surface (Fig. 5b). The morphology demonstrated an arrangement of smaller crystallite regions at 850 °C (Fig. 5c) with similar crystallographic features (Fig. 5f). These observations of ex situ heated materials may also be perceived as a reconstructed LDH due to its exposure to aqueous environments, that is, the LDH ‘memory effect’.47 Moreover, the effect of rehydration due to exposure to air conditions can also be viewed from the analysis of the in situ samples, 40 days after experimentation (Figure S7). The hexagonal symmetry in the associated SAED patterns highlighted a partial revert to the LDH structure (Fig. 5f and Figure S7B). However, the existence of trevorite phases remain, as shown by HRTEM analysis in both ex situ and in situ experiments (Figure S10B and Fig. 4, respectively). Combinative EFTEM images of the ex situ heated sample up to 850 °C highlighted an even distribution of Ni and Fe throughout the material. This was evidenced by the ubiquitous EFTEM intensities of Ni (green) and Fe (red), respectively (Figure S9A). The non-specific location of either Ni or Fe contrasts the EFTEM findings from in situ experiments (Fig. 3a). This is acknowledged to be the effect of rehydration of the heated LDH material due to its exposure to moist air conditions as a consequence of sample preparation, where the Ni and Fe sites could revert back to an LDH structure. A potential reconstruction to an LDH structure was also verified by STEM-EDX, demonstrating a greater Ni content in the analysed region (areas 3 and 4 in Figure S8B and Table 2 in Supporting Information).

Despite maintaining the focus of interest on the nanoscale features responsible for the thermal evolution of these LDH samples, it is acknowledged that the above findings may not have reflected their bulk material counterparts. Hence, we compared these behaviours via the application of in situ and ex situ XRD analysis (Figure S11A and B, respectively). Powder samples were measured at intermediate temperature stages of aforementioned heating protocols (Figure S2). In particular, XRD was strategically conducted firstly at room temperature and then samples were heated to 250 °C for 2 h followed by 450 °C for 2 h and finally to 850 °C for 2 h and post-analysed at each stage (Figure S11A(a)–(d) and Figure S11B(a)–(d), respectively). At room temperature, the in situ XRD analysis revealed typical LDH (003), (006), (101), (015) and (110) crystallographic planes at 2θ angles 16°, 26.6°, 39.4°, 45.7° and 70.9°, respectively.48 Upon thermal treatment to 250 °C, a lateral shift of the (003) and loss of (006) basal planes suggests an expansion of the unit cell and a reduction in basal plane periodicity. This is attributed to the dehydration of the associated H2O content from interlayer LDH galleries. Moreover, the retention of the LDH (110) plane is observed in both in situ TEM and in situ XRD approaches (Fig. 2g and Figure S11A(b)), validating how single nanoparticle behaviours can affect bulk material characteristics. As the samples were subject to 450 °C in situ, a phase transition took place. The loss of the LDH diffraction peaks and the appearance of the (111), (200) and (220) of NiO occurred at 43.4°, 50.6° and 74.1°, respectively.49 This behaviour was also viewed via the in situ TEM experiments, where the birth of the spherical NiO particles was observed at this temperature (Fig. 2c). At 850 °C, the (111), (200) and (220) diffraction peaks of NiO became more pronounced. This is believed to be due to the growth and nucleation of further NiO particles, as observed in our in situ TEM experiments (Fig. 2d). It is also possible that these diffraction peaks could be attributed to (222), (400) and (440) of the NiFe2O4 phase, but this is difficult to distinguish due to their ambiguity with the NiO peaks. Parallel ex situ XRD experiments were also carried out on the Ni-Fe LDH samples. At room temperature, the XRD peaks at 2θ angles 5.3°, 10.32°, 15.7°, 17.6° and 20.7° corresponded to the (003), (006), (101), (104) and (018) planes of the Ni-Fe LDH crystallographic planes (Figure S11B(a)).50 Similar traits relative to in situ XRD approaches were evidenced at 250 °C. The loss of the (006), (101), (104) and (018) peaks indicate a breakdown in the Ni-Fe LDH crystal structure. At this temperature, the XRD patterns also showed a decreasing and broadening of the (003) peaks indicating an alteration in basal spacings (Figure S11B (b)). The evolution of the NiO (111), (200) and (220) planes and the NiFe2O4 (220), (311), (222), (400), (511), (440), (622) and (444) planes were evidenced at 450 °C and 850 °C, respectively (Figure S11B(c) and (d) respectively). The evolution of these NiO phases is also mirrored via our in situ XRD approach. Moreover, the NiO {200} and NiFe2O4 {220} planar families were observed both in the SAED patterns recorded during in situ TEM experiments (Figs. 2h and 4f, respectively). However, ambiguities were also seen in selected ex situ XRD peaks of the calcined samples as they could correspond to either oxide or trevorite structures. The XRD peaks at 2θ angles 17.0°, 19.7° and 27.9° could be hypothesised to derive from the respective NiO (111), (200) and (220) or NiFe2O4 (222), (400) and (440) family of planes. This has also been observed by previous work.22 Furthermore, this indistinctness in the possible source of XRD peaks validates our motivation of applying the techniques of SAED and HRTEM. The observed correlations between XRD macroscopic techniques and EM analysis indicates that the bulk features derived from the LDH nanoscale transformations. Hence, our findings highlight the importance of understanding the nanoscale thermal degradation properties for the design and fabrication of LDH-based devices.

In light of such findings related to the Ni-Fe LDH material, we applied an identical experimental approach and analysis to a similar Mg-Al LDH structure. This yielded a comparison of both size and composition of various LDH structures with regards to their associated thermal degradations. Figure 6 displays the TEM findings from in situ heating of an Mg-Al LDH material, where the aforementioned heating ramp was imposed (Figure S2). Similar to the Ni-Fe LDH, the material appeared to have a well-defind platelet structures; however, they portrayed a width of approximately 350 nm and rounded morphology, as shown by BFTEM (Fig. 6a). Moreover, the crystallographic resemblances of these hydrotalcite platelets to the Ni-Fe compositions were seen via the existence of well-defined {100}, {110}, {120} and {330} planar families of the LDH structure (Fig. 6e).51

An overall examination of the LDH platelets as they are subject to the in situ heating ramp revealed an inhomogeneous contrast and a porous structure develops within the platelets (Fig. 6d and Movie V2 Supplementary Information). These porous structures evolved in an irregular, somewhat random, fashion on the hydrotalcite materials. The borders of the platelets were not observed to have the same behaviours, possibly due to their different crystallographic environments when compared the central domains of the materials. In contrast, they developed a denser frame on the outer regions of the heated material. Indeed, it is indicative that the composition and dimension of LDH materials has an influential effect on its thermal degradative properties. Similar features of the Mg-Al LDH degradation, in particular the evolution of a porous matrix, are also evidenced in the recent independent studies of Warringham et al.52 In contrast to the Mg-Al LDH thermal degradation, the Ni-Fe LDH does not develop a morphological variation along the borders of the platelets (Figs. 6d and 2d, respectively). Analogously, the development of a porous structure of the Ni-Fe LDH is not as pronounced as the Mg-Al counterpart. The Mg-Al LDH morphology and crystalline structures are maintained as they are subject to in situ heating up to 250 °C, as demonstrated by the TEM micrographs and electron diffraction patterns in Fig. 6b, f. An intensity attenuation in the {100} and {110} diffraction spots occurs (Fig. 6f). The loss in periodicity of these planes suggests a crystallographic rearrangement of the LDH structures and is accredited to the loss of H2O from the interstitial layers and a dehydroxlisation of the octahedral layers.23,53 Interestingly, this transformation was also paralleled in the Ni-Fe LDH structures at this temperature (Fig. 2f).

As the continuation of the in situ heating ramp regime reached 450 °C, Mg-Al platelets demonstrated increased porosity with an evident rearrangement taking place, as shown in BFTEM (Fig. 6c). The overall morphology of the LDH remained intact but an amorphization occurred. This was evidenced by in situ SAED patterns via the loss of intensity and broadening of the {100}, {110} and {120} LDH diffraction spots (Fig. 6f, g). Moreover, there are parallels in this Mg-Al LDH crystallographic transition, however, as amorphization was found to occur at approximately 450 °C when compared to the Ni-Fe degradations, which displayed an amorphization tendencies at 250 °C (Fig. 2f). Hence, it is believed that the Mg-Al LDH structural transformations are less susceptible to degradations at higher temperatures, suggesting that LDH dehydration procedures exhibit a compositional dependence. This feature was also corroborated in our TGA, where greater mass losses were observed in the Ni-Fe case (Figure S12).

At the 850 °C step, the emergence of new diffraction spots in the associated SAED patterns provided direct experimental confirmation of a crystallographic transformation from an LDH material to a combination of α-Al2O3 and MgAl2O4 spinel structures (Fig. 6h). This was evidenced by the existence of the associated {111} and {022} reflections of the α-Al2O3 and {422}, {531} and {444} of MgAl2O4 spinel.54 This evolution to mixed metal oxides was paralleled in the Ni-Fe degradations at this temperature (Fig. 2h) and was also similar to previous work.24,55 These in situ methods showed that a complete decomposition did not occur via the existence of the {110}, {120} and {300} LDH planes (Fig. 6h).

The size dependence of the LDH platelets to their calcination behaviours was also investigated (Figure S14). We applied our in situ heating experiment conditions to Mg-Al LDH platelets of relatively larger size (approximately 3 μm in lateral dimension). In comparison to the smaller Mg-Al LDH material (Fig. 6), the larger platelets were seen to have very similar degradation procedures. The reduction in crystallinity up to 450 °C (Figure S14E–G) and evolution of pores across the material at 850 °C (Figure S14D) compares to that of their smaller counterparts (Fig. 6). Overall, it was seen that the degradations were not platelet size dependent, rather the materials composition having a more profound effect on such mechanistic behaviours.

Our results yield important impacts in relation to the employment of LDHs as dye adsorbents, flame retardants and catalysts. For example, our findings potentially impart further information into the role of LDH platelets and their calcined derivatives have on the thermal stability, toxicity and structural morphology of when incorporated into polymers or resins.11 Moreover, the distribution of calcined products could be of interest for the design and testing of future nanocatalysts in relation to the dispersion and confinement of catalytic sites with enhanced catalytic activity or selectivity. The impacts of our analysis may further be extended to dye sorbents, possibly revealing insights as to how sorption capacities are affected by the extent of LDH calcinations. More generally, the application of in situ techniques elucidated the progressions of the structural alterations of the nanoplatelets. We have revealed a direct observation of the nanoscale nucleation behaviours and mechanisms involved in Ni-Fe LDH and Mg-Al LDH thermal degradation.

This study reports the application of in situ TEM to investigate and compare the thermal degradation behaviours of LDH materials. The application of in situ heating techniques in the TEM revealed the morphological and crystallographic integrity of individual LDH platelets as well as the nanoscale behaviours and progression mechanisms involved in Mg-Al LDH thermal degradation. Ni-Fe and Mg-Al LDH compounds exhibited similar behaviours but also had unique signatures in their own right. The in situ heating of Ni-Fe LDH material resulted in the transformation into a coexistence of Ni oxide-type particles arranged throughout a NiFe2O4 trevorite matrix, confirmed by TEM, HRTEM and STEM-EDX. In particular, the nucleation processes of NiO particles was directly visualised by EFTEM methods. The calcinations were also paralleled via ex situ and XRD methods, exhibiting identical phase transformations, verifying the observed behaviours by TEM are due to the inherent properties of the LDH materials. It was found that the calcined Mg-Al LDH evolved into Al2O3 and MgAl2O4 structures, effectively characterised by HRTEM and SAED. However, the transition of the Mg-Al material occurred via a different mechanism, with a development of a porous matrix as opposed to the generation of spherical particles. The application of SAED in situ effectively characterised the crystallographic transitions involved in the degradations of both Ni-Fe and Mg-Al of LDH materials.

The applicability of our experimental methodologies was confirmed by our analysis of LDH materials of a different compositions, establishing a suitable in situ approach to characterise the LDH thermal transformation behaviours. Ultimately, our successful nanoscale assessment of such mechanisms has an important bearing on their further development and strategic designs of catalysts and flame retardants as well as many additional current areas of nanotechnology.

## Methods

### Ni-Fe LDH synthesis

Ni(NO3)2·6H2O, Fe(NO3)3·9H2O, TEA and urea were dissolved in 80 ml of deionized water giving a final concentration of 7.5, 2.5, 10 and 17.5 mM, respectively, and stirred at room temperature for 24 h. Then, the reaction mixture was heated to 100 °C in an oil bath for 48 h and the obtained yellow precipitate was washed with deionized water several times.

### Mg-Al LDH synthesis

To prepare a 3:1 Mg-Al LDH with intercalated carbonate, 343 g (1.68 mol) MgCl2·6H2O and 136 g (0.56 mol) AlCl3·6H2O were dissolved in 100 ml deionized water as reactant mixture A. A second reactant mixture B was prepared by dissolving 180 g (4.5 mol) NaOH pellets and 30 g (0.28 mol) Na2CO3 in 100 ml deionized water. Subsequently, the reactant mixtures A and B were pumped together within 30 min with a syringe pump into a beaker, containing another 150 ml deionized water, under heavy shear-mixing using an ultraturrax rotor-stator-mixer. The reaction product was subsequently washed five times with deionized water via dispersing the particles in fresh deionized water and solid–liquid separating the particles via gravitational sedimentation (centrifugation). Eventually, the whole product of the washed LDH particles was redispersed in 1,500 ml of deionized water and heated at 80 °C for 4 days to yield well-ordered crystallites.

To synthesise large platelets of carbonate containing Mg-Al LDH with Mg:Al ratio of 3, Mg(NO3)2·6H2O, Al(NO3)3·9H2O and urea were dissolved in 60 ml of deionised water giving a final concentration of 0.125, 0.042 and 1.67 M, respectively. Then, the solution was transferred to 100 ml round bottom flask and immersed in an oil bath previously heated to 90 °C and continuously stirred for 48 h under reflux. After this time, the flask was cooled in a water bath for 30 min, the precipitate was separated by centrifugation (3,000 rpm/10 min) and washed three times with water.

### Experimental methods

#### Heating conditions of LDH materials for in situ and ex situ experiments

The LDH samples were initially exposed to a temperature of 30 °C for 10 min. This was followed by a ramp to 250 °C at 10 °C/min and kept at this temperature for 120 min. Subsequently, the same sample was heated to 450 °C at 10 °C/min and kept at 450 °C for 120 min. Finally, the sample was elevated to 850 °C at 10 °C/min and maintained at 850 °C for 2 h.

### TEM sample preparation

Samples were prepared for in situ TEM experiments by suspending the synthesised LDH materials in deionized water and sonicating for 30 min. Subsequently, 5 μl of the associated samples were subsequently placed DENS Solutions Nano-chipsTM for in situ experiments, respectively. Excess sample was wicked using filter paper and then air dried for approximately 30 min each before TEM analysis. Ex situ heated Ni-Fe samples were prepared for TEM analysis by directly depositing the powder forms of the materials onto lacey Carbon TEM Cu grids (TED Pella Inc., USA).

### Characterisation methods

Powder XRD was used to characterise the precursor LDH bulk sample as well as LDH samples subject to various thermal conditions. A Bruker Advance Powder X-ray diffractometer (Bruker, MA, USA) operating with a molybdenum K-α emission source ($${\mathrm{\lambda }} = 0.7107$$ Å−1) was used to record the associated diffractograms. In situ XRD experiments were conducted using a PANalytical EMRYREAN XRD with θθ Bragg–Brentano geometry. A cobalt K-α1,2 emission source ($${\mathrm{\lambda }} = 1.789$$ Å−1). To exclude any possibility of reconstruction to an LDH structure due to water exposure, we performed XRD using samples in powder form.

TGA was also conducted as a measure of the mass losses when the LDH was subject to thermal environments. TGA studies were performed using a PerkinElmer Pyris 1 TGA (PerkinElmer, MA, USA).

TEM was used to characterise the overall morphology and crystallographic structure of the associated LDH samples. TEM and SAED were conducted on a FEI Titan300 (FEI, Oregon, USA), operated at 300 kV. For comparative purposes of the in situ heating experiments, TEM magnifications and electron diffraction camera lengths were kept constant. In situ heating TEM experiments were conducted using a DENS Solutions WildfireTM TEM holder (DENS solutions, Delft, The Netherlands). This experimental in situ TEM sample holder set-up consists of a microelectromechanical systems chip design with electron transparent silicon nitride windows onto which the sample is deposited. The heating of the chip is conducted via a metallic spiralling system with suitable contacts and interpolated with an electron transparent SiN window support film, onto which the sample is directly deposited. Hence this set-up exposes the sample to thermal environments, and as such, the nanoscale properties and features of the LDH materials can be characterised as they are subject to specific elevated temperatures. Furthermore, the TEM set-up was carefully monitored to reduce the beam effects discussed above. As the sample was subject to heating, the electron beam was blanked to avoid unwarranted interactions. By doing this, we can be reassured that the discussed findings are indeed effect of the applied heating treatment rather than undesired beam-induced effects. EDX and EFTEM studies were carried out using a EDAX single SiLi EDX detector and GIF Tridiem energy filter (Gatan, USA) with a spectrometer, respectively. Our SAED experimental set-up utilised a selected area aperture to assist in measuring the relative diffraction patterns, allowing us to analyse crystallographic properties from regions much smaller than the individual hydrotalcites themselves. An analysis of such materials exposed to in situ heating environments yielded a direct visualisation of how these elevated temperatures affect the properties of LDH materials. In addition, the accompanying SAED methods permitted an investigation into the crystallographic features that occur at the various thermal stages in real time. This approach was used to directly observe the thermal degradations of hydrotalcites and assess the nanoscale features that play a significant role in these processes such as size, morphology and localised composition.

TEM images and SAED patterns were analysed using Digital Micrograph® (Gatan Inc., California, USA). The EFTEM experimental technique extends from EELS in the electron microscope. Through collection and analysis of the inelastically scattered electrons transmitted through the sample, an EEL spectrum can be acquired. At higher energies of these spectra, core-loss signals in EELS originate from the ionisation of inner shells (e.g. K, L or M) due to energy transfer from the incident electron beam, and are characteristic of which element they come from. By selecting certain energies using an energy slit width to contain such ionisation edge signals, the unwanted electrons are filtered out, resulting in images containing vital elemental information.56

### Data availability

The data related to the findings of this work are available from the corresponding author subject to reasonable request.


text:
# Group Homomorphism Preserves Inverses

## Theorem

Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.

Let $\phi: \struct {G, \circ} \to\struct {H, *}$ be a group homomorphism.

Let:

$e_G$ be the identity of $G$
$e_H$ be the identity of $H$

Then:

$\forall x \in G: \map \phi {x^{-1} } = \paren {\map \phi x}^{-1}$

Hence the notation $\map \phi x^{-1}$ can be used unambiguously.

This can be illustrated using the following commutative diagram:

$\begin{xy} \[email protected][email protected]+1em{ G \ar[r]^*{\phi} \ar[d]_*{\iota_G} & H \ar[d]^*{\iota_H} \\ G \ar[r]_*{\phi} & H }\end{xy}$

where $\iota_G$ and $\iota_H$ are the inversion mappings on $G$ and $H$ respectively.

## Proof 1

Let $x \in G$.

Then:

 $\ds \map \phi x * \map \phi {x^{-1} }$ $=$ $\ds \map \phi {x \circ x^{-1} }$ Definition of Group Homomorphism $\ds$ $=$ $\ds \map \phi {e_G}$ Definition of Inverse Element $\ds$ $=$ $\ds e_H$ Group Homomorphism Preserves Identity

So, by definition, $\map \phi {x^{-1} }$ is the right inverse of $\map \phi x$.

$\Box$

Similarly:

 $\ds \map \phi {x^{-1} } * \map \phi x$ $=$ $\ds \map \phi {x^{-1} \circ x}$ Definition of Group Homomorphism $\ds$ $=$ $\ds \map \phi {e_G}$ Definition of Inverse Element $\ds$ $=$ $\ds e_H$ Group Homomorphism Preserves Identity

So, again by definition, $\map \phi {x^{-1} }$ is the left inverse of $\map \phi x$.

$\Box$

Finally, as $\map \phi {x^{-1} }$ is both:

a left inverse of $\map \phi x$

and:

a right inverse of $\map \phi x$

it is by definition an inverse.

From Inverse in Group is Unique, $\map \phi {x^{-1} }$ is the only such element.

Hence the result.

$\blacksquare$

## Proof 2

A direct application of Homomorphism to Group Preserves Inverses.

$\blacksquare$

## Proof 3

From Group Homomorphism of Product with Inverse, we have:

$\forall x, y \in G: \map \phi {x \circ y^{-1} } = \map \phi x * \paren {\map \phi y}^{-1}$

Putting $x = e_G$ and $y = x$ we have:

 $\ds \map \phi {x^{-1} }$ $=$ $\ds \map \phi {e_G \circ x^{-1} }$ $\ds$ $=$ $\ds \map \phi {e_G} * \paren {\map \phi x}^{-1}$ $\ds$ $=$ $\ds e_H * \paren {\map \phi x}^{-1}$ Group Homomorphism Preserves Identity $\ds$ $=$ $\ds \paren {\map \phi x}^{-1}$

$\blacksquare$

format: plain


text:
Math Help - Word Problems

1. Word Problems

Hey Guys, I'm having a lot of problems trying to answer a word problem and could use some help thanks.
....its a long one so sorry

A point on the outside of the axle of a truck has a circular motion that can be modelled by a since curve.
If you measure distance from the axle's centre to the bottom of the truck, that distance remains constant as long as the truck is on a level road. If the truck goes over a bump, the springs absorb the shock but the truck bounces for a while.
A truck is moving so that the axle, which is 6cm in diameter, rotates at 1 rotation per second. As the truck hits a bump, the spring depresses by 20cm and then continues to depress by 80% of the previous bounce as it bounces every half-second.
If the middle of the axle is 30cm from the bottom of the truck, construct a graph and equation illustrating the distance from a point on the outside of the axle to the bottom of the truck.

The question is only worth 2 marks but I would still like to try to get it....you don't need to make the graph but helping with the equation would be nice....

Thanks

2. Superposition

First, the equation: $h(t)=30+20(.80)^{2t}\cos(4\pi t)+3\sin(2\pi t)$

Now, the explanation, term by term.

"30" represents the "natural" height of the center of the axle above the ground.
"20" represents the initial amplitude of the bump.
" $(.80)^{2t}$" means that every half second, the amplitude of the bump decreases by 80%
" $\cos()$" is used because when t=0, the truck has just hit the bump and is 20cm above its natural height, beginning its decline, hence the entire term $20(.80)^{2t}\cos(4\pi t)$=+20 at t=0.
" $4\pi t$" is used because the period of the bounce is half a second and the argument should be $\frac{2\pi t}{P}$ for period P.
"3" represents the amplitude of the point spinning around the axle. (half the diameter)
"sin" is used arbitrarily, as the problem does not state the orientation of this point at the exact moment the truck hits the bump.
" $2\pi$" is the argument used because the axle has a period of 1 second.

Graph these three terms separately and it is easy to see. The straight line at h=30 represents the trucks "natural" height above the ground, the equilibrium of the system. The cosine term is a half second period bounce steadily decreasing in amplitude, and the sine term represents the point on the axle happily spinning round.

Lastly, we appeal to the Law of Superposition, which despite it's scary name, simply means that by adding all the separate terms governing this motion together, we get the end result of this motion, albeit a complicated-looking graph.

3. Thank you so much!
There is no way I would have gotten this without you!

4. Question

Hi,

I would like to know why does 2t in 20(0.8)^2t means "half second".

Thank you


text:
Chemical Kinetics Test 4

Total Questions:50 Total Time: 75 Min

Remaining:

Questions 1 of 50

Question:The rate of a gaseous reaction is given by the expression $$K\,[A]\,[B]$$. If the volume of the reaction vessel is suddenly reduced to 1/4th of the initial volume, the reaction rate relating to original rate will be

1/10

1/8

8

16

Questions 2 of 50

Question:A catalyst increases the rate of reaction because it

Increases the activation energy

Decreases the energy barrier for reaction

Decreases the collision diameter

Increases the temperature coefficient

Questions 3 of 50

Question:For the reaction $${N_2}(g) + 3{H_2}(g) \to 2N{H_3}(g)$$ under certain conditions of temperature and partial pressure of the reactants, the rate of formation of $$N{H_3}$$ is $$0.001\,kg\,{h^{ - 1}}$$. The rate of conversion of $${H_2}$$ under the same conditions is

$$1.82\, \times {10^{ - 4}}\,kg/hr$$

$$0.0015\,kg/hr$$

$$1.52 \times {10^4}kg/hr$$

$$1.82 \times {10^{ - 14}}kg/hr$$

Questions 4 of 50

Question:A catalyst increases the rate of a chemical reaction by

Increasing the activation energy

Decreasing the activation energy

Reacting with reactants

Reacting with products

Questions 5 of 50

Question:Velocity constant of a reaction at 290 K was found to be $$3.2 \times {10^{ - 3}}$$ . At 310 K it will be about

$$1.28 \times {10^{ - 2}}$$

$$9.6 \times {10^{ - 3}}$$

$$6.4 \times {10^{ - 3}}$$

$$3.2 \times {10^{ - 4}}$$

Questions 6 of 50

Question:The temperature coefficient of a reaction is

Specific reaction rate at $${25^o}C$$

Rate of the reaction at $${100^o}C$$

Ratio of the rate constants at temperatures $${35^o}C$$and $${25^o}C$$

Ratio of the rate constants at two temperatures differing by $${1^o}C$$

Questions 7 of 50

Question:The rate law for the reaction Sucrose + Water $$\to$$ Glucose + Fructose is given by

Rate $$= K$$[sucrose] [water]

Rate $$= K$$[sucrose] [water]$$^0$$

Rate $$= K$$[sucrose]$$^0$$[water]

Rate $$= K$$[sucrose]$$^{1/2}$$[water]$$^{1/2}$$

Questions 8 of 50

Question:$$A + 2B \to C + D$$. If $$- \frac{{d[A]}}{{dt}} = 5 \times {10^{ - 4}}mol\,{l^{ - 1}}{s^{ - 1}}$$1, then $$- \frac{{d[B]}}{{dt}}$$ is

$$2.5 \times {10^{ - 4}}mol\,{l^{ - 1}}{s^{ - 1}}$$

$$5.0 \times {10^{ - 4}}mol\,{l^{ - 1}}{s^{ - 1}}$$

$$2.5 \times 10 - 3mol\,{l^{ - 1}}{s^{ - 1}}$$

$$1.0 \times {10^{ - 3}}mol\,{l^{ - 1}}{s^{ - 1}}$$

Questions 9 of 50

Question:The reaction $$2NO(g) + {O_2}(g)$$$$\leftrightarrow$$$$2N{O_2}(g)$$is of first order. If volume of reaction vessel is reduced to 1/3, the rate of reaction would be

1/3 times

2/3 times

3 times

6 times

Questions 10 of 50

Question:The reaction $${N_2}{O_5}\,({\rm{in}}\,CC{l_4}\,{\rm{solution}}) \to 2N{O_2}({\rm{solution}}) + \frac{1}{2}{O_2}(g)$$ is of first order in $${N_2}{O_5}$$ with rate constant $$6.2 \times {10^{ - 1}}{s^{ - 1}}.$$ What is the value of rate of reaction when $$[{N_2}{O_5}] = 1.25\,mole\,{l^{ - 1}}$$

$$7.75 \times {10^{ - 1}}mole\,{l^{ - 1}}{s^{ - 1}}$$

$$6.35 \times {10^{ - 3}}mole\,{l^{ - 1}}{s^{ - 1}}$$

$$5.15 \times {10^{ - 5}}mole\,{l^{ - 1}}{s^{ - 1}}$$

$$3.85 \times {10^{ - 1}}mole\,{l^{ - 1}}\,{s^{ - 1}}$$

Questions 11 of 50

Question:A reaction that is of the first order with respect to reactant A has a rate constant $$6{\min ^{ - 1}}.$$ If we start with $$[A] = 0.5\,mol\,\,{l^{ - 1}}$$, when would $$[A]$$reach the value 0.05mol $${l^{ - 1}}$$

0.384 min

0.15 min

min

3.84 min

Questions 12 of 50

Question:The rate constant for the reaction, $$2{N_2}{O_5} \to 4N{O_2}$$$$+ {O_2}$$ is $$3 \times {10^{ - 5}}{\sec ^{ - 1}}$$. If the rate is $$2.40 \times {10^{ - 5}}mol\,\,litr{e^{{\rm{ - 1}}}}{\sec ^{ - 1}}$$. Then the concentration of $${N_2}{O_5}$$(in mol litre-1) is

1.4

1.2

0.04

0.8

Questions 13 of 50

Question:For a chemical reaction $$A \to B$$ it is found that the rate of reaction doubles, when the concentration of A is increased four times. The order in A for this reaction is

Two

One

Half

Zero

Questions 14 of 50

Question:The hydrolysis of ethyl acetate is a reaction of $$C{H_3}COOEt + {H_2}O \to C{H_3}COOH + EtOH$$

First order

Second order

Third order

Zero order

Questions 15 of 50

Question:The rate of reaction between $$A$$ and $$B$$ increases by a factor of 100, when the concentration of $$A$$ is increased 10 folds. The order of reaction with respect to $$A$$ is

10

1

4

2

Questions 16 of 50

Question:A first order reaction requires 30 minutes for 50% completion. The time required to complete the reaction by 75% will be

45 minutes

15 minutes

60 minutes

None of these

Questions 17 of 50

Question:Inversion of canesugar in dilute acid (conversion into glucose and fructose) is a

Unimolecular reaction

Bimolecular reaction

Trimolecular reaction

Pseudo-unimolecular reaction

Questions 18 of 50

Question:The half life period of a first order reaction

$$\frac{{0.693}}{t}$$

$$\frac{{0.693}}{K}$$

$$\frac{{2.303}}{t}$$

$$\frac{{0.303}}{{{K_1}}}$$

Questions 19 of 50

Question:Decomposition of nitrogen pentoxide is known to be a first order reaction 75 percent of the oxide had decomposed in the first 24 minutes. At the end of an hour, after the start of the reaction, the amount of oxide left will be

Nil

Questions 20 of 50

Question:A reaction $$2A \to$$ products is found to follow zero oder kinetics, then

$$\frac{{dx}}{{dt}} = k{[A]^2}$$

$$\frac{{dx}}{{dt}} = k{[A]^0}$$

$$\frac{{dx}}{{dt}} = k[A]$$

$$\frac{{dx}}{{dt}} = k[2A]$$

Questions 21 of 50

Question:The alkaline hydrolysis of ethyl acetate is represented by the equation $$C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH$$ Experimentally it is found that for this reaction $$\frac{{dx}}{{dt}} = k[C{H_3}COO{C_2}{H_5}]\,[NaOH]$$ Then the reaction is

Bimolecular and of first order

Bimolecular and of second order

Pseudo–bimolecular

Pseudo–unimolecular

Questions 22 of 50

Question:In the reaction $$A + B \to$$ Products, if $$B$$ is taken in excess, then it is an example of

Second order reaction

Zero order reaction

Pseudounimolecular reaction

First order reaction

Questions 23 of 50

Question:The half life of a first order reaction is $$69.35\,\sec$$. The value of the rate constant of the reaction is

$$1.0\,{s^{ - 1}}$$

$$0.1\,{s^{ - 1}}$$

$$0.01\,{s^{ - 1}}$$

$$0.001\,{s^{ - 1}}$$

Questions 24 of 50

Question:The half life for the reaction $${N_2}{O_5}$$$$\leftrightarrow$$ $$2N{O_2} + \frac{1}{2}{O_2}$$ in $$24\,hrs$$ at $${30^o}C$$. Starting with $$10\,g$$ of $${N_2}{O_5}$$ how many grams of $${N_2}{O_5}$$ will remain after a period of 96 hours

$$1.25\,g$$

$$0.63\,g$$

$$1.77\,g$$

$$0.5\,g$$

Questions 25 of 50

Question:For a first order reaction, the half-life period is independent of

Initial concentration

Cube root of initial concentration

First power of final concentration

Square root of final concentration

Questions 26 of 50

Question:Order of a reaction can have

$$+ ve$$ values

Whole number values

Fractional values

All of the above

Questions 27 of 50

Question:The order of the reaction occurring by following mechanism should be (i)$${A_2} \to A + A$$ (fast) (ii)$$A + {B_2} \to AB + B$$ (slow) (iii)$$A + B \to$$ (fast)

$$1\,\frac{1}{2}$$

$$3\frac{1}{2}$$

2

None of these

Questions 28 of 50

Question:After how many seconds will the concentration of the reactants in a first order reaction be halved, if the decay constant is $${\rm{1}}{\rm{.155}} \times {\rm{1}}{{\rm{0}}^{{\rm{ - 3}}}}{\sec ^{ - 1}}$$

100 sec

200 sec

400 sec

600 sec

Questions 29 of 50

Question:What is the order of a reaction which has a rate expression rate $$= K{[A]^{3/2}}{[B]^{ - 1}}$$

2-Mar

2-Jan

0

None of these

Questions 30 of 50

Question:Which of the following expression is correct for first order reaction? $$(CO)$$ refers to initial concentration of reactant

$${t_{1/2}} \propto CO$$

$${t_{1/2}} \propto C{O^{ - 1}}$$

$${t_{1/2}} \propto C{O^{ - 2}}$$

$${t_{1/2}} \propto C{O^0}$$

Questions 31 of 50

Question:Units of rate constant of first and zero order reactions in terms of molarity M unit are respectively

$$se{c^{ - 1}},\,\,\,M\,se{c^{ - 1}}$$

$$se{c^{ - 1}},\,\,\,M$$

$$M\,se{c^{ - 1}},\,\,\,se{c^{ - 1}}$$

$$M,\,se{c^{ - 1}}$$

Questions 32 of 50

Question:The reaction $$2{N_2}{O_5}$$ $$\leftrightarrow$$ $$2{N_2}{O_4} + {O_2}$$ is

Bimolecular and second order

Unimolecular and first order

Bimolecular and first order

Bimolecular and zero order

Questions 33 of 50

Question:The half-life period for a first order reaction is 693 seconds. The rate constants for this reaction would be

$$0.1se{c^{ - 1}}$$

$$0.01se{c^{ - 1}}$$

$$0.001se{c^{ - 1}}$$

$$0.0001se{c^{ - 1}}$$

Questions 34 of 50

Question:For an elementary reaction, 2A + B$$\to$$ C + D the molecularity is

Zero

One

Two

Three

Questions 35 of 50

Question:Which of the following is an example of pseudo unimolecular reaction

$$C{H_3}COOC{H_3} + {H_2}O \to C{H_3}COOH + C{H_3}OH$$

$$C{H_3}COOC{H_3} + {H_2}O \to C{H_3}COOH + C{H_3}OH$$

$$2FeC{l_3} + SnC{l_2} \to SnC{l_4} + 2FeC{l_2}$$

$$NaOH + HCl \to NaCl + {H_2}O$$

Questions 36 of 50

Question:Hydrolysis of DDT is a first order reaction, its half life is 10 years. Time to hydrolyse 10 g DDT to half is

100 years

50 years

years

10 years

Questions 37 of 50

Question:In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is

7.5 minutes

15 minutes

30 minutes

60 minutes

Questions 38 of 50

Question:In the first order reaction, the concentration of the reactant is reduced to 25% in one hour. The half life period of the reaction is

hr

4 hr

1/2 hr

1/4 hr

Questions 39 of 50

Question:$${t_{\frac{1}{4}}}$$ can be taken as the time taken for the concentration of a reactant to drop to $$\frac{3}{4}$$ of its initial value. If the rate constant for a first order reaction is $$K$$ , the $${t_{\frac{1}{4}}}$$ can be written as

$$0.10/K$$

$$0.29/K$$

$$0.69/K$$

$$0.75/K$$

Questions 40 of 50

Question:For a first order reaction $$A \to B$$ the reaction rate at reactant concentration of $$0.01M$$ is found to be $$2.0 \times {10^{ - 5}}mol\;{L^{ - 1}}{s^1}$$ . The half life period of the reaction is

220 s

30 s

300 s

347 s

Questions 41 of 50

Question:The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is

1

2

1

2

Questions 42 of 50

Question:If a substance with half life 3 days is taken at other place in 12 days. What amount of substance is left now

1/4

1/8

1/16

1/32

Questions 43 of 50

Question:The energy of activation is

The energy associated with the activated molecules

Threshold energy – energy of normal molecules

Threshold energy + energy of normal molecules

Energy of products – energy of reactants

Questions 44 of 50

Question:Which one of the following does not represent Arrhenius equation

$$k = A{e^{ - E/RT}}$$

$${\log _e}k = {\log _e}A - \frac{E}{{RT}}$$

$${\log _{10}}k = {\log _{10}}A - \frac{E}{{2.303\,RT}}$$

$$k = A{E^{ - RT}}$$

Questions 45 of 50

Question:On increasing the temperature, the rate of the reaction increases because of

Decrease in the number of collisions

Decrease in the energy of activation

Decrease in the number of activated molecules

Increase in the number of effective collisions

Questions 46 of 50

Question:Energy of activation of a reactant is reduced by

Increased temperature

Reduced temperature

Reduced pressure

Increased pressure

Questions 47 of 50

Question:Collision theory is applicable to

First order reactions

Zero order reactions

Bimolecular reactions

Intra molecular reactions

Questions 48 of 50

Question:The rate constant of a reaction at temperature 200K is 10 times less than the rate constant at 400 K. What is the activation energy $$({E_a})$$of the reaction (R = gas constant)

1842.4 R

921.2 R

460.6 R

230.3 R

Questions 49 of 50

Question:In respect of the equation $$k = A{e^{ - {E_a}/RT}}$$ in chemical kinetics, which one of the following statement is correct

k is equilibrium constant

$${E_a}$$is energy of activation

R is Rydberg’s constant

Questions 50 of 50

Question:The rate constant is doubled when temperature increases from 27$$^\circ$$C to 37$$^\circ$$C. Activation energy in kJ is


text:
• This event has passed.

# Michał Pilipczuk, Structural properties of powers of sparse graphs

## April 7 Wednesday @ 5:00 PM - 6:00 PM KST

Zoom ID: 934 3222 0374 (ibsdimag)

### Speaker

Michał Pilipczuk
Institute of Informatics, University of Warsaw
https://www.mimuw.edu.pl/~mp248287/

For a graph G and an integer d, the dth power of G is the graph $G^d$ on the same vertex set as G where two vertices are considered adjacent if and only if they are at distance at most d in G. Assuming that G is sparse, what can we say about the structure in $G^d$? Certainly $G^d$ can be dense, as the square of a star is a complete graph, but $G^d$ still retains many properties that can be derived from the sparsity of G. We will present some recent results in this spirit, in particular connected to colorings and to the Erdős-Hajnal property, assuming that G is drawn from a fixed class of bounded expansion or from a fixed nowhere dense class. The talk will be based on the recent papers: “Clustering Powers of Sparse Graphs” (with J. Nešetřil, P. Ossona de Mendez, and X. Zhu) and “Erdős-Hajnal properties for powers of sparse graphs” (with M. Briański, P. Micek, and M. Seweryn).

## Details

Date:
April 7 Wednesday
Time:
5:00 PM - 6:00 PM KST
Event Category:
Event Tags:

## Venue

Zoom ID: 934 3222 0374 (ibsdimag)

## Organizer

O-joung Kwon (권오정)
기초과학연구원 수리및계산과학연구단 이산수학그룹
대전 유성구 엑스포로 55 (우) 34126
IBS Discrete Mathematics Group (DIMAG)
Institute for Basic Science (IBS)
55 Expo-ro Yuseong-gu Daejeon 34126 South Korea
E-mail: dimag@ibs.re.kr, Fax: +82-42-878-9209
Copyright © IBS 2018. All rights reserved.


text:
VectorRotation CoordinatedGeometry VectorMethod Intermediate
2016

Problem - 3954

The points $(0,0)$, $(a,11)$, and $(b,37)$ are the vertices of an equilateral triangle. Find the value of $ab$.

The solution for this problem is available for $0.99. You can also purchase a pass for all available solutions for$99.


text:
The font that deserves the very first mention is none other than Very nice Sans Serif. I have personally used this font many times in WPF and it worked wonders. The font is a bit square and thus makes the UI look a bit more futuristic. However, since the font comes in many different weights, you must manage the weights properly.

The next that deserves mention is Lightweight font. It is a very versatile font and works somewhat similar to the updated light futura. One thing to note is that it might look very thin for display in SilverLight and WPF. This is because SL and WPF tend to thin out the typefaces.

For making a Bold and noticeable display or headline, you may want to go with a very unique sans serif.  APS (American Philological Society) released this font which supports Unicode well and has some additional open-type features.


text:
# Chapter 9 Temporal consistency of categorical variables

Unlike continuous variables for which averages, standard deviations, and ranges convey statistical and socioeconomical meaning, categorical variables may be analyzed by tabulating the frequency of their values. Figure 9.1 allows to plot the relative frequencies of categorical variables over time for a single country. For example, the user may tabulate the absolute and relative frequencies for the values in categorical variables such as relationharm, marital, urban, and educat7 for Pakistan as presented below.

Figure 9.2 does the same, but presents the results for all eight countries at the same time. For example, the frequency of values “Yes”" and “No”" for the harmonized variable ownhouse is presented below.

This tool is useful to evaluate whether categorical variables have been harmonized properly. A large change in the relative frequency of values in a categorical variable could indicate that the harmonization process has been inconsistent. For example, if someone mistakenly exchanges the value labels for urban (i.e., rural=1 urban=0 instead of rural=0 urban=1), the inconsistency with previous survey rounds could be easily detected in these dashboards. For example, as of today, Jun/19/2019, variable computer in Pakistan (in category Assets) presents a weird trend from 2013 to 2015. This clearly indicates an error either in the harmonization or in the raw data.


text:
# “Too many math alphabets” error

I just got the error

Too many math alphabets used in version normal.

on a document I'm editing. What can I do to fix this, other than stop using an alphabet? All I've found online is instructions on what to do if you're using the bm.sty package, which I'm not calling (but maybe it's hidden in the AMS stuff?). I'll note, the thing that broke it was adding mathrsfs and using \mathscr.

Anyways, as you'll see in my preamble, I tried what they suggested anyways, and it didn't work.

My preamble:

\documentclass[11pt]{amsart}
\newcommand{\bmmax}{0}
\newcommand{\hmmax}{0}
\usepackage{amsmath, amssymb, amsthm, latexsym, amscd, enumerate, MnSymbol,bbm, etex,nicefrac,mathrsfs}

I'm using

\mathbf
\mathcal
\mathfrak
\mathscr
\mathsf
\mathbb
\mathbbm %(for a lower case blackboard bold letter)

I recognize this is a lot of math fonts, but doesn't seem unreasonable.

EDIT: getting rid of MnSymbol fixed things...

-

It looks like you've got some redundant packages in there which are using up math family slots (of which there are only 16 total). (Note that amsart also loads the amsfonts package which uses up several more for \mathfrak, \mathbb, etc.) For example, in the latexsym docs:

These fonts are not automatically included in the NFSS2/LATEX2ε since they take up important space and aren’t necessary if one makes use of the packages amsfonts or amssymb.

Furthermore, if you're using MnSymbol, doesn't that override much of the amssymb package? I could be wrong about that.

Finally, it looks the bbm family takes up five (or is it only three?) on its own—if you don't need that many you can copy out just the definitions you need, such as

\DeclareMathAlphabet{\mathbbm}{U}{bbm}{m}{n}

Bearing all this in mind, here's a minimal example that at least compiles: (I've removed a couple of unrelated packages)

\documentclass[11pt]{amsart}
\usepackage{amsmath, amssymb, amscd, MnSymbol,mathrsfs}
\DeclareMathAlphabet{\mathbbm}{U}{bbm}{m}{n}% from bbm.sty
\begin{document}
$\mathbf A \mathcal A \mathfrak A \mathscr A \mathsf A \mathbb A \mathbbm a$
\end{document}

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Honestly, I have no idea why MnSymbol was there; my coauthor added it at some point. –  Ben Webster Oct 2 '10 at 7:51
Interestingly, I must have forgotten at least one math alphabet I was using, since your example compiles on my computer, but using your trick in the actual file doesn't. As I mentioned in the edit to the question, removing MnSymbol does, though. –  Ben Webster Oct 2 '10 at 7:53
MnSymbol might be used for an obscure glyph you otherwise can't access. It would be nice, huh, if LaTeX provided better logging of where all the math alphabets got used up :) –  Will Robertson Oct 2 '10 at 8:05
A question, since both XeTeX and LuaTeX can have 256 math families (right?), is this LaTeX error still relevant or can we get ride of it? –  Khaled Hosny Oct 2 '10 at 9:54
Funny, I never thought of that. I do increase the number of math families that can be allocated in unicode-math, but I haven't done any testing on it. I should add a test file that just defines, say, 50 math families to see what happens. If it works then perhaps this should be added to xltxtra & luatextra as well. –  Will Robertson Oct 2 '10 at 10:01
show 6 more comments

I recently encountered this error when compiling a proceedings volume. Each paper by itself could be typeset without throwing this error, but when all of them were included in the same LaTeX document, the math alphabet limit was exceeded.

We were able to solve the issue by declaring a separate math version for one of the more alphabet-hungry papers. Concretely, we put the command \DeclareMathVersion{normal2} at the end of the preamble, and the commands \mathversion{normal2} and \mathversion{normal} just before and after respectively the \include for the alphabet-hungry paper. This gave that paper an alphabet allocation (managed by the \c@mv@normal2 counter) that is separate from that of the others (managed by the \c@mv@normal counter), and this allowed both to stay below the limit. One thing this trick relies upon is that new math versions are created with all settings of the old one in place, so all declarations previously made for math version normal will be in place also in math version normal2 (and normal3, and so on, if one needs to repeat the trick).

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Welcome to TeX.SX! You can have a look at our starter guide to familiarize yourself further with our format. –  Martin Schröder Jan 21 at 14:24


text:
### Archive

Archive for September, 2011

## Contour plots on R

In one of the homework problems, you are asked to do a Bayesian sensitivity analysis for a posterior probability with respect to the prior.   Specifically, suppose we are interested in learning about a proportion and we wish to compute the posterior probability $P(\theta > 0.5)$.  We observe the data (25 successes, 40 failures) and we want to know the sensitivity of the posterior probability with respect to the prior mean $\eta = a/(a+b)$ and the prior precision parameter $K = a + b$ where $a$ and $b$ are the beta shape parameters.

It is easy to construct a contour plot to illustrate this sensitivity analysis on R.

1.  First we set up a grid of values of $\eta$ and $K$:

eta = seq(.1, .9, by=.1)
K = c(1, 2, 8, 16, 32)

2.  Next we write a short function that computes the posterior probability $P(\theta > 0.5)$ as a function of  $\eta$ and $K$.

myfun=function(eta, K)
1 - pbeta(.5, K*eta+ 25, K*(1-eta) + 40)

3.  We now use the outer function to compute this probability on all combinations of  $\eta$ and $K$ — the result is a matrix where the rows correspond to $\eta$ and the columns to $K$.

PROB = outer(eta, K, myfun)

4.  Last, we use the contour function with inputs the two vectors and the matrix of probabilities.  We can specify which levels to draw on the contour graph.

contour(eta, K, PROB,
levels=c(0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.95),
xlab="eta", ylab="K", main="P(theta > 0.5)")

Categories: R work

## Prediction problem

In class, I described a simple Bayesian prediction problem.  One observes the number of fire calls in a small California city for 30 days — the total number of fire calls in this period is 158.  An administrator is interested in predicting the total number of calls in the next seven days.

We assume that y, the number of calls $y_i$ in a day, is Poisson with mean $\lambda$.  Suppose our prior beliefs about $\lambda$ are described by a normal density with mean 10 and standard deviation 2.

I first write a simple function compute.post that computes and plots the posterior (using the likelihood times prior recipe) on a grid between lo and hi.
compute.post=function(lo, hi,...){ lambda = seq(lo, hi, length=100) prior = dnorm(lambda, 10, 2) like = lambda^158 * exp(-30*lambda) post = prior * like plot(lambda, post,...) list(L = lambda, P = post) }

With some trial and error, the interval (3.3, 7.5) seems to bracket the posterior well.

S=compute.post(3.3, 7.5, type="l", col="red")

Next I take a simulated sample from the posterior by using the sample function.

sim.lambda = sample(S$L, 1000, replace=TRUE, prob=S$P)

Last, I simulate a sample from the posterior predictive density of z, the number of fire calls in seven days.  Since [z | $\lambda$] is Poisson(7 $\lambda$, we simulate z by (1) simulating $\lambda$ from the posterior and (2) simulating z from a Poisson(7 $\lambda$) distribution.  This is easy to do by using the vector properties of the random simulation functions.

sim.z = rpois(1000, 7*sim.lambda)

Here is a graph of the predictive distribution (by tabulating and plotting the simulated values of z).

Categories: Single parameter

## Brute force Bayes for one parameter

Although we talk a lot about conjugate analyses, one doesn’t need to restrict oneself to the use of conjugate priors.  Here we illustrate learning about a Poisson mean using a normal prior on the mean $\lambda$.

Suppose I observe the following number of fire calls for a particular community each week:  0, 0, 1, 3, 2, 2.  If we assume $y_i$, the number of fire calls in week $i$, is Poisson($\lambda$), then the likelihood function is given by

$L(\lambda) = \exp(- 6 \lambda) \lambda^8, \, \, \lambda>0$

Suppose I represent my prior beliefs about $\lambda$ by a normal curve with mean 3 and standard deviation 1.  Then the posterior is given by (up to an unknown proportionality constant) by

$g(\lambda | data) = L(\lambda) \times \exp(-(\lambda - 3)^2)$

Here is a simple brute-force method of summarizing this posterior.

1.  Choose a grid of $\lambda$ values that covers the region where the posterior is concentrated (this might take some trial and error).

2.  On this grid, compute the prior, likelihood, and posterior.

3.  Using the R sample function, take a large sample from the grid where the probabilities of the points are proportional to the like x prior values.

4.  Summarize this posterior simulated sample to learn about the location of the posterior.

Here’s some R code for this example.  I use the plot function to make sure the grid does cover the posterior.  The vector L contains 1000 draws from the posterior.

lambda = seq(0, 5, by=0.1)
like = exp(-6*lambda)*lambda^8
prior = dnorm(lambda, 3, 1)
post = like * prior
plot(lambda, post)
L = sample(lambda, size=1000, prob=post, replace=TRUE)
plot(density(L))  # density graph of simulated draws

Categories: Single parameter


text:
Complex Numbers in Polar and Exponential Form. 0. Summary:: Hello, my question asks if the complex exponential equation 4e^(-j) is equal to 4 ∠-180°. See . θ (theta) must be expressed in unit radians. this still outputs an answer in complex number form, what i need from matlab is my answer to be outputted in the exponential form i.e (8-7i) would be converted to '10.6e^-0.7i' if at all possible Have to multiple by 180/pi. Practice: Polar & rectangular forms of complex numbers. Convert the complex number 8-7j into exponential and polar form. Edited: KALYAN ACHARJYA on 4 Sep 2019 Accepted Answer: KALYAN ACHARJYA. However, I would like to convert it to polar form with this kind of arrangements. Not only numbers and fractions are allowed as inputs, but also the symbols (parameters) are accepted. (Complex Exponential Form) 10 September 2020. The exponential form of log, 1024 = 5 is. This exponential to polar form conversion calculator converts a number in polar form to its equivalent value in rectangular form. R cannot be a negative number. convert complex exponential number to rectangular representation? Based on this command, the angle we got is not right. Complex number forms review . See the answer. So far we have look at different ways to represent either a rotating vector or a stationary vector using complex numbers to define a point on the complex plane. To use the calculator, one need to choose representation form of complex number and input data to the calculator. Follow 641 views (last 30 days) Michael Andrews on 24 Mar 2015. Then the relationship between, Exponential, Polar and Rectangular form in defining a complex number is given as. Polar & rectangular forms of complex numbers. $$\mathbb{V} = -3 + j4$$ We will first convert the phasor from rectangular form to exponential form. In this section we’ll look at both of those as well as a couple of nice facts that arise from them. To find the modulus value of given complex number in the form of a + ib, we have to use the formula √ a ² + b ². ; The absolute value of a complex number is the same as its magnitude. Here, a0 is called the real part and b0 is called the imaginary part. Polar form of complex numbers. So, this is our imaginary axis and that is our real axis. r (cos θ + is sin θ) is known as polar form. Another factor to consider is whether you want the answers to be in rectangular form or in polar/exponential form. Converting polar form complex numbers into rectangular form. Complex Number Forms. Answers (1) Matt J on 15 Jan 2018. Show transcribed image text. Follow 16 views (last 30 days) Aditya Mehta on 15 Jan 2018 ... Vote. Sign in to comment. Exponential forms of numbers take on the format, re jθ, where r is the amplitude of the expression and θ is the phase of the expression. cosh [i T/2 -ln 3] sinh[ln 2 i] Problem 2 What is the difference of the hyperbolic functions squared, cosh2z - sinh2z? Review the different ways in which we can represent complex numbers: rectangular, polar, and exponential forms. Convert the following complex number to its rectangular form (a+bj) and to its exponential form rel®and then illustrate the number graphically on xy coordinate system: 5(cos 125° + jsin 1259) Polar & Exponential Form. You may already be familiar with complex numbers written in their rectangular form: a0 +b0j where j = √ −1. Follow 420 views (last 30 days) Dalton Houghton-Schaffer on 4 Sep 2019. Thank you. Expert Answer . 1 ⋮ Vote. Free exponential equation calculator - solve exponential equations step-by-step This website uses cookies to ensure you get the best experience. Answered: Matt J on 15 Jan 2018 how to i convert a complex number in exponential form z=5*exp^i*60°? z = + i . Lesson Video 14:59. So first let's think about where this is on the complex plane. Let me space that out a little bit more evenly. What are some examples of quaternions in rectangular and exponential form? Arrangement (in polar form) S11 S21 (S11 AND S22 IN SAME LINE) S12 S22(S12 AND S22 IN NEXT LINE) On the other hand, "[theta, rho] = cart2pol(real(a), imag(a))". Google Classroom Facebook Twitter. How to use the Euler Formula to convert a complex number from the exponential form to rectangular form. For example, you can convert complex number from algebraic to trigonometric representation form or from exponential back to algebraic, ect. is called the rectangular form, to refer to rectangular coordinates.. We will now extend the definitions of algebraic operations from the real numbers to the complex numbers. What is their sum, cosh2z + sinh z? It is the distance from the origin to the point: See and . 2. Sign in to answer this question. 2 Exponential forms of numbers take on the format, re jθ, where r is the amplitude of the expression and θ is the phase of the expression.The amplitude r must be expressed in absolute value form. Division . Show Hide all comments. Phasor Notation. (Work it out by hand using the exponential forms, don't just state it.) Exponential Form 32 = 9 3-2 = 1/9 Logarithmic Form log39 = 2 ... View question - Write 81 in exponential form. Let's say that I have the complex number z and in rectangular form we can write it as negative three plus two i. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Exponential form, polar form, Cartesian form for complex numbers question. Rectangular to Exponential Form Conversion Calculator. 1. This is the Page of Learning about Electronics where you will find a wide range and assortment of calculators. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. This is the currently selected item. 17. Write each of the following in rectangular (x+iy) form by hand, using the exponential forms to help you. complex numbers trigonometric form Euler formula e sine cosine modulus argument. Multiplying and … [2 marks] I know already. Here r stands for modulus and θ stands for argument. See . I'm not fimular with MATLAB keywords but need to … And let's see, the real part is negative three, so we could go one, two, three to the left of the origin. Label the x-axis as the real axis and the y-axis as the imaginary axis. Previous question Next question Transcribed Image Text from this Question. 0. Polar → Rectangular Conversion. Solution Ineachcasecomparethegivennumberwiththestandardformz =rejθ toidentifythemodulus r andtheargumentθ. The amplitude r must be expressed in absolute value form. Exponential Form of Complex Numbers; Euler Formula and Euler Identity interactive graph; 6. Question: B) (10 Points) Find The Rectangular And Exponential Form Of The Following --;2(5 - 5)(-1-1) -je-This problem has been solved! Geometric Interpretation . Solving a Logarithmic Equation Algebra 2 Inverse, Exponential and Logarithmic Functions. Exponential form (2) — The Davidson Group. z = a + ib = r e iθ, Exponential form with r = √ (a 2 + b 2) and tan(θ) = b / a , such that -π < θ ≤ π or -180° < θ ≤ 180° Use Calculator to Convert a Complex Number to Polar and Exponential Forms Enter the real and imaginary parts a and b and the number of decimals desired and press "Convert to Polar and Exponential". Polar Form of Complex Numbers; Convert polar to rectangular using hand-held calculator; Polar to Rectangular Online Calculator; 5. It is the distance from the origin to the point: See and . (That is, we know r ... mathonweb.com. Vote. Vote. Most people are familiar with complex numbers in the form $$z = a + bi$$, however there are some alternate forms that are useful at times. Any idea on this? Complex Numbers Complex numbers consist of real and imaginary parts. Label the x-axis as the real axis and the y-axis as the imaginary axis. Products and Quotients of Complex Numbers; Graphical explanation of multiplying and dividing complex numbers; 7. Unlike the polar form, which is expressed in unit degrees, a complex exponential number is expressed in unit radians. Write complex numbers in polar exponential form (Dec 12, 2020) )=re^jθ (Rectangular) (Polar) (Trigonometric) (Exponential, θ radians) Q1. We will often represent these numbers using a 2-d space we’ll call the complex plane. 0 Comments. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … 0 ⋮ Vote. Key Concepts. Statistica helps out parents, students & researchers for topics including SPSS through personal or group tutorials. (V3+ )500 Then express the result in the rectangular form. This representation is known as rectangular form or argand plane of a complex number. I am having trouble converting polar form complex numbers into rectangular form by writing a MATLAB script file. convert complex numbers between exponential and Cartesian (algebraic/rectangular) forms, convert complex numbers between exponential form and polar form, perform operations (addition, subtraction, multiplication, and division) on complex numbers in exponential form, use the properties of complex numbers in exponential form to solve problems. Vote. For example, if the two numbers are in rectangular form but you would like their product in polar form, it makes sense to convert them immediately and then multiply them. Check with Mathematica. How to solve a logarithmic equation by putting it into exponential form. Key Concepts. By using this website, you agree to our Cookie Policy. Subscribe to this blog. Next lesson. 0. For two complex numbers $$z_1 = x_1 + i y_1$$ and $$z_2 = x_2 + i y_2\text{,}$$ we define 4. Commented: Star Strider on 25 Mar 2015 This is the question I have and I have no idea how to write the code! Email. Looking for maths or statistics tutors in Perth? ; The absolute value of a complex number is the same as its magnitude. Suppose we have a complex number expressed in polar form and we want to express it in rectangular form. Use the exponential form to calculate the following power. 9 3-2 = 1/9 Logarithmic form log39 = 2... View question - write in... 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Numbers using a 2-d space we ’ ll call the complex number is the same as its.! Polar form complex numbers trigonometric rectangular to exponential form Euler Formula e sine cosine modulus argument and polar form to form! ; the absolute value of a complex exponential number is the same as its magnitude complex... Products and Quotients of complex number from the exponential form ( 2 ) — the group. From rectangular form + sinh z rectangular, polar form of log, 1024 = 5 is calculator, need... Group tutorials j4  we will first convert the complex number here, a0 is called real! ; 7 Strider on 25 Mar 2015 this is the question I have and I no... Cos θ + is sin θ ) is known as polar form with this of! Its magnitude on 24 Mar 2015 and b0 is called the imaginary part we know...! But also the symbols ( parameters ) are accepted from them KALYAN ACHARJYA 4. Including SPSS through personal or group tutorials conversion calculator converts a number in form! Their sum, cosh2z + sinh z ) Dalton Houghton-Schaffer on 4 Sep 2019 accepted Answer KALYAN! Converts a number in polar form with this kind of arrangements as negative three plus two I rectangular,,! And I have and I have no idea how to solve a Logarithmic equation putting... Exponential number is expressed in absolute value form the distance from the origin the... Exponential and polar form, Cartesian form for complex numbers ; Euler Formula Euler. Have the complex plane the origin to the point: See and similar! In absolute value of a complex exponential number is the distance from origin... Mehta on 15 Jan 2018... Vote from rectangular form ) Matt J on Jan... Or group tutorials Electronics where you will find a wide range and assortment of.... Students & researchers for topics including SPSS through personal or group tutorials ( )... Electronics where you will find a wide range and assortment of calculators using this website, you agree to Cookie... And the y-axis as the imaginary part idea how to write the code fractions! In their rectangular form and in rectangular form: a0 +b0j where J = √ −1 a number in form... Question Next question Transcribed Image Text from this question is their sum, cosh2z + sinh z 2-d we! What is their sum, cosh2z + sinh z plane similar to calculator! As well as a couple of nice facts that arise from them are examples. Its magnitude same as its magnitude I am having trouble converting polar.... Couple of nice facts that arise from them putting it into exponential and Logarithmic Functions as well a. Davidson group θ ) is known as polar form and we want to express it in rectangular exponential! Express the result in the rectangular plane the symbols ( parameters ) are accepted ; convert to! I am having trouble converting polar form of log, 1024 = 5 is n't. I am having trouble converting polar form to calculate the following power of a complex number expressed in unit,! And fractions are allowed as inputs, but also the symbols ( parameters ) are accepted the from. Last 30 days ) Aditya Mehta on 15 Jan 2018 I convert a complex number expressed in degrees. To express it in rectangular form J on 15 Jan 2018: See and 8-7j exponential! Two I Next question Transcribed Image Text from this question form Euler Formula rectangular to exponential form sine cosine modulus argument the. Log, 1024 = 5 is numbers: rectangular, polar form of complex numbers in the form... 15 Jan 2018 into exponential and polar form, which is expressed in absolute of. That out a little bit more evenly exponential forms, do n't just state it. $we will represent. 15 Jan 2018 so, this is our real axis and the y-axis as the part! X-Axis as the imaginary axis plus two I interactive graph ; 6 and assortment of calculators j4! To express it in rectangular form the way rectangular coordinates are plotted in the complex plane number 8-7j exponential. For argument look at both of those as well as a couple of nice facts that arise from.. Theta ) must be expressed in absolute value of a complex number from the origin the. And I have and I have no idea how to use the Euler Formula and Euler Identity interactive ;. You agree to our Cookie Policy Logarithmic Functions calculator converts a number in exponential form 2018 Vote! 'S say that I have the complex plane similar to the point: See.!, one need to choose representation form of complex numbers ; convert to. Distance from the exponential form of complex number rectangular to exponential form input data to the point: See.! A wide range and assortment of calculators equation by putting it into exponential and Logarithmic Functions Algebra. Origin to the point: See and like to convert a complex number z and in rectangular.. Numbers complex numbers ; convert polar to rectangular form we can write it as negative plus... The way rectangular coordinates are plotted in the rectangular form number from the origin to the way coordinates. A little bit more evenly writing a MATLAB script file ; the value! Our Cookie Policy θ + is sin θ ) is known as polar form complex numbers in the rectangular.! Value in rectangular and exponential forms, do n't just state it. three two. Will first convert the phasor from rectangular form by writing a MATLAB script file, would! Interactive graph ; 6 absolute value of a complex number in polar.. Use the Euler Formula and Euler Identity interactive graph ; 6 Online calculator ; 5 Euler Formula e cosine... Numbers consist of real and imaginary parts — the Davidson group already be with... ( Work it out by hand using the exponential form z=5 * exp^i * 60° point: and! Polar & rectangular forms of complex numbers written in their rectangular form 2... View question - 81. The rectangular plane need to choose representation form of log, 1024 = 5 is this website, you to. Of a complex number is expressed in absolute value of a complex number in exponential form negative plus..., one need to choose representation form of log, 1024 = 5 is θ + is sin θ is! Z and in rectangular form we can write it as negative three plus I! This exponential to polar form complex numbers into rectangular form to its equivalent value in form! Uses cookies to ensure you get the best experience express it in rectangular form or argand plane a... Houghton-Schaffer on 4 Sep 2019 accepted Answer: KALYAN ACHARJYA on 4 Sep.! The form are plotted in the complex plane the question I have the complex z! Cosine modulus argument: a0 +b0j where J = √ −1 b0 is the! Symbols ( parameters ) are accepted the Page of Learning about Electronics where you find. The absolute value form z=5 * exp^i * 60° from rectangular rectangular to exponential form by writing a script... Ll call the complex plane similar to the calculator calculator ; 5 to I convert complex!, the angle we got is not right θ + is sin θ ) is known as form. } = -3 + j4$ \$ we will first convert the complex plane θ + sin... 81 in exponential form to convert a complex number and that is, we r. Calculator converts a number in exponential form 32 = 9 3-2 = Logarithmic. Kalyan ACHARJYA on 4 Sep 2019 accepted Answer: KALYAN ACHARJYA and polar form of complex numbers ; Euler to. ; 5 in rectangular form as a couple of nice facts that from..., exponential and Logarithmic Functions Strider on 25 Mar 2015 this is the question I have no idea to. In rectangular form: a0 +b0j where J = √ −1 arise from them complex plane to... = √ −1 of real and imaginary parts the Davidson group the absolute value form in! Form complex numbers question 2-d space we ’ ll call the complex number or group tutorials form this. And Euler Identity interactive graph ; 6 Algebra 2 Inverse, exponential Logarithmic. Script file parameters ) are accepted to convert a complex number is the same as its magnitude a couple nice. And Quotients of complex numbers question plotted in the rectangular plane command, the angle we got is not....

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text:
# DBL_EPSILON Macro

The difference between 1.0 and the next larger representable double precision floating-point value.

Attention: The default size of the MPLAB XC16 double type is 32 bits, but this can be changed to 64 bits with the -fno-short-double option.

Include

<float.h>

Definition

Compiler Definition
MPLAB XC8 1.1920928955078125e-7F
MPLAB XC16 1.1920928955078125e-7F/2.2204460492503131e-16L (See attention note)
MPLAB XC32 2.2204460492503131e-16L


text: 3 −(aq) −691.1 S2−(aq) +41.8 H+(aq) 0.0 Sn2+(aq) −10.0 SO. What is the enthalpy change for the reaction of 1 mole of H 2 (g) with 1 mole of Cl 2 (g) if both the reactants and products are at standard state conditions? If you're seeing this message, it means we're having trouble loading external resources on our website. &\mathrm{692\:g\:\ce{C8H18}⟶−3.31×10^4\:kJ} of heat from the reaction. terms of standard enthalpy's of formation. 1/2 H2 (g) + 1/2 Cl2 (g) → HCl (aq) Δ r G°(298.15 K) = -31.320 ± 0.020 (×1.067) kcal/mol: 3.1: HCl (aq, 200 H2O) → HCl (aq) Δ r H°(298.15 K) = -0.212 ± 0.004 kcal/mol: 2.3: HCl (aq, 200 H2O) → HCl (aq, 600 H2O) Δ r H°(298.15 K) = -0.083 ± 0.002 kcal/mol: 2.1: HCl (g) → H+ (g) + Cl- … It's a base. The combustion of 1.00 L of isooctane produces 33,100 kJ of heat. These values are especially useful for computing or predicting enthalpy changes for chemical reactions that are impractical or dangerous to carry out, or for processes for which it is difficult to make measurements. The distances traveled would differ (distance is not a state function) but the elevation reached would be the same (altitude is a state function). can be used as a conversion factor to find the heat produced when 1 mole of HCl reacts: $ΔH =\mathrm{1\; \cancel{mol\; HCl} \times \dfrac{ −2.9\; kJ}{0.0500 \;\cancel{ mol\; HCl}} =−58\; kJ} \[\ce{HCl (aq) + NaOH(aq) \rightarrow NaCl (aq) + H2O(l)} all have the same amount of each product as well. The thermometer tracks AP® is a registered trademark of the College Board, which has not reviewed this resource. like 20.8 degrees celsius. based on temperature change and since the temperature increased, we end up with a positive result. \nonumber$. $\ce{H2}(g)+\ce{Cl2}(g)⟶\ce{2HCl}(g)\hspace{20px}ΔH^\circ_{298}=\mathrm{−184.6\:kJ} Algae can produce biodiesel, biogasoline, ethanol, butanol, methane, and even jet fuel. \nonumber_{4}=+90.2\: \ce{kJ}\:[1×(\ce{NO})]$. What is the standard enthalpy change for the reaction: \ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g)\hspace{20px}ΔH°=\:? Summing these reaction equations gives the reaction we are interested in: \[\ce{3NO2}(g)+\ce{H2O}(l)⟶\ce{2HNO3}(aq)+\ce{NO}(g) molecules can absorb energy. The distance you traveled to the top of Kilimanjaro, however, is not a state function. or released by a reaction. magnitude that we wanna know. Ozone, O3(g), forms from oxygen, O2(g), by an endothermic process. This can be obtained by multiplying reaction (iii) by $$\frac{1}{2}$$, which means that the ΔH° change is also multiplied by $$\frac{1}{2}$$: \[\ce{ClF}(g)+\frac{1}{2}\ce{O2}(g)⟶\frac{1}{2}\ce{Cl2O}(g)+\frac{1}{2}\ce{OF2}(g)\hspace{20px} ΔH°=\frac{1}{2}(205.6)=+102.8\: \ce{kJ} The process used to produce algal fuel is as follows: grow the algae (which use sunlight as their energy source and CO2 as a raw material); harvest the algae; extract the fuel compounds (or precursor compounds); process as necessary (e.g., perform a transesterification reaction to make biodiesel); purify; and distribute (Figure $$\PageIndex{5}$$). Naturally, it could be very useful to know how much heat a chemical a lot of heat to cool down. the amount of heat that leaks in or out of the system so that we can be fairly confident that any heat transfer heat capacity of water is due to the breaking in formation of hydrogen bonds that are associated with relatively small If we plug these into Hess's Many readily available substances with large enthalpies of combustion are used as fuels, including hydrogen, carbon (as coal or charcoal), and hydrocarbons (compounds containing only hydrogen and carbon), such as methane, propane, and the major components of gasoline. Some strains of algae can flourish in brackish water that is not usable for growing other crops. \[\ce{N2}(g)+\ce{2O2}(g)⟶\ce{2NO2}(g) Write the balanced chemical equation for the enthalpy of formation of nickel (II) oxide, and include the phases. The following conventions apply when we use $$ΔH$$: \[\begin {align*} &\textrm{(two-fold increase in amounts)}\label{5.4.7}\\ &\ce{2H2}(g)+\ce{O2}(g)⟶\ce{2H2O}(l)\hspace{20px}ΔH=\mathrm{2×(−286\:kJ)=−572\:kJ}\\ &\textrm{(two-fold decrease in amounts)}\\ &\frac{1}{2}\ce{H2}(g)+\dfrac{1}{4}\ce{O2}(g)⟶\frac{1}{2}\ce{H2O}(l)\hspace{20px}ΔH=\mathrm{\frac{1}{2}×(−286\:kJ)=−143\:kJ} \end {align*} \label{5.4.6B}, $\ce{ H2(g) + 1/2 O2(g) ⟶ H2O(g)} \;\;\; ΔH=\ce{−242\:kJ} \label{5.4.7B}$, Example $$\PageIndex{1}$$: Measurement of an Enthalpy Change. I'm gonna put 100 According to the US Department of Energy, only 39,000 square kilometers (about 0.4% of the land mass of the US or less than $$\dfrac{1}{7}$$ of the area used to grow corn) can produce enough algal fuel to replace all the petroleum-based fuel used in the US. If you stand on the summit of Mt. (This amount of energy is enough to melt 99.2 kg, or about 218 lbs, of ice.). We have 28.2 degrees celsius. the specific heat capacities of hundreds of substances Temperature, or the speed at Next, we see that $$\ce{F_2}$$ is also needed as a reactant. Before we further practice using Hess’s law, let us recall two important features of ΔH. \nonumber\]. &\ce{C}(s)+\frac{1}{2}\ce{O2}(g)⟶\ce{CO}(g)&&ΔH^\circ_{298}=\mathrm{−111\:kJ}\\ Have questions or comments? Enthalpy changes are typically tabulated for reactions in which both the reactants and products are at the same conditions. When 0.0500 mol of HCl(aq) reacts with 0.0500 mol of NaOH(aq) to form 0.0500 mol of NaCl(aq), 2.9 kJ of heat are produced. own pressure and volume. \nonumber\]. Now there's a simple In the two-step process, first carbon monoxide is formed: $\ce{C}_{(s)}+\dfrac{1}{2}\ce{O}_{2(g)}⟶\ce{CO}_{(g)}\;\;\;ΔH^∘_{298}=\mathrm{−111\;kJ} \label{ 5.4.12}$. \end {align*}\]. If we have values for the appropriate standard enthalpies of formation, we can determine the enthalpy change for any reaction, which we will practice in the next section on Hess’s law. Just like a ball at the top of a hill, molecules always move toward a lower energy state if they can.