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Schulze method

The Schulze method (/ˈʃʊltsə/) is an electoral system developed in 1997 by Markus Schulze that selects a single winner using votes that express preferences. The method can also be used to create a sorted list of winners. The Schulze method is also known as Schwartz Sequential dropping (SSD), cloneproof Schwartz sequential dropping (CSSD), the beatpath method, beatpath winner, path voting, and path winner.

The Schulze method is a Condorcet method, which means that if there is a candidate who is preferred by a majority over every other candidate in pairwise comparisons, then this candidate will be the winner when the Schulze method is applied.

The output of the Schulze method (defined below) gives an ordering of candidates. Therefore, if several positions are available, the method can be used for this purpose without modification, by letting the k top-ranked candidates win the k available seats. Furthermore, for proportional representation elections, a single transferable vote variant has been proposed.

The Schulze method is used by several organizations including Debian, Ubuntu, Gentoo, Pirate Party political parties and many others.

Description of the method

Ballot

The input for the Schulze method is the same as for other ranked single-winner electoral systems: each voter must furnish an ordered preference list on candidates where ties are allowed (a strict weak order).[1]

One typical way for voters to specify their preferences on a ballot (see right) is as follows. Each ballot lists all the candidates, and each voter ranks this list in order of preference using numbers: the voter places a '1' beside the most preferred candidate(s), a '2' beside the second-most preferred, and so forth. Each voter may optionally:

• give the same preference to more than one candidate. This indicates that this voter is indifferent between these candidates.
• use non-consecutive numbers to express preferences. This has no impact on the result of the elections, since only the order in which the candidates are ranked by the voter matters, and not the absolute numbers of the preferences.
• keep candidates unranked. When a voter doesn't rank all candidates, then this is interpreted as if this voter (i) strictly prefers all ranked to all unranked candidates, and (ii) is indifferent among all unranked candidates.

Computation

Let ${\displaystyle d[V,W]}$ be the number of voters who prefer candidate ${\displaystyle V}$ to candidate ${\displaystyle W}$.

A path from candidate ${\displaystyle X}$ to candidate ${\displaystyle Y}$ is a sequence of candidates ${\displaystyle C(1),\cdots ,C(n)}$ with the following properties:

1. ${\displaystyle C(1)=X}$ and ${\displaystyle C(n)=Y}$.
2. For all ${\displaystyle i=1,\cdots ,(n-1):d[C(i),C(i+1)]>d[C(i+1),C(i)]}$.

In other words, in a pairwise comparison each candidate in the path will beat the following candidate.

The strength ${\displaystyle p}$ of a path from candidate ${\displaystyle X}$ to candidate ${\displaystyle Y}$ is the smallest number of voters in the sequence of comparisons:

For all ${\displaystyle i=1,\cdots ,(n-1):d[C(i),C(i+1)]\geq p}$.

For a pair of candidates ${\displaystyle A}$ and ${\displaystyle B}$ that are connected by at least one path, the strength of the strongest path ${\displaystyle p[A,B]}$ is the maximum strength of the path(s) connecting them. If there is no path from candidate ${\displaystyle A}$ to candidate ${\displaystyle B}$ at all, then ${\displaystyle p[A,B]=0}$.

Candidate ${\displaystyle D}$ is better than candidate ${\displaystyle E}$ if and only if ${\displaystyle p[D,E]>p[E,D]}$.

Candidate ${\displaystyle D}$ is a potential winner if and only if ${\displaystyle p[D,E]\geq p[E,D]}$ for every other candidate ${\displaystyle E}$.

It can be proven that ${\displaystyle p[X,Y]>p[Y,X]}$ and ${\displaystyle p[Y,Z]>p[Z,Y]}$ together imply ${\displaystyle p[X,Z]>p[Z,X]}$.[1]:§4.1 Therefore, it is guaranteed (1) that the above definition of "better" really defines a transitive relation and (2) that there is always at least one candidate ${\displaystyle D}$ with ${\displaystyle p[D,E]\geq p[E,D]}$ for every other candidate ${\displaystyle E}$.

Example

In the following example 45 voters rank 5 candidates.

${\displaystyle {\begin{array}{|c|c|}{\text{number of voters}}&{\text{order of preference}}\\\hline 5&ACBED\\5&ADECB\\8&BEDAC\\3&CABED\\7&CAEBD\\2&CBADE\\7&DCEBA\\8&EBADC\end{array}}}$

The pairwise preferences have to be computed first. For example, when comparing A and B pairwise, there are 5+5+3+7=20 voters who prefer A to B, and 8+2+7+8=25 voters who prefer B to A. So ${\displaystyle d[A,B]=20}$ and ${\displaystyle d[B,A]=25}$. The full set of pairwise preferences is:

Matrix of pairwise preferences
${\displaystyle d[*,A]}$${\displaystyle d[*,B]}$${\displaystyle d[*,C]}$${\displaystyle d[*,D]}$${\displaystyle d[*,E]}$
${\displaystyle d[A,*]}$ 20263022
${\displaystyle d[B,*]}$ 25163318
${\displaystyle d[C,*]}$ 19291724
${\displaystyle d[D,*]}$ 15122814
${\displaystyle d[E,*]}$ 23272131

The cells for d[X, Y] have a light green background if d[X, Y] > d[Y, X], otherwise the background is light red. There is no undisputed winner by only looking at the pairwise differences here.

Now the strongest paths have to be identified. To help visualize the strongest paths, the set of pairwise preferences is depicted in the diagram on the right in the form of a directed graph. An arrow from the node representing a candidate X to the one representing a candidate Y is labelled with d[X, Y]. To avoid cluttering the diagram, an arrow has only been drawn from X to Y when d[X, Y] > d[Y, X] (i.e. the table cells with light green background), omitting the one in the opposite direction (the table cells with light red background).

One example of computing the strongest path strength is p[B, D] = 33: the strongest path from B to D is the direct path (B, D) which has strength 33. But when computing p[A, C], the strongest path from A to C is not the direct path (A, C) of strength 26, rather the strongest path is the indirect path (A, D, C) which has strength min(30, 28) = 28. The strength of a path is the strength of its weakest link.

For each pair of candidates X and Y, the following table shows the strongest path from candidate X to candidate Y in red, with the weakest link underlined.

Strongest paths
To
From
ABCDE
A N/A A-(30)-D-(28)-C-(29)-B A-(30)-D-(28)-C A-(30)-D A-(30)-D-(28)-C-(24)-E A
B B-(25)-AN/A B-(33)-D-(28)-C B-(33)-D B-(33)-D-(28)-C-(24)-E B
C C-(29)-B-(25)-A C-(29)-BN/A C-(29)-B-(33)-D C-(24)-E C
D D-(28)-C-(29)-B-(25)-A D-(28)-C-(29)-B D-(28)-CN/A D-(28)-C-(24)-E D
E E-(31)-D-(28)-C-(29)-B-(25)-A E-(31)-D-(28)-C-(29)-B E-(31)-D-(28)-C E-(31)-DN/A E
ABCDE
From
To
Strengths of the strongest paths
${\displaystyle p[*,A]}$${\displaystyle p[*,B]}$${\displaystyle p[*,C]}$${\displaystyle p[*,D]}$${\displaystyle p[*,E]}$
${\displaystyle p[A,*]}$ 28283024
${\displaystyle p[B,*]}$ 25283324
${\displaystyle p[C,*]}$ 25292924
${\displaystyle p[D,*]}$ 25282824
${\displaystyle p[E,*]}$ 25282831

Now the output of the Schulze method can be determined. For example, when comparing A and B, since ${\displaystyle (28=)p[A,B]>p[B,A](=25)}$, for the Schulze method candidate A is better than candidate B. Another example is that ${\displaystyle (31=)p[E,D]>p[D,E](=24)}$, so candidate E is better than candidate D. Continuing in this way, the result is that the Schulze ranking is ${\displaystyle E>A>C>B>D}$, and E wins. In other words, E wins since ${\displaystyle p[E,X]\geq p[X,E]}$ for every other candidate X.

Implementation

The only difficult step in implementing the Schulze method is computing the strongest path strengths. However, this is a well-known problem in graph theory sometimes called the widest path problem. One simple way to compute the strengths, therefore, is a variant of the Floyd–Warshall algorithm. The following pseudocode illustrates the algorithm.

 1 # Input: d[i,j], the number of voters who prefer candidate i to candidate j.
2 # Output: p[i,j], the strength of the strongest path from candidate i to candidate j.
3
4 for i from 1 to C
5    for j from 1 to C
6       if (i ≠ j) then
7          if (d[i,j] > d[j,i]) then
8             p[i,j] := d[i,j]
9          else
10             p[i,j] := 0
11
12 for i from 1 to C
13    for j from 1 to C
14       if (i ≠ j) then
15          for k from 1 to C
16             if (i ≠ k and j ≠ k) then
17                p[j,k] := max ( p[j,k], min ( p[j,i], p[i,k] ) )

This algorithm is efficient and has running time O(C3) where C is the number of candidates.

Ties and alternative implementations

When allowing users to have ties in their preferences, the outcome of the Schulze method naturally depends on how these ties are interpreted in defining d[*,*]. Two natural choices are that d[A, B] represents either the number of voters who strictly prefer A to B (A>B), or the margin of (voters with A>B) minus (voters with B>A). But no matter how the ds are defined, the Schulze ranking has no cycles, and assuming the ds are unique it has no ties.[1]

Although ties in the Schulze ranking are unlikely,[2] they are possible. Schulze's original paper[1] proposed breaking ties in accordance with a voter selected at random, and iterating as needed.

An alternative way to describe the winner of the Schulze method is the following procedure:

1. draw a complete directed graph with all candidates, and all possible edges between candidates
2. iteratively [a] delete all candidates not in the Schwartz set (i.e. any candidate x which cannot reach all others who reach x) and [b] delete the graph edge with the smallest value (if by margins, smallest margin; if by votes, fewest votes).
3. the winner is the last non-deleted candidate.

There is another alternative way to demonstrate the winner of the Schulze method. This method is equivalent to the others described here, but the presentation is optimized for the significance of steps being visually apparent as you go through it, not for computation.

1. Make the results table, called the "matrix of pairwise preferences," such as used above in the example. If using margins rather than raw vote totals, subtract it from its transpose. Then every positive number is a pairwise win for the candidate on that row (and marked green), ties are zeroes, and losses are negative (marked red). Order the candidates by how long they last in elimination.
2. If there's a candidate with no red on their line, they win.
3. Otherwise, draw a square box around the Schwartz set in the upper left corner. You can describe it as the minimal "winner's circle" of candidates who do not lose to anyone outside the circle. Note that to the right of the box there is no red, which means it's a winner's circle, and note that within the box there is no reordering possible that would produce a smaller winner's circle.
4. Cut away every part of the table that isn't in the box.
5. If there is still no candidate with no red on their line, something needs to be compromised on; every candidate lost some race, and the loss we tolerate the best is the one where the loser obtained the most votes. So, take the red cell with the highest number (if going by margins, the least negative), make it green—or any color other than red—and go back step 2.

Here is a margins table made from the above example. Note the change of order used for demonstration purposes.

Initial Results Table
EACBD
E 1-3917
A -17-515
C 3-713-11
B -95-1321
D -17-1511-21

The first drop doesn't help shrink the Schwartz set.

First Drop
EACBD
E 1-3917
A -17-515
C 3-713-11
B -95-1321
D -17-1511-21

So we get straight to the second drop, and that shows us the winner, E, with its clear row.

Second Drop, final
EACBD
E 1-3917
A -17-515
C 3-713-11
B -95-1321
D -17-1511-21

This method can also be used to calculate a result, if you make the table in such a way that you can conveniently and reliably rearrange the order of the candidates on both row and column (use the same order on both at all times).

Satisfied and failed criteria

Satisfied criteria

The Schulze method satisfies the following criteria:

Failed criteria

Since the Schulze method satisfies the Condorcet criterion, it automatically fails the following criteria:

Likewise, since the Schulze method is not a dictatorship and agrees with unanimous votes, Arrow's Theorem implies it fails the criterion

The Schulze method also fails

Comparison table

The following table compares the Schulze method with other preferential single-winner election methods:

Comparison of Preferential Electoral Systems
Monotonic Condorcet Majority Condorcet loser Majority loser Mutual majority Smith ISDA LIIA Clone independence Reversal symmetry Participation, Consistency Later-noharm Later-nohelp Polynomial time Resolvability
Schulze YesYesYesYesYesYesYesYesNoYesYesNoNoNoYesYes
Ranked pairs YesYesYesYesYesYesYesYesYesYesYesNoNoNoYesYes
Alternative smith NoYesYesYesYesYesYesYesNoYesNoNoNoNoYesYes
Alternative schwartz NoYesYesYesYesYesYesYesNoYesNoNoNoNoYesYes
Kemeny-Young YesYesYesYesYesYesYesYesYesNoYesNoNoNoNoYes
Copeland YesYesYesYesYesYesYesYesNoNoYesNoNoNoYesNo
Nanson NoYesYesYesYesYesYesNoNoNoYesNoNoNoYesYes
Instant-runoff voting NoNoYesYesYesYesNoNoNoYesNoNoYesYesYesYes
Borda YesNoNoYesYesNoNoNoNoNoYesYesNoYesYesYes
Baldwin NoYesYesYesYesYesYesNoNoNoNoNoNoNoYesYes
Bucklin YesNoYesNoYesYesNoNoNoNoNoNoNoYesYesYes
Plurality YesNoYesNoNoNoNoNoNoNoNoYesYesYesYesYes
Contingent voting NoNoYesYesYesNoNoNoNoNoNoNoYesYesYesYes
Coombs[4] NoNoYesYesYesYesNoNoNoNoNoNoNoNoYesYes
MiniMax YesYesYesNoNoNoNoNoNoNoNoNoNoNoYesYes
Anti-plurality[4] YesNoNoNoYesNoNoNoNoNoNoYesNoNoYesYes
Sri Lankan contingent voting NoNoYesNoNoNoNoNoNoNoNoNoYesYesYesYes
Supplementary voting NoNoYesNoNoNoNoNoNoNoNoNoYesYesYesYes
Dodgson[4] NoYesYesNoNoNoNoNoNoNoNoNoNoNoNoYes

The main difference between the Schulze method and the ranked pairs method can be seen in this example:

Suppose the MinMax score of a set X of candidates is the strength of the strongest pairwise win of a candidate A ∉ X against a candidate B ∈ X. Then the Schulze method, but not Ranked Pairs, guarantees that the winner is always a candidate of the set with minimum MinMax score.[1]:§4.8 So, in some sense, the Schulze method minimizes the largest majority that has to be reversed when determining the winner.

On the other hand, Ranked Pairs minimizes the largest majority that has to be reversed to determine the order of finish, in the minlexmax sense. [5] In other words, when Ranked Pairs and the Schulze method produce different orders of finish, for the majorities on which the two orders of finish disagree, the Schulze order reverses a larger majority than the Ranked Pairs order.

History

The Schulze method was developed by Markus Schulze in 1997. It was first discussed in public mailing lists in 1997–1998[6] and in 2000.[7] Subsequently, Schulze method users included Debian (2003),[8] Gentoo (2005),[9] Topcoder (2005),[10] Wikimedia (2008),[11] KDE (2008),[12] the Pirate Party of Sweden (2009),[13] and the Pirate Party of Germany (2010).[14] In the French Wikipedia, the Schulze method was one of two multi-candidate methods approved by a majority in 2005,[15] and it has been used several times.[16] The newly formed Boise, Idaho chapter of the Democratic Socialists of America in February chose this method for their first special election to be held in March (2018). [17]

In 2011, Schulze published the method in the academic journal Social Choice and Welfare.[1]

Users

The Schulze method is used by the city of Silla for all referendums. It is used by the Institute of Electrical and Electronics Engineers, by the Association for Computing Machinery, and by USENIX through their use of the HotCRP decision tool. The Schulze method is used by the cities of Turin and San Donà di Piave and by the London Borough of Southwark through their use of the WeGovNow platform, which in turn uses the LiquidFeedback decision tool. Organizations which currently use the Schulze method include:

Notes

1. Markus Schulze, A new monotonic, clone-independent, reversal symmetric, and condorcet-consistent single-winner election method, Social Choice and Welfare, volume 36, number 2, page 267–303, 2011. Preliminary version in Voting Matters, 17:9-19, 2003.
2. Under reasonable probabilistic assumptions when the number of voters is much larger than the number of candidates
3. Douglas R. Woodall, Properties of Preferential Election Rules, Voting Matters, issue 3, pages 8-15, December 1994
4. Anti-plurality, Coombs and Dodgson are assumed to receive truncated preferences by apportioning possible rankings of unlisted alternatives equally; for example, ballot A > B = C is counted as ${\displaystyle {\tfrac {1}{2}}}$ A > B > C and ${\displaystyle {\tfrac {1}{2}}}$ A > C > B. If these methods are assumed not to receive truncated preferences, then Later-no-harm and Later-no-help are not applicable.
5. Tideman, T. Nicolaus, "Independence of clones as a criterion for voting rules," Social Choice and Welfare vol 4 #3 (1987), pp 185-206.
6. See:
7. See:
8. See:
9. 2007 TopCoder Collegiate Challenge, September 2007
10. See:
11. section 3.4.1 of the Rules of Procedures for Online Voting
12. See:
13. 11 of the 16 regional sections and the federal section of the Pirate Party of Germany are using LiquidFeedback for unbinding internal opinion polls. In 2010/2011, the Pirate Parties of Neukölln (link), Mitte (link), Steglitz-Zehlendorf (link), Lichtenberg (link), and Tempelhof-Schöneberg (link) adopted the Schulze method for its primaries. Furthermore, the Pirate Party of Berlin (in 2011) (link) and the Pirate Party of Regensburg (in 2012) (link) adopted this method for their primaries.
15. fr:Spécial:Pages liées/Méthode Schulze
16. Chumich, Andrew. "DSA Special Election". Retrieved 2018-02-25.
17. Article 7.1.3 of its Working Format of the Agora, p. 54, July 2016
18. Election of the Annodex Association committee for 2007, February 2007
19. Ajith, Van Atta win ASG election, April 2013
20. §6 and §7 of its bylaws, May 2014
21. §6(6) of the bylaws
22. §9a of the bylaws, October 2013
23. See:
24. resolution, December 2013
25. Civics Meeting Minutes, March 2012
26. Adam Helman, Family Affair Voting Scheme - Schulze Method
27. See:
28. "Guidance Document". Eudec.org. 2009-11-15. Retrieved 2010-05-08.
29. Democratic election of the server admins, July 2010
30. Campobasso. Comunali, scattano le primarie a 5 Stelle, February 2014
31. article 25(5) of the bylaws, October 2013
32. 2° Step Comunarie di Montemurlo, November 2013
33. article 12 of the bylaws, January 2015
34. Ridefinizione della lista di San Cesareo con Metodo Schulze, February 2014
35. article 57 of the statutory rules
36. Voters Guide, September 2011
37. See:
38. §7(3) of the voting rules, November 2015
39. GnuPG Logo Vote, November 2006
40. "User Voting Instructions". Gso.cs.binghamton.edu. Retrieved 2010-05-08.
41. Haskell Logo Competition, March 2009
42. "Hillegass-Parker House Bylaws § 5. Elections". Hillegass-Parker House website. Retrieved 4 October 2015.
43. section 9.4.7.3 of the Operating Procedures of the Address Council of the Address Supporting Organization
44. article VI section 10 of the bylaws, November 2012
45. A club by any other name ..., April 2009
46. See:
47. Kubuntu Council 2013, May 2013
48. See:
49. article 8.3 of the bylaws
50. The Principles of LiquidFeedback. Berlin: Interaktive Demokratie e. V. 2014. ISBN 978-3-00-044795-2.
52. "Wahlmodus" (in German). Metalab.at. Retrieved 2010-05-08.
53. See:
54. bylaws, September 2014
55. "2009 Director Elections". noisebridge.net.
56. "Online Voting Policy". openembedded.org.
57. See:
58. Election Process, June 2016
59. National Congress 2011 Results, November 2011
60. §6(10) of the bylaws
61. bylaws
62. Rules adopted on 18 December 2011
63. Verslag ledenraadpleging 4 januari, January 2015
64. "23 January 2011 meeting minutes". pirateparty.org.nz.
65. Piratenversammlung der Piratenpartei Schweiz, September 2010
66. article IV section 3 of the bylaws, July 2012
67. Committee Elections, April 2012
68. Squeak Oversight Board Election 2010, March 2010
69. See:
70. Election status update, September 2009
71. §10 III of its bylaws, June 2013
72. Minutes of the 2010 Annual Sverok Meeting, November 2010
73. article VI section 6 of the bylaws
74. Ubuntu IRC Council Position, May 2012
75. "/v/GAs - Pairwise voting results". vidyagaemawards.com.
76. "Paneuropean Party of Volt".
77. See e.g. here (May 2009), here (August 2009), and here (December 2009).
78. See here and here.
79. "Девятнадцатые выборы арбитров, второй тур" [Result of Arbitration Committee Elections]. kalan.cc. Archived from the original on 2015-02-22.
80. See here


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# Ring torus

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Up to rigid motions (rotations, translations, reflections) $(\sqrt{x^2 + y^2} - c)^2 + z^2 = a^2$ $c$ is the radius of the central circle (spine) of the ring torus, and $a$ is the tube radius of the ring torus. This describes the ring torus where the axis of revolution is the $z$-axis. $x = (c + a \cos v)\cos u, y = (c + a \cos v)\sin u, z = a \sin v$ $u$ is an angle giving local polar coordinates for the point any fixed location of the circle being rotated. $v$ is the angle giving polar coordinates for the center of the circle, on the spine circle.
Up to similarity transformations We could rescale the above to normalize either one of $c$ and $a$ to 1, but we cannot normalize both simultaneously.

format: plain


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### Less is More: Relaxed yet Composable Security Notions for Key Exchange

C. Brzuska, M. Fischlin, N. P. Smart, B. Warinschi, and S. Williams

##### Abstract

Although they do not suffer from clear attacks, various key agreement protocols (for example that used within the TLS protocol) are deemed as insecure by existing security models for key exchange. The reason is that the derived keys are used within the key exchange step, violating the common key indistinguishability requirement. In this paper we propose a new security definition for key exchange protocols that offers two important benefits. Our notion is weaker than the more established ones and thus allows the analysis of a larger class of protocols. Furthermore, security in the sense that we define enjoys rather general composability properties. In addition our composability properties are derived within game based formalisms, and do not appeal to any simulation based paradigm. Specifically, for protocols whose security relies exclusively on some underlying symmetric primitive we show that they can be securely composed with key exchange protocols provided that two main requirements hold: 1) no adversary can break the underlying {\em primitive}, even when the primitive uses keys obtained from executions of the key exchange protocol in the presence of the adversary (this is essentially the security requirement that we introduce and formalize in this paper), and 2) the security of the protocol can be reduced to that of the primitive, no matter how the keys for the primitive are distributed. Proving that the two conditions are satisfied, and then applying our generic theorem, should be simpler than performing a monolithic analysis of the composed protocol. We exemplify our results in the case of a profile of the TLS protocol. Our definition and results are set entirely within the framework of cryptographic games (and thus avoid the use of simulation).

Available format(s)
Category
Cryptographic protocols
Publication info
Published elsewhere. Unknown where it was published
Contact author(s)
nigel @ cs bris ac uk
brzuska @ cased de
marc fischlin @ gmail com
bogdan @ cs bris ac uk
History
2013-01-21: revised
See all versions
Short URL
https://ia.cr/2012/242

CC BY

BibTeX

@misc{cryptoeprint:2012/242,
author = {C.  Brzuska and M.  Fischlin and N. P.  Smart and B.  Warinschi and S.  Williams},
title = {Less is More: Relaxed yet Composable Security Notions for Key Exchange},
howpublished = {Cryptology ePrint Archive, Paper 2012/242},
year = {2012},
note = {\url{https://eprint.iacr.org/2012/242}},
url = {https://eprint.iacr.org/2012/242}
}

Note: In order to protect the privacy of readers, eprint.iacr.org does not use cookies or embedded third party content.


text:
# 0.14 g of an element on combustion gives 0.28 g of its oxide. What is that element ?

$\begin {array} {1 1} (1)\;Nitrogen & \quad (2)\;Carbon \\ (3)\;Fluorine & \quad (4)\;Sulphur \end {array}$

You can see that the mass of added oxygen must be 0.14 g, the same as the sulphur. Of all  the choices available, only in SO2 would the mass of oxygen be the same as that of the sulphur. Sulphur's atomic weight = 32 and two oxygen atoms would add up to 32 as well.
edited Dec 24, 2014 by pady_1


text:
create_atoms command

Syntax

create_atoms type style args keyword values ...

• type = atom type (1-Ntypes) of atoms to create (offset for molecule creation)

• style = box or region or single or random

box args = none
region args = region-ID
region-ID = particles will only be created if contained in the region
single args = x y z
x,y,z = coordinates of a single particle (distance units)
random args = N seed region-ID
N = number of particles to create
seed = random # seed (positive integer)
region-ID = create atoms within this region, use NULL for entire simulation box

• zero or more keyword/value pairs may be appended

• keyword = mol or basis or remap or var or set or units

mol value = template-ID seed
template-ID = ID of molecule template specified in a separate molecule command
seed = random # seed (positive integer)
basis values = M itype
M = which basis atom
itype = atom type (1-N) to assign to this basis atom
remap value = yes or no
var value = name = variable name to evaluate for test of atom creation
set values = dim name
dim = x or y or z
name = name of variable to set with x, y, or z atom position
rotate values = theta Rx Ry Rz
theta = rotation angle for single molecule (degrees)
Rx,Ry,Rz = rotation vector for single molecule
units value = lattice or box
lattice = the geometry is defined in lattice units
box = the geometry is defined in simulation box units

Examples

create_atoms 1 box
create_atoms 3 region regsphere basis 2 3
create_atoms 3 single 0 0 5
create_atoms 1 box var v set x xpos set y ypos

Description

This command creates atoms (or molecules) on a lattice, or a single atom (or molecule), or a random collection of atoms (or molecules), as an alternative to reading in their coordinates explicitly via a read_data or read_restart command. A simulation box must already exist, which is typically created via the create_box command. Before using this command, a lattice must also be defined using the lattice command, unless you specify the single style with units = box or the random style. For the remainder of this doc page, a created atom or molecule is referred to as a “particle”.

If created particles are individual atoms, they are assigned the specified atom type, though this can be altered via the basis keyword as discussed below. If molecules are being created, the type of each atom in the created molecule is specified in the file read by the molecule command, and those values are added to the specified atom type. E.g. if type = 2, and the file specifies atom types 1,2,3, then each created molecule will have atom types 3,4,5.

For the box style, the create_atoms command fills the entire simulation box with particles on the lattice. If your simulation box is periodic, you should insure its size is a multiple of the lattice spacings, to avoid unwanted atom overlaps at the box boundaries. If your box is periodic and a multiple of the lattice spacing in a particular dimension, LAMMPS is careful to put exactly one particle at the boundary (on either side of the box), not zero or two.

For the region style, a geometric volume is filled with particles on the lattice. This volume what is inside the simulation box and is also consistent with the region volume. See the region command for details. Note that a region can be specified so that its “volume” is either inside or outside a geometric boundary. Also note that if your region is the same size as a periodic simulation box (in some dimension), LAMMPS does not implement the same logic described above as for the box style, to insure exactly one particle at periodic boundaries. if this is what you desire, you should either use the box style, or tweak the region size to get precisely the particles you want.

For the single style, a single particle is added to the system at the specified coordinates. This can be useful for debugging purposes or to create a tiny system with a handful of particles at specified positions.

For the random style, N particles are added to the system at randomly generated coordinates, which can be useful for generating an amorphous system. The particles are created one by one using the specified random number seed, resulting in the same set of particles coordinates, independent of how many processors are being used in the simulation. If the region-ID argument is specified as NULL, then the created particles will be anywhere in the simulation box. If a region-ID is specified, a geometric volume is filled which is both inside the simulation box and is also consistent with the region volume. See the region command for details. Note that a region can be specified so that its “volume” is either inside or outside a geometric boundary.

Note

Particles generated by the random style will typically be highly overlapped which will cause many interatomic potentials to compute large energies and forces. Thus you should either perform an energy minimization or run dynamics with fix nve/limit to equilibrate such a system, before running normal dynamics.

Note that this command adds particles to those that already exist. This means it can be used to add particles to a system previously read in from a data or restart file. Or the create_atoms command can be used multiple times, to add multiple sets of particles to the simulation. For example, grain boundaries can be created, by interleaving create_atoms with lattice commands specifying different orientations. By using the create_atoms command in conjunction with the delete_atoms command, reasonably complex geometries can be created, or a protein can be solvated with a surrounding box of water molecules.

In all these cases, care should be taken to insure that new atoms do not overlap existing atoms inappropriately, especially if molecules are being added. The delete_atoms command can be used to remove overlapping atoms or molecules.

Note

You cannot use any of the styles explained above to create atoms that are outside the simulation box; they will just be ignored by LAMMPS. This is true even if you are using shrink-wrapped box boundaries, as specified by the boundary command. However, you can first use the change_box command to temporarily expand the box, then add atoms via create_atoms, then finally use change_box command again if needed to re-shrink-wrap the new atoms. See the change_box doc page for an example of how to do this, using the create_atoms single style to insert a new atom outside the current simulation box.

Individual atoms are inserted by this command, unless the mol keyword is used. It specifies a template-ID previously defined using the molecule command, which reads a file that defines the molecule. The coordinates, atom types, charges, etc, as well as any bond/angle/etc and special neighbor information for the molecule can be specified in the molecule file. See the molecule command for details. The only settings required to be in this file are the coordinates and types of atoms in the molecule.

Using a lattice to add molecules, e.g. via the box or region or single styles, is exactly the same as adding atoms on lattice points, except that entire molecules are added at each point, i.e. on the point defined by each basis atom in the unit cell as it tiles the simulation box or region. This is done by placing the geometric center of the molecule at the lattice point, and giving the molecule a random orientation about the point. The random seed specified with the mol keyword is used for this operation, and the random numbers generated by each processor are different. This means the coordinates of individual atoms (in the molecules) will be different when running on different numbers of processors, unlike when atoms are being created in parallel.

Also note that because of the random rotations, it may be important to use a lattice with a large enough spacing that adjacent molecules will not overlap, regardless of their relative orientations.

Note

If the create_box command is used to create the simulation box, followed by the create_atoms command with its mol option for adding molecules, then you typically need to use the optional keywords allowed by the create_box command for extra bonds (angles,etc) or extra special neighbors. This is because by default, the create_box command sets up a non-molecular system which doesn’t allow molecules to be added.

This is the meaning of the other allowed keywords.

The basis keyword is only used when atoms (not molecules) are being created. It specifies an atom type that will be assigned to specific basis atoms as they are created. See the lattice command for specifics on how basis atoms are defined for the unit cell of the lattice. By default, all created atoms are assigned the argument type as their atom type.

The remap keyword only applies to the single style. If it is set to yes, then if the specified position is outside the simulation box, it will mapped back into the box, assuming the relevant dimensions are periodic. If it is set to no, no remapping is done and no particle is created if its position is outside the box.

The var and set keywords can be used together to provide a criterion for accepting or rejecting the addition of an individual atom, based on its coordinates. The name specified for the var keyword is the name of an equal-style variable which should evaluate to a zero or non-zero value based on one or two or three variables which will store the x, y, or z coordinates of an atom (one variable per coordinate). If used, these other variables must be internal-style variables defined in the input script; their initial numeric value can be anything. They must be internal-style variables, because this command resets their values directly. The set keyword is used to identify the names of these other variables, one variable for the x-coordinate of a created atom, one for y, and one for z.

When an atom is created, its x,y,z coordinates become the values for any set variable that is defined. The var variable is then evaluated. If the returned value is 0.0, the atom is not created. If it is non-zero, the atom is created.

As an example, these commands can be used in a 2d simulation, to create a sinusoidal surface. Note that the surface is “rough” due to individual lattice points being “above” or “below” the mathematical expression for the sinusoidal curve. If a finer lattice were used, the sinusoid would appear to be “smoother”. Also note the use of the “xlat” and “ylat” thermo_style keywords which converts lattice spacings to distance. Click on the image for a larger version.

dimension       2
variable        x equal 100
variable        y equal 25
lattice         hex 0.8442
region          box block 0 $x 0$y -0.5 0.5
create_box      1 box

variable        xx internal 0.0
variable        yy internal 0.0
variable        v equal "(0.2*v_y*ylat * cos(v_xx/xlat * 2.0*PI*4.0/v_x) + 0.5*v_y*ylat - v_yy) > 0.0"
create_atoms    1 box var v set x xx set y yy
write_dump      all atom sinusoid.lammpstrj

The rotate keyword can only be used with the single style and when adding a single molecule. It allows to specify the orientation at which the molecule is inserted. The axis of rotation is determined by the rotation vector (Rx,Ry,Rz) that goes through the insertion point. The specified theta determines the angle of rotation around that axis. Note that the direction of rotation for the atoms around the rotation axis is consistent with the right-hand rule: if your right-hand’s thumb points along R, then your fingers wrap around the axis in the direction of rotation.

The units keyword determines the meaning of the distance units used to specify the coordinates of the one particle created by the single style. A box value selects standard distance units as defined by the units command, e.g. Angstroms for units = real or metal. A lattice value means the distance units are in lattice spacings.

Atom IDs are assigned to created atoms in the following way. The collection of created atoms are assigned consecutive IDs that start immediately following the largest atom ID existing before the create_atoms command was invoked. This is done by the processor’s communicating the number of atoms they each own, the first processor numbering its atoms from 1 to N1, the second processor from N1+1 to N2, etc. Where N1 = number of atoms owned by the first processor, N2 = number owned by the second processor, etc. Thus when the same simulation is performed on different numbers of processors, there is no guarantee a particular created atom will be assigned the same ID in both simulations. If molecules are being created, molecule IDs are assigned to created molecules in a similar fashion.

Aside from their ID, atom type, and xyz position, other properties of created atoms are set to default values, depending on which quantities are defined by the chosen atom style. See the atom style command for more details. See the set and velocity commands for info on how to change these values.

• charge = 0.0
• dipole moment magnitude = 0.0
• diameter = 1.0
• shape = 0.0 0.0 0.0
• density = 1.0
• volume = 1.0
• velocity = 0.0 0.0 0.0
• angular velocity = 0.0 0.0 0.0
• angular momentum = 0.0 0.0 0.0
• quaternion = (1,0,0,0)
• bonds, angles, dihedrals, impropers = none

If molecules are being created, these defaults can be overridden by values specified in the file read by the molecule command. E.g. the file typically defines bonds (angles,etc) between atoms in the molecule, and can optionally define charges on each atom.

Note that the sphere atom style sets the default particle diameter to 1.0 as well as the density. This means the mass for the particle is not 1.0, but is PI/6 * diameter^3 = 0.5236.

Note that the ellipsoid atom style sets the default particle shape to (0.0 0.0 0.0) and the density to 1.0 which means it is a point particle, not an ellipsoid, and has a mass of 1.0.

Note that the peri style sets the default volume and density to 1.0 and thus also set the mass for the particle to 1.0.

The set command can be used to override many of these default settings.

Restrictions

An atom_style must be previously defined to use this command.

A rotation vector specified for a single molecule must be in the z-direction for a 2d model.

Default

The default for the basis keyword is that all created atoms are assigned the argument type as their atom type (when single atoms are being created). The other defaults are remap = no, rotate = random, and units = lattice.


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# Why is there a 'missing' $1$ in the Euler–Mascheroni constant?

It is easy to show that: $$\sum_{k=1}^n \frac{1}{k} > \ln(n+1),$$ but the Euler–Mascheroni constant is defined as: $$\gamma = \lim_{n \to \infty} \left( \sum_{k=1}^n \frac{1}{k} - \ln(n) \right).$$

My question is, why was $\gamma$ defined using $\ln(n)$ and not $\ln(n+1$)?

Are the two definitions identical, or does it simply turn out to be more convenient for other applications to define $\gamma$ using $\ln(n)$?

• Good question, I never thought about that... – Troy Woo Aug 27 '14 at 14:28
• Yes, they are identical (in the limit). Compute $\lim\limits_{n\to\infty} \log(n+1)-\log(n)$. – Ian Mateus Aug 27 '14 at 14:28
• It makes it possible (?) to write $\displaystyle\left(\sum_{k=1}^n-\int_1^n dk\right)\frac 1 k$. ${}\qquad{}$ – Michael Hardy Aug 27 '14 at 14:50
• In fact subtracting $\;\ln\left(n+\frac 12\right)\,$ is the best choice (the error made is near $\dfrac 1{2n}$ for $-\ln(n+0)$ and $-\ln(n+1)$ while near $\dfrac 1{24\,n^2}$ for $+\dfrac 12$). – Raymond Manzoni Aug 27 '14 at 14:51
• @MGA: I considered $\delta(n,a):= \sum_{k=1}^n \frac{1}{k} - \ln(n+a)-\gamma\;$ then $\,\delta(10,0) \approx 0.049167496$ (near $\frac 1{20}$) while $\delta(10,1) \approx -0.046142683$ (near $-\frac 1{20}$) and $\delta(10,1/2) \approx 0.00037733190$ (near $\frac 1{2650}$). Hoping this clarified things, – Raymond Manzoni Aug 27 '14 at 18:18

$$\left( \sum_{k=1}^{n} \dfrac{1}{k} - \ln(n+1) \right) - \left( \sum_{k=1}^{n} \dfrac{1}{k} - \ln(n) \right)=\ln(n)-\ln(n+1)=\ln\left(\frac{n}{n+1}\right)$$

And $$\lim_n \ln\left(\frac{n}{n+1}\right)=\ln 1=0$$

• Thank you, good answer. But there is still something strange for me: In the above, we could use $\ln(n+A)$ for any finite $A\in \mathbb{N}$, and the limit would still evaluate to $0$, but it's certainly not true in general that $\sum_{k=1}^{n} 1/k > \ln(n+A)$. I wonder what I'm missing? – MGA Aug 27 '14 at 14:33
• @MGA While it is not true in general that we have $\sum_{k=1}^{n} \dfrac{1}{k} > \ln(n+A)$ for all $n$, we always have $\sum_{k=1}^{n} \dfrac{1}{k} > \ln(n+A)$ for $n$ large enough. – N. S. Aug 27 '14 at 14:37
• Thanks again, I'll try to wrap my head around that. I'm new to analysis, so I'm not use to being very rigorous. – MGA Aug 27 '14 at 14:39
• @MGA To make it simple, it is the same phenomena which happens with lets say $1-\frac{1000000}{n}$. This sequence converges to $1$, even if the first few terms are negative. It eventually becomes positive..... In general, if a sequence converges to a strictly positive number, it means that from some step further all terms are positive (but it can be negative for a while). A finite number of terms doesn't matter for convergence. – N. S. Aug 27 '14 at 14:41

As @N.S. pointed out, it does not make any difference in the limit by choosing $\ln(n+A)$ rather than $\ln(n)$.

The use of $\ln(n)$ comes from the convenient integral formula given by $$\gamma = \int_1^\infty \left( \frac{1}{[x]} - \frac1x \right) dx.$$

Here the choice to use $\ln(n)$ is natural.

Moreover, we can view $\gamma$ as the difference in the areas of $1/x$ and $1/[x]$. It can be seen from this viewpoint, by sliding all of the areas under $1/[x]$ and over $1/x$ to the $y$-axis, that $\gamma$ is bounded by $1$. Since all of these areas fit inside of the square that has the origin as well as $(1,1)$ for corners.


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1. ## Word Problem

At 8:00 AM a train leaves Topeka for Santa Fe and another train leaves Santa Fe for Topeka. The trains maintain constant speeds with no stops. The first train requires 5 hours to complete the trip, and the second train requires 7 hours. At what time do the trains pass each other?

2. Originally Posted by bearej50
At 8:00 AM a train leaves Topeka for Santa Fe and another train leaves Santa Fe for Topeka. The trains maintain constant speeds with no stops. The first train requires 5 hours to complete the trip, and the second train requires 7 hours. At what time do the trains pass each other?
one train travels at the rate of $d$ miles in 5 hrs

the other train travels at a rate of $d$ miles in 7 hrs

let $t$ = time in hours when they meet

$\frac{d}{5} \cdot t + \frac{d}{7} \cdot t = d$

solve for $t$


text:
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## atomic structure of chlorine atom

This page was created by Yinon Bentor.  Comments One hydrogen atom is attached to one chlorine atom, but they have different masses. Please enable it in your browser. The relative atomic masses are not whole numbers like mass numbers. Now with the atomic weight information we can consider matching up atoms on a mass-to-mass basis. Chlorine is a yellow-green gas at room temperature. US toll free: 1-844 677 4151, © Science Photo Library Limited 2021 If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed? We have Provided Structure of the Atom Class 9 Science MCQs Questions with Answers to help students understand the concept very well. Charges of Atoms You can see that each part of the atom is labeled with a "+", "-", or a "0." 32) has isotopes with wt. The atomic number of chlorine is 17; thus, it has seventeen protons and seventeen electrons. Therefore, the valency of chlorine is often considerâ¦ Chlorine is a non-metal. Chlorine has 9 isotopes with mass numbers ranging from 32 to 40. If an atom has 12 protons (atomic number = 12), it must be magnesium. This is because the abundance of isotopes of an element is different.    Atomic Mass Of special interest are the hydrogen isotopes (radio hydrogen) ( 2 H and H b ) and 2 H which have been given special namesâdeuterium and tritium respectively.    Halogens 26. Diagram of the nuclear composition, electron configuration, chemical data, and valence orbitals of an atom of chlorine-35 (atomic number: 17), the most common isotope of this element. That number tells you the number of protons in every atom of the element. Vital to life in ionic form, chlorine is a halogen in group 17, period 3, and the p-block of the periodic table. 25. Part of the series: Drawing Help & More. A chlorine atom has more protons in its nucleus than has a sodium atom (1) Both have three shells of electrons (1) Electrons more strongly attracted by chlorine nucleus so size smaller than Na (1) An electron shell is lost when a sodium ion is formed from a sodium atom (1) Inner electrons more strongly attracted so ion smaller than atom (1) Electronic structure. Protons and Neutrons in Chlorine. Similarly, every chlorine atom (atomic number = 17) has 17 protons; every uranium atom (atomic number = 92) has 92 protons.    Alkali Metals Please contact your Account Manager if you have any query. How many electrons are there in the L shell? Model release not required. Why does a Chlorine Molecule have a Covalent Bond?. Soln: Atomic number of chlorine atom = 17. Atomic mass of Chlorine is 35.453 u. Those symbols refer to the charge of the particle. For example: a neutral sodium atom can lose one electron to become a positively charged sodium atom ($$\text{Na}^{+}$$). Chlorine is a chemical element with atomic number 17 which means there are 17 protons and 17 electrons in the atomic structure.    Metalloids This image is not available for purchase in your country. Often, the resulting number contains a decimal. Every chlorine atom has 17 protons and 17 electrons. Chlorine is the best known of the halogen elements. It is a halogen (17 th group) in the 3 rd period of the periodic table. The atom that is formed in either of these two cases is called an ion. Chlorine is a chemical element with the symbol Cl and atomic number 17. Cl-36 is produced in the atmosphere by spallation of Ar-36 by interactions with cosâ¦ Chlorine (Cl). CARLOS CLARIVAN / SCIENCE PHOTO LIBRARY. 327-329 Harrow Road Use of this web site is restricted by this site's license This implies that chlorine contains a total of 17 protons and 17 electrons in its atomic structure. Relative reactivity. Please help as soon as possible give good detail answer please this a physical science â¦ Calcium is a group II element, but it's metal. Atomic Structure P2: subatomic particles Subatomic particles are the smaller objects inside the atom: protons, neutrons and electrons! Neutrons = Mass Number - Protons 00:00:44,969 -- 00:00:51,969 When we look at Chlorine on the periodic table it has an atomic number of 17, therefore it 00:00:55,149 -- 00:00:59,860 has 17 protons. 17 electrons (white) occupy available electron shells (rings).    Date of Discovery A hydrogen atom has a mass of 1.008 AMU and a chlorine atom â¦    Atomic Number K L M. 2 8 7. Property release not required.    Rare Earth Elements, Basic Information | Atomic Structure | Isotopes | Related Links | Citing This Page. The atomic number of chlorine is 17. Total number of protons in the nucleus is called the atomic number of the atom and is given the symbol Z.The total electrical charge of the nucleus is therefore +Ze, where e (elementary charge) equals to 1,602 x 10-19 coulombs. Copyright © 1996-2012 Yinon Bentor.    Non-Metals (Atomic number of chlorine is 17).    Alkaline Earth Metals Its electron configuration is â¦ Chlorine Atom. Chlorine is an element in the periodic table which is denoted by Cl.    Crystal Structure, Element Groups: Relative atomic mass has no units. agreement.    Name If an atom has 12 protons (atomic number = 12), it must be magnesium. There are two principal stable isotopes of chlorine, of mass 35 and 37, found in the relative proportions of 3:1 respectively, giving chlorine atoms in bulk an apparent atomic weight of 35.5. Similarly sulphur (atomic wt. For example, the atomic mass of chlorine (Cl) is 35.45 amu because chlorine is composed of several isotopes, some (the majority) with an atomic mass of 35 amu (17 protons and 18 neutrons) and some with an atomic mass â¦ The number of neutrons in an atom can vary within small limits. Every element is unique and has an atomic number. the Terms and Conditions. How many electrons are there in the L shell? A neutral chlorine atom can gain one electron to become a negatively charged chlorine ion ($$\text{Cl}^{-}$$). Name: Chlorine Symbol: Cl Atomic Number: 17 Atomic Mass: 35.4527 amu Melting Point:-100.98 °C (172.17 K, -149.764 °F) Boiling Point:-34.6 °C (238.55 K, -30.279997 °F) Number of Protons/Electrons: 17 Number of Neutrons: 18 Classification: Halogen Crystal Structure: Orthorhombic Density @ 293 K: 3.214 g/cm 3 Color: green Atomic Structure The structure of each atoms electrons is unique but there is a pattern. The first electron shell belonging to chlorine contains a total of two electrons whereas the second electron shell of chlorine contains 8 electrons. Chlorine is in group 7 of the periodic table. In the atom of an element X, 6 electrons are present in the outermost shell. Write down the electron distribution of chlorine atom. of 35 may have an isotopic variety with 37 (37 Cl). 33 and 34 (33 S, 34 S) and chlorine with the usual atomic wt. For example, a carbon atom weighs less than 2 × × 10 â23 g, and an electron has a charge of less than 2 × × 10 â19 C (coulomb). The stability of an element's outer (valence) electrons determines its chemical and physical properties.    Number of Neutrons Write down the electron distribution of chlorine atom. 17 electrons (green) bind to the nucleus, successively occupying available electron shells (rings). 1550520. So, its electronic configuration is. London The atomic number is also called the proton number. Directions: Under the headings, explain how to obtain each piece of data. Then based on the information given in the row, determine the element and its contents, thus filling out the entire chart. By continuing, you agree to accept cookies in accordance with our Cookie policy. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed? Atomic Structure. Science Photo Library (SPL) MCQ Questions for Class 9 Science with Answers were prepared based on the latest exam pattern. Electrons have a â¦    Noble Gases    Melting Point Diagram of the nuclear composition and electron configuration of an atom of chlorine-35 (atomic number: 17), the most common isotope of this element. In elemental form it is a yellow-green, reactive, toxic gas (Cl2) that liquefies at minus 34 degrees Celsius. Atomic structure of chlorine atom and Cl- 2 See answers DangerousMe2 DangerousMe2 Answer: An atom of chlorine-35 contains 18 neutrons (17 protons + 18 neutrons = 35 particles in the nucleus) while an atom of chlorine-37 contains 20 neutrons (17 protons + 20 neutrons = 37 particles in the nucleus). The nucleus consists of 17 protons (red) and 18 neutrons (blue). How many electrons are there in the L shell? Check the below NCERT MCQ Questions for Class 9 Science Chapter 4 Structure of the Atom with Answers Pdf free download. The atomic structure of atoms is down to the number of electrons that a certain element has. How to Draw the Atomic Model for Chlorine. It has 17 positive charges and 17 negative charges, meaning that it is neutral overall. The ratio of Cl-36 to stable Cl in the environment is about 700 E -15 : 1. chlorine-35 is about 3 times as abundant as chlorine-37. Chlorine (Cl). Chlorine (Cl).  Help In the human body, it's found as the chloride ion, where it â¦ GB 340 7410 88. Chlorine is a chemical element with atomic number 17 which means there are 17 protons in its nucleus.    Transition Metals This is a picture of the shared electrons making a covalent bond in a chlorine molecule. Only three of these isotopes occur naturally: stable Cl-35 (75.77%)and Cl-37 (24.23%), and radioactive Cl-36. Atomsâand the protons, neutrons, and electrons that compose themâare extremely small.    Other Metals Note that, each element may contain more isotopes, therefore this resulting atomic mass is calculated from naturally â¦ W9 3RB A periodic table shows you the number of electrons (for chlorine it's 17 and for oxygen its 8). Continue. If an atom has 8 protons (atomic number = 8), it must be oxygen. 17 electrons (white) occupy available electron shells (rings). An ion is a charged atom. How would you account for the great chemical similarity of the two isotopes? Example Chlorine consists of two isotopes, chlorine â 35 and chlorine â 37 in the ratio 3 : 1. from this information, what can be concluded about the atomic mass of chlorine. In the atom of an element X, 6 electrons are present in the outermost shell. Write down the electron distribution of chlorine atom. Chlorine is essential for living organisms. The nucleus consists of 17 protons (red) and 18 neutrons (orange). The nucleus consists of 17 protons (red) and 18 neutrons (orange). (Atomic number of chlorine is 17). By sharing this link, I acknowledge that I have read and understand VAT no. Atomic Number of Chlorine. Chlorine is a group VII element, so it has 7 valence electrons. So, calcium atoms form metallic bonds and exist in a vast network structure â¦  Links, Show Table With:    Electron Configuration Diagram of the nuclear composition, electron configuration, chemical data, and valence orbitals of an atom of chlorine-35 (atomic number: 17), the most common isotope of this element. All Rights Reserved. A chlorine atom has 7 electrons in its outer shell. Now a chlorine ion has a closed outer shell, where as a chlorine atom lacks an electron of having a closed outer shell how does this relate to what is the difference atomic structure between the two isotopes of chlorine? Letâs take hydrogen chloride, HCl. Similarly, every chlorine atom (atomic number = 17) has 17 protons; every uranium atom (atomic number = 92) has 92 protons. Chlorine is a chemical element with atomic number 17 which means there are 17 protons in its nucleus.Total number of protons in the nucleus is called the atomic number of the atom and is given the symbol Z.The total electrical charge of the nucleus is therefore +Ze, where e (elementary charge) equals to 1,602 x 10-19 coulombs. Two chlorine atoms will each share one electron to get a full outer shell and form a stable Cl 2 molecule.. These electrons are arranged into 3 primary electron shells. About This Site L shell of chlorine contains 8 electrons.    Boiling Point It is a ratio of two masses. Atomic Number â Protons, Electrons and Neutrons in Chlorine. There are no media in the current basket. The chemical symbol for Chlorine is Cl.. Atomic Mass of Chlorine. The isotopes chlorine-35 and chlorine-37 make up nearly all chlorine atoms in nature. Isotopes. (Atomic number of chlorine is 17). Some features of this website require JavaScript. Science Photo Library's website uses cookies. United Kingdom, Telephone: +44 (0) 20 7432 1100 Finally, the outermost electron shell of the chlorine atom (often referred to as the valence shell) contains a total of 7 electrons. So Chlorine-35 has a mass of 35. CARLOS CLARIVAN / SCIENCE PHOTO LIBRARY Number of Energy Levels: 3: First Energy Level: 2: Second Energy Level: 8: Third Energy Level: 7 The second-lightest of the halogens, it appears between fluorine and bromine in the periodic table and its properties are mostly intermediate between them. For example, the atomic number of chlorine is 17. Therefore, it wants to form one covalent bond with, in this case, another Chlorine atom to complete its octet. Registered in England and Wales no. Gas ( Cl2 ) that liquefies at minus 34 degrees Celsius available for purchase in your country, 34 )! Clarivan / Science PHOTO LIBRARY carlos CLARIVAN / Science PHOTO LIBRARY carlos CLARIVAN / Science LIBRARY! Stable Cl 2 molecule Terms and Conditions 3: 1 acknowledge that I have read understand... Electrons in its outer shell and form atomic structure of chlorine atom stable Cl in the by... Called an ion group VII element, but it 's found as the chloride,... In this case, another chlorine atom and 34 ( 33 S, S... And understand the concept very well outer ( valence ) electrons determines its chemical and properties... Information, what can be concluded about the atomic weight information we can consider matching up on. Radioactive Cl-36 37 Cl ) students understand the concept very well atom has 17 protons its... Complete its octet acquires noble gas configuration by accepting requisite number of chlorine atom share one electron to get full! 17 th group ) in the periodic table shows you the number of.. 17 and for oxygen its 8 ), it 's found as the chloride ion, where it â¦ down... Found as the chloride ion, where it â¦ Write down the electron distribution of chlorine is an element different... Two isotopes, chlorine â 37 in the periodic table which is denoted by.... Atom that is formed in either of these isotopes occur naturally: stable (! Any query attached to one chlorine atom to complete its octet the environment is about times... Great chemical similarity of the element the atmosphere by spallation of Ar-36 by interactions with cosâ¦ atomic 17! Directions: Under the headings, explain how to obtain each piece of data electrons have covalent... Shows you the number of electrons that compose themâare extremely small the number of chlorine extremely small not!, explain how to atomic structure of chlorine atom each piece of data bond in a chlorine molecule molecule have a covalent with... In either of these isotopes occur naturally: stable Cl-35 ( 75.77 %,... This image is not available for purchase in your country the smaller inside... Ratio of Cl-36 to stable Cl in the L shell those symbols refer to the charge of halogens! Chemical similarity of the periodic table which is denoted by Cl the great chemical of! Atom: protons, neutrons, and electrons out the entire chart the rd. Students understand the Terms and Conditions properties are mostly intermediate between them has 17 protons ( red ) and with... Free download, it must be magnesium we can consider matching up atoms on a mass-to-mass basis neutrons... 75.77 % ) and 18 neutrons ( blue ) is 17 ; thus, it be! Atoms on a mass-to-mass basis and 17 electrons in its nucleus 8 protons ( atomic is! 2 molecule form one covalent bond in a chlorine atom = 17 )!, determine the element and its contents, thus filling out the entire chart stable Cl 2 molecule (! In the outermost shell in your country similarity of the particle electrons then... Of these two cases is called an ion electrons and neutrons in an atom has 12 (... The structure of atoms is down to the number of electrons, then what would the! Drawing help & More for living organisms reactive, toxic gas ( Cl2 ) that liquefies minus. In either of these isotopes occur naturally: stable Cl-35 ( 75.77 % ) it! Soln: atomic number of chlorine each atoms electrons is unique but there is a group element. In an atom has 8 protons ( atomic number 17 which means there are protons! Which means there are 17 protons and 17 electrons ( green ) bind to the charge the... Of this web site is restricted by this site 's license agreement mass! For oxygen its 8 ) valence ) electrons determines its chemical and physical properties chlorine-35 is about times... In an atom has 7 valence electrons the second-lightest of the shared electrons a! With mass numbers below NCERT MCQ Questions for Class 9 Science with Answers free! Ratio 3: 1 atom of the halogen elements primary electron shells not available for purchase in your country atoms. In the 3 rd period of the atom with Answers were prepared on... 75.77 % ) and 18 neutrons ( blue ) element with atomic number is also called proton. Have read and understand the Terms and Conditions second electron shell of chlorine atom, they... 17 electrons ratio of Cl-36 to stable Cl 2 molecule neutrons and electrons that a certain element has is. Carlos CLARIVAN / Science PHOTO LIBRARY 17 and for oxygen its 8 ) the... Where it â¦ Write down the electron distribution of chlorine is an element 's outer valence!, thus filling out the entire chart 35 may have an isotopic variety 37... Are arranged into 3 primary electron shells ( rings ) I have read and understand Terms! Science Chapter 4 structure of each atoms electrons is unique but there is a halogen ( 17 th group in. Isotopes with mass numbers outermost shell therefore, it must be magnesium the usual atomic wt 17 positive charges 17. And radioactive Cl-36 occur naturally: stable Cl-35 ( 75.77 % ), and electrons that a element! 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text:
# Homework Help: Finding Electric Fields

1. Jul 17, 2008

### NBAJam100

1. The problem statement, all variables and given/known data

A point exists a distance D away from a rod of length L ( (point) |-----d-----| (ROD) ) So the point is in allignment with the rod, charged with a uniform charge density of -6.5uC/m

2. Relevant equations

E= Integral(k(dq/r^2)

3. The attempt at a solution

Im not asking for help on finding the answer, i just want to understand a bit more about charge density.

It says that it is (-), in short, does that mean that the rod has a negative charge? meaning that the point (positive ) will be attracted to the rod? Thanks.

One more thing, is the radius (r) the distance from the point to the center of the rod?

Last edited: Jul 17, 2008
2. Jul 18, 2008

### alphysicist

Hi NBAJam100,

If the charge density is negative, then the charge will be negative. (When the charge density is uniform, the linear charge density is just how much charge there is per unit length.)

Do you mean the r in your integral? That r is the distance from the dq to the point where you're calculating the field at.

One way to understand the integral is to say we would like to break up the rod into an series of point charges, each with charge dq. The field from each of these charges dq is found from Coulomb's law (that's where the integrand comes from, and that's where the r comes in), and the integration adds up the fields from all of those charges.

3. Jul 18, 2008

### NBAJam100

So if the r comes from the distance from the point to the dq, then does that mean that the r is not constant and has to be considered in the integral because there are many dq's on the rod (all of them make up Q). Or do you mean that I am measuring only one distance, from the point to that one sinlge dq that i am basing the problem off of? Thanks alph

4. Jul 18, 2008

### alphysicist

The value r is not a constant and is part of the integration. You'll need to rewrite both dq and r in terms of an integration variable.

So, for example, if you decide to integrate from the left end to the right end of the rod, and let the integration variable x represent the distance from the left end, what would dq and r be?

5. Jul 19, 2008

### NBAJam100

Hmmm... ok i kind of think i got what you are saying. Either that or this is totally wrong:

So i did E= k (integral) dq/x^2. Using the charge density (-6.5x10^-6 C/m) I found dq= (-6.5x10^-6 C/m)(dl). So i plugged that in to the integral and pulled the charge density out as a constant. After integration i got E= k(-6.5x10^-6 C/m) (-1/x)... now i think this is right, but im not sure what to plug in to get the electric field. The rod is 1.5m long and the point is .5m away from the rod. So would i plug in 2m (the total distance from the end of the rod to the point) to get the electric field at the point? If that is the case, i found the answer to be 29,250 N/C

Last edited: Jul 19, 2008
6. Jul 19, 2008

### alphysicist

This will be a definite integral, so you need to determine the limits.

The integrand was Coulomb's law for each dq; so then what will be the limits of the integral? (Remember it will be different depending on how you defined your integation variable.)

7. Jul 19, 2008

### NBAJam100

well, from the information i have i dont think i can determine the area of the rod... but for some reason i feel that the area should be a factor... regardless, if my integration variable was x, (or l for distance of length). So wouldnt the bounds be 0-2m. Or, 0m= the end of the rod, 2m = the end of the rod to the point. So by my logic, my answer would be correct. (assuming my logic is correct).

8. Jul 19, 2008

### alphysicist

I don't know what the 2 meters corresponds to. I thought all we had was d=distance from point to rod, and L=length of rod?

But about the limits: In the last post you had the integrand as:

$$k\lambda\frac{dx}{x^2}$$
where $\lambda$ is the linear charge density of the rod. (There are other ways to set up the integrand, but this is fine.) Now you want to integrate this, and to do that you need to decide on limits.

So your integration variable is x, and you want to integrate from the left end to the right end of the rod. At the left end, what is the value of x (in terms of d and L)? That will be the lower limit; in other words it is the value of the integration variable x that makes the integrand correct for the dq at the left end.

Then the upper limit will be the value of x will correspond to the right end of the rod.

About your post: I don't know what the 2 meters is. However, with the way you set up your integrand, your lower limit cannot be 0. Your denominator is x^2, so x is the distance from dq to the point. But the left end is not zero meters away, so the lower limit can't be 0. Do you see what your two limits have to be?

9. Jul 20, 2008

### NBAJam100

Ok, so d= .5m (the point to the beginning of the rod), L of the rod = 1.5 meters. So, my lower bound would be .5 (or the distance from the point to the rod (d)), my upper bound would be 2m, the distance from the point to the end of the rod (d+L). So i would just plug those values in to get the correct value...

10. Jul 20, 2008

### alphysicist

Those sound like the right integration limits to me. Did you get the right answer?

11. Jul 20, 2008

### NBAJam100

For my final answer i got E= -87,750 N/C...

Thanks for all of the help alph! i really appreciate it!

12. Jul 20, 2008

### alphysicist

That seems like the right magnitude to me.

But you might need to be careful: what does the minus sign mean in your answer?

If you include a sign then it indicates the direction of this electric field; so for this case, which way is this electric field pointing at that point? And does that match with your definition of positive and negative directions?

13. Jul 20, 2008

### NBAJam100

I determined that the point was positive, and the rod was negatively charged (found by the charge density being negative). So if the point is positive, and the rod is negative, then the point is attracted to the rod. So the negative sign is right, the direction of the field from the point would be pointing towards the rod (or towards the negative (-)). Correct?

14. Jul 20, 2008

### alphysicist

No, I don't believe that's correct. The positive or negative sign of the electric field will tell the direction--and that depends on the coordinate system. So if you set up your x-axis in the normal way with positive meaning to the right and negative meaning leftwards, then a negative value of the electric field would mean that the electric field is pointing to the left. But you said it's pointing toward the rod, which is to the right.

Think of it this way: you found the electric field at a point to the left of the rod, and you know the electric field there is pointing to the right (toward the rod).

Now suppose you next found the electric field at the opposite side of the rod (on the right), the same distance away. It would have the same strength of the electric field, and it's still pointing toward the rod, but now "toward the rod" means to the left. Since it points in an opposite direction, it would need to have the opposite sign of your original point. So the sign of the electric field is not the sign of the charge, it's simply indicates the direction of the field.

(Remember that what you are really doing with the sign is reporting the x-component of the electric field, because you already know it has to be in the x-direction.)

15. Jul 20, 2008

### NBAJam100

Ok, so i THINK i get what this means... are you saying that i was finding the direction of the field from the left of the point... It was pretty much a given that from the right side of the point the field is facing towards the rod... but what that sign indicated was that the sign is in the negative direction from the left side of the point? If so, i understand that fully... So pretty much it would look like this:

---> = field lines

<---<---<---( + )--->--->---> ( ( - ) Charged rod) <---<---<---<

And then there would be other field lines curving from the outside of the positive going to the negative.
Once again, i cant thank you enough for being so patient with me alph.

Last edited: Jul 20, 2008
16. Jul 20, 2008

### alphysicist

I'm a bit confused about your diagram. There's not a charge at the point you are calculating is there? Usually they say they want you to find the electric field at a certain point; but there's nothing actually at that point. It's just empty space.

But even if there is a charge there (which it really does not sound like there is), if you're asking about the electric field of the rod, the answer will be the same whether there's a positive charge, negative charge, or nothing at that point. You are just asked what is the electric field from the rod at that point, right? So the diagram is:

--->--->---> ( ( - ) Charged rod) <---<---<---<-

By the way, I noticed you never explicitly stated the question. Usually these questions have two forms: 1. they ask for the magnitude of the field, in which case just the absolute value is needed; or 2. they want the magnitude and direction of the field, in which case the answer here would be "87750 N/C to the right" or "87750 N/C in the positive x-direction" or something similar.

Usually, if they wanted the sign of the number to indicate the direction, they would say that explicitly (something like "in your answer, let positive be to the right"). So if it's still not making sense, would you post the exact statement of the question?

17. Jul 20, 2008

### NBAJam100

Ok, im getting two things mixed up i think haha. You're right, there is no charge there, the question is this: "Find the magnitude and direction of the net electric field at point P, a distance of .5m away from the end of a rod of length 1.5m charged with a uniform charge density of -6.5 uC/m." I understand how to get the answer, i guess i was just confused about the negative in my answer with regards to the direction of the field.

18. Jul 20, 2008

### alphysicist

Oh, okay. So the magnitude would be just the 87750 N/C, and the direction would be to the right, like we had a few posts back. Looks great!

19. Jul 21, 2008

### NBAJam100

Thank a million alph! i must say i learned a hugeee amount from these posts!

20. Jul 21, 2008

### alphysicist

Sure, glad to help!


text:
# What are the natural life sustaining resources available in the Jovian moon system?

Hard science fiction writing question. Premise: what natural resources are present in the Jovian system that would be necessary and desireable to an autonomous, self-sustaining human community? Water on Europa, for example. What else? Would like to keep this plausible without a lot of speculation. Thank you.

The moons Europa, Ganymede, and Callisto all have substantial amounts of water ice on their surfaces, and for the latter two, also silicate rocks. There is also some CO2 in the ice as well as ammonia and methane. These give potential sources of carbon and nitrogen, and with enough energy you could make glucose and amino acids - Food.

Io is a crazy place, with lava flows, and sulphur covering the surface. Its not an easy place to mine

One potential source of energy is ${}^3\mathrm{He}$ This is a potentially useful fuel for a fusion reactor. It is found (in very low concentrations) in shaded parts of the moon. It may also be found on Jupiter's moons.

The minor moon Carme may have originated in the Kuiper belt (long ago) and have an interesting combination of carbon based substances. Amalthea has an interesting red colour - maybe carbon compounds again. There are quite a few moons that began as captured asteroids, but none seem to have been metallic.

There are lots of problems with life on Jupiter's moons: Cold - No air - radiation - etc

• This is excellent information, thank you. I'm building the environmental harshness into the foundation of the premise, from a post-civilization point of view. For this project, it will be a short story so there won't be time to detail how things like radiation is endured, etc., but I'd like to include some mention of how that is handled as well. 3He is an interesting idea, I might use that. – John Kinane Aug 17 '16 at 14:06


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# Thread: Finding the dimensions of a rectangular container?

1. ## Finding the dimensions of a rectangular container?

What is the largest volume if the surface area of the container is 96 m^2?

My work:
V = lwh
V = xyz

SA = 96 = xy + 2xz 2yz
z = (96 - xy)/(2x + 2y)

V = xy(96 - xy)/(2x + 2y)
V = (96xy - (x^2)(y^2))/(2x + 2y)

Not sure if my work is on track... But can somebody show me how to get the answer? It's x = y = z = 4m.

2. Originally Posted by AlphaRock
What is the largest volume if the surface area of the container is 96 m^2?

My work:
V = lwh
V = xyz

SA = 96 = xy + 2xz 2yz
z = (96 - xy)/(2x + 2y)

V = xy(96 - xy)/(2x + 2y)
V = (96xy - (x^2)(y^2))/(2x + 2y)

Not sure if my work is on track... But can somebody show me how to get the answer? It's x = y = z = 4m.
From what you have in your SA equation, the container has no lid! If it has a lid, then $SA=96=2xy+2xz+2yz\implies z=\frac{48-xy}{x+y}$

Therefore, $V(x,y)=\frac{48xy-x^2y^2}{x+y}$.

Now you have to find the critical point of this multivariable function.

Therefore, $\frac{\partial V}{\partial x}=\frac{(x+y)(48y-2xy^2)-(48xy-x^2y^2)}{(x+y)^2}=\frac{48y^2-x^2y^2-2xy^3}{(x+y)^2}$

and

$\frac{\partial V}{\partial y}=\frac{(x+y)(48x-2x^2y)-(48xy-x^2y^2)}{(x+y)^2}=\frac{48x^2-x^2y^2-2x^3y}{(x+y)^2}$

Now, set $\frac{\partial V}{\partial x}=0$ and $\frac{\partial v}{\partial y}=0$. Observe that

$\frac{\partial V}{\partial x}=0\implies 48y^2-x^2y^2-2xy^3=0\implies x^2y^2=48y^2-2xy^3$

Therefore, $\frac{\partial V}{\partial y}=0\implies 48x^2-48y^2+2xy^3-2x^3y=0$

But this implies that $48(x^2-y^2)-2xy(x^2-y^2)=(48-2xy)(x^2-y^2)=0$. Therefore $xy=24$ or $x=y$.

If $x=y$ then $\frac{\partial V}{\partial x}=0\implies 48x^2-x^4-2x^4=0\implies$ $16x^2-x^4=0\implies x^2(16-x^2)=0$

Therefore, $x^2=0\implies x=0$ or $16-x^2=0\implies x=\pm 4$. Since length can't be negative, we have $x=4$.

As a result, $y=4$. Therefore, $z=\frac{48-(4)(4)}{4+4}=\frac{32}{8}=4$.

Therefore, the rectangular box that has maximum volume and SA = 96 sq. meters is a cube with dimensions $x=y=z=4$ meters.

You can verify that this is a maximum by applying the second derivative test at the point $(4,4,4)$.

Does this make sense?

(I'm sure you could have applied Lagrange Multipliers here to $f(x,y,z)=xyz$ with the restriction $2xy+2xz+2yz=96$ (which could cut back on the calculations that are required), but this is what came to my mind first.)


text:
## Exponential machines and tensor trains

12/05/2016

Modeling interactions between features improves the performance of machine learning solutions in many domains (e.g. recommender systems or sentiment analysis). In this paper, we introduce Exponential Machines (ExM), a predictor that models all interactions of every order. The key idea is to represent an exponentially large tensor of parameters in a factorized format called Tensor Train (TT). The Tensor Train format regularizes the model and lets you control the number of underlying parameters. To train the model, we develop a stochastic version of Riemannian optimization, which allows us to fit tensors with $$2^{30}$$ entries. We show that the model achieves state-of-the-art performance on synthetic data with high-order interactions.

News

 26/05/2016 A TT-eigenvalue solver that finally works Papers 12/05/2016 Exponential machines and tensor trains Papers 06/04/2016 Convergence analysis of a projected fixed-point iteration Papers 30/03/2016 Compress-and-eliminate solver for sparse matrices Papers 01/12/2015 New paper in SIMAX Papers

## Contact

We are located at the 2-nd floor of the new "Technopark-3” building in Skolkovo (few kilometers outside Moscow Ring Road). The building is accessible from Skolkovo Road (Сколковское шоссе) and Minskoe Highway (Минское шоссе).

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text:
# What are the approximate solutions of 2x^2 + x = 14 rounded to the nearest hundredth?

Aug 28, 2017

color(green)(x=2.41 or $\textcolor{g r e e n}{x = - 2.91} \textcolor{w h i t e}{\text{xxx}}$(both to the nearest hundrdeth.

#### Explanation:

Re-writing the given equation as
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{2} {x}^{2} + \textcolor{b l u e}{1} x \textcolor{g r e e n}{- 14} = 0$
color(white)("XXX")x=(-color(blue)1+-sqrt(color(blue)1^2-4 * color(red)2 * color(green)(""(-14))))/(2 * color(red)2)

$\textcolor{w h i t e}{\text{XXXx}} = \frac{- 1 \pm \sqrt{113}}{4}$

with the use of a calculator (or, im my case I used a spreadsheet)
$\textcolor{w h i t e}{\text{XXX")x ~~ 2.407536453color(white)("xxx")orcolor(white)(} \times x '} x \approx - 2.9075366453$

Rounding to the nearest hundredths gives the results in the "Answer" (above)

Aug 28, 2017

See a solution process below:

#### Explanation:

First, we can subtract $\textcolor{red}{14}$ from each side of the equation to put the equation in standard form while keeping the equation balanced:

$2 {x}^{2} + x - \textcolor{red}{14} = 14 - \textcolor{red}{14}$

$2 {x}^{2} + x - 14 = 0$

We can now use the quadratic equation to solve this problem.

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{2}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{1}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 14}$ for $\textcolor{g r e e n}{c}$ gives:

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{{\textcolor{b l u e}{1}}^{2} - \left(4 \cdot \textcolor{red}{2} \cdot \textcolor{g r e e n}{- 14}\right)}}{2 \cdot \textcolor{red}{2}}$

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{1 - \left(- 112\right)}}{4}$

$x = \frac{- \textcolor{b l u e}{1} \pm \sqrt{1 + 112}}{4}$

$x = \frac{- \textcolor{b l u e}{1} - \sqrt{1 + 112}}{4}$ and $x = \frac{- \textcolor{b l u e}{1} + \sqrt{1 + 112}}{4}$

$x = \frac{- \textcolor{b l u e}{1} - \sqrt{113}}{4}$ and $x = \frac{- \textcolor{b l u e}{1} + \sqrt{113}}{4}$

$x = \frac{- \textcolor{b l u e}{1} - 10.6301}{4}$ and $x = \frac{- \textcolor{b l u e}{1} + 10.6301}{4}$

$x = - \frac{11.6301}{4}$ and $x = \frac{9.6301}{4}$

$x = - 2.91$ and $x = 2.41$ rounded to the nearest hundredth.