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strojnjashka [21]
3 years ago
5

A. A main group element with the valence electron configuration 5s25p4 is in periodic group . It forms a monatomic ion with a ch

arge of . B. A main group element with the valence electron configuration 5s1 is in periodic group . It forms a monatomic ion with a charge of ____ .
Chemistry
2 answers:
tino4ka555 [31]3 years ago
8 0
It’s part A correct not part B
pashok25 [27]3 years ago
4 0

Correct Question:

A. A main group element with the valence electron configuration 5s25p4 is in periodic group _____ . It forms a monatomic ion with a charge of __. B. A main group element with the valence electron configuration 5s1 is in periodic group ___ . It forms a monatomic ion with a charge of ____.

Answer:

A. 16, -2

B. 1, +1

Explanation:

The electronic configuration is the distribution of the electrons at the shells and subshells following the Linus Pauling's diagram of the crescent energy of the subshells. The sell is represented by the numbers, and the subshell by letters. The maximum number presented in the distribution is the period of the element in the periodic table.

If the valence subshell is "s", then the element is at groups 1 or 2, if there's only one electron at it, it's from group 1, and if there are two electrons, it's from group 2.

If the valence subshell is "p", then the element is at groups 13 to 18. The numbers of electrons of p must be added to the number of electrons of the last s subshell, and then summed with 10 to determine the group.

If the valence subshell is "d", then the element is an external transition metal, and the groups can be at 3 to 12. To determine it, the number of electrons of "d" must be added to the number of electrons of the last "s".

If the valence subshell is "f", then the element is an internal transition metal, and it is from group 3.

A. The valence subshell is p, so the group is 16, or 6A (4 of p + 2 of s + 10). So, the element has 6 valence electrons, and to be stable as the octet rule, it must have 8 electrons, so it must gain 2 electrons, and will have charge -2.

B. The valence subshell is s, so the group is 1. It has only one valence electron, so, to be stable, it must lose this electron (the other shell is completed with 8 electrons), so the charge will be +1.

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Answer : The concentration of A,B\text{ and }C at equilibrium are 0.132 M, 0.232 M  and 0.168 M  respectively.

Explanation :

The given chemical reaction is,

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First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

\Delta G^o=-RT \ln k

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Now put all the given values in the above relation, we get:

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Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

                          A(aq)+B(aq)\rightleftharpoons C(aq)

Initially               0.30      0.40         0  

At equilibrium  (0.30-x) (0.40-x)     x

The expression of equilibrium constant will be,

k=\frac{[C]}{[A][B]}

5.45=\frac{x}{(0.30-x)\times (0.40-x)}

By solving the term x, we get

x=0.168\text{ and }0.716

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

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The concentration of B at equilibrium = (0.40-x) = 0.40 - 0.168 = 0.232 M

The concentration of C at equilibrium = x = 0.168 M

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