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3 years ago

If an ionic compound has a 2- charge how does that affect the total number of electrons in a Lewis structure

1 answer:

7 0

In the situation where an ion has a charge of -2, the Lewis structure is affected because it must add two additional electrons.

The Lewis structure is a term to refer to the graphical representation that shows the pairs of electrons in dashes or points of bonds between:

The atoms of a molecule
The lone electron pairs that may exist

Additionally, the charge of an ionic compound is calculated using the difference in electrical charge between the electrons. On the other hand, an ion is formed by adding or removing electrons, so the magnitude of the resulting charge depends on the number of electrons added or removed.

According to the above, if an ionic compound has a -2 charge, the Lewis structure must add two electrons.

Learn more in: brainly.com/question/1488567

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A toxicologist studying mustard gas, S(CH2CH2Cl)2, a blistering agent, prepares a mixture of 0.675 M SCl2and 0.973 M C2H4 and al

Musya8 [376]

Answer:

The value of $K_p$ is 0.02495.

Explanation:

Initial concentration of $SCL_2$ gas = 0.675 M

Initial concentration of $C_2H_4$ gas = 0.973 M

Equilibrium concentration of mustard gas = 0.35 M

$SCl_2 (g) + 2 C_2H_4 (g)\rightleftharpoons S(CH_2CH_2Cl)_2(g)$

initially

0.675 M 0.973 M 0

At equilibrium ;

(0.675-0.35) M (0.973-2 × 0.35) M 0.35 M

The equilibrium constant is given as :

$K_c=\frac{[S(CH_2CH_2Cl)_2]}{[SCl_2][C_2H_4]^2}$

$=\frac{0.35 M}{(0.675-0.35) M\times ((0.973-2 × 0.35) M)^2}$

$K_c=14.45$

The relation between $K_p$ and $K_c$ are :

$K_p=K_c\times (RT)^{\Delta n}$

where,

$K_p$ = equilibrium constant at constant pressure = ?

$K_c$ = equilibrium concentration constant =14.45

R = gas constant = 0.0821 L⋅atm/(K⋅mol)

T = temperature = 20.0°C =20.0 +273.15 K=293.15 K

$\Delta n$ = change in the number of moles of gas = [(1) - (1 + 2)]=-2

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

$K_p=14.45\times (0.0821L.atm/K.mol\times 293.15 K)^{-2}$

$K_p=6.2\times 10^{4}$

$K_p=0.02495$

The value of $K_p$ is 0.02495.

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Answer:

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