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3 years ago

Which of the following symbols represents a chlorine ion with a stable arrangement of eight valence electrons?

Chemistry

1 answer:

frozen [14]3 years ago

3 0

A stable arrangement of eight valence electrons : ³⁵Cl⁻¹

<h3>Further explanation</h3>

Chlorine is a halogen gas, located in group 17, p block

Chlorine has an atomic number of 17 and an atomic mass of 35

Electron configuration: [Ne] 3s²3p⁵

If we look at the electron configuration, then Cl will bind 1 more electron so that the configuration is stable like Argon (atomic number 18)

So by binding this one electron, chlorine forms negative ions (anions)

³⁵Cl⁻¹

B. Cl⁻² binds 2 electrons, exceeding the octet rule

C. Cl⁺¹, releases 1 electron, remains unstable

D. Cl, the neutral form of Cl, is still unstable with a 7-electron valence configuration

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Answer:

Percentage recovery of acetylferrocene = 81.6%

Explanation:

Mass of the sample mixture = 22.092 g

Ratio of mixture of acetylferrocene and ferrocene = 1 : 1

This means that the sample conatains equal amounts of acetylferrocene and ferrocene.

Therefore the mass of each sample in the mixture = 22.092 g / 2 = 11.046 g

Mass of acetylferrocene recovered = 9.017 g

Percentage recovery of acetylferrocene = (mass of recovered/ mass in sample) * 100%

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Maru [420]

Answer : The mass of $CS_2$ is, 555.028 grams

Explanation :

First er have to calculate the concentration of $S_2$ .

$\text{Concentration of }S_2=\frac{\text{Moles of }S_2}{\text{Volume of solution}}=\frac{8.08mole}{5.35L}=1.51mole/L$

Now we have to calculate the concentration of $CS_2$ .

The given balanced chemical reaction is,

$S_2(g)+C(s)\rightleftharpoons CS_2(g)$

Initial conc. 1.51 0 0

At eqm. conc. (1.51-x) (x) (x)

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[CS_2]}{[S_2]}$

Now put all the given values in this expression, we get :

$9.40=\frac{x}{(1.51-x)}$

By solving the term 'x', we get :

x = 1.365 M

Concentration of $CS_2$ = x M = 1.365 M

Now we have to calculate the moles of $CS_2$ .

$\text{Moles of }CS_2=\text{Concentration of }CS_2}\times \text{Volume of solution}=1.365mole/L\times 5.35L=7.303mole$

Now we have to calculate the mass of $CS_2$ .

Molar mass of $CS_2$ = 76 g/mole

$\text{Mass of }CS_2=\text{Moles of }CS_2}\times \text{molar mass of }CS_2}=7.303mole\times 76g/mole=555.028g$

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