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Studentka2010 [4]
3 years ago
7

+|++that this atom is neutral? ​

Chemistry
2 answers:
Svet_ta [14]3 years ago
6 0

Answer:

I think yes....

Explanation:

I think yes

3241004551 [841]3 years ago
3 0
Yes that looks right to me
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A 1.00 kg sample of Sb2S3 (s) and a 10.0 g sample of H2 (g) are allowed to react in a 25.0 L container at 713 K. At equilibrium,
Scorpion4ik [409]

<u>Answer:</u> The value of K_c is coming out to be 0.412

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For Sb_2S_3</u>

Given mass of Sb_2S_3 = 1.00 kg = 1000 g   (Conversion factor: 1 kg = 1000 g)

Molar mass of Sb_2S_3 = 339.7 g/mol

Putting values in equation 1, we get:

\text{Moles of }Sb_2S_3=\frac{1000g}{339.7g/mol}=2.944mol

  • <u>For hydrogen gas:</u>

Given mass of hydrogen gas = 10.0 g

Molar mass of hydrogen gas = 2 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen gas}=\frac{10.0g}{2g/mol}=5mol

  • <u>For hydrogen sulfide:</u>

Given mass of hydrogen sulfide = 72.6 g

Molar mass of hydrogen sulfide = 34 g/mol

Putting values in equation 1, we get:

\text{Moles of hydrogen sulfide}=\frac{72.6g}{34g/mol}=2.135mol

The chemical equation for the reaction of antimony sulfide and hydrogen gas follows:

                  Sb_2S_3(s)+3H_2(g)\rightarrow 2Sb(s)+3H_2S(g)

Initial:            2.944      5

At eqllm:      2.944-x     5-3x         2x        3x

We are given:

Equilibrium moles of hydrogen sulfide = 2.135 moles

Calculating for 'x', we get:

\Rightarrow 3x=2.135\\\\\Rightarrow x=\frac{2.135}{3}=0.712

Equilibrium moles of hydrogen gas = (5 - 3x) = (5 - 3(0.712)) = 2.868 moles

Volume of the container = 25.0 L

Molarity of a solution is calculated by using the formula:

\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}

The expression of K_c for above equation, we get:

K_c=\frac{[H_2S]^3}{[H_2]^3}

The concentration of solids and liquids are not taken in the expression of equilibrium constant.

K_c=\frac{(\frac{2.135}{25})^3}{(\frac{2.868}{25})^3}\\\\K_c=0.412

Hence, the value of K_c is coming out to be 0.412

3 0
3 years ago
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