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SVETLANKA909090 [29]
3 years ago
7

B. The equilibrium constant for the reaction 2H2(g) + S2(g) 2H2S(g) is Keq = . (3 points)

Chemistry
1 answer:
eimsori [14]3 years ago
4 0

Answer:

i. Keq=4157.99.

ii. More hydrogen sulfide will be produced.

Explanation:

Hello,

i. In this case, for the concentrations at equilibrium on the given chemical reaction, the equilibrium constant results:

Keq=\frac{[H_2S]^2}{[H_2]^2[S_2]} =\frac{(0.97M)^2}{(0.051M)^2(0.087)} =4157.99

ii. Now, by means of the Le Chatelier's principle, the addition of a reactant shifts the reaction towards products, it means that more hydrogen sulfide will be produced in order to reach equilibrium.

Best regards.

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graphite is a form of carbon in which carbon atoms are arranged in a layer.

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12.2 scientific notation
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This would be 1.22 x 10^1

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3 0
2 years ago
What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?
bekas [8.4K]

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

6 0
2 years ago
Read 2 more answers
1: At which temperature would a reaction withΔH = -102 kJ/mol, ΔS = -0.188 kJ/(mol×K) be spontaneous? 2: At which temperature wo
Naddik [55]

Answer:

1: At temperatures below 542.55 K

2: At temperatures above 660 K

Explanation:

Hello there!

In this case, according to the thermodynamic definition of the Gibbs free energy, it is possible to write the following expression:

\Delta G=\Delta H-T\Delta S

Whereas ΔG=0 for the spontaneous transition. In such a way, we proceed as follows:

1:

0=\Delta H-T\Delta S\\\\T=\frac{-102kJ/mol}{-0.188kJ/mol-K} \\\\T=542.55K

It means that at temperatures lower than 542.55 K the reaction will be spontaneous.

2:

0=\Delta H-T\Delta S\\\\T=\frac{132kJ/mol}{0.200kJ/mol-K} \\\\T=660K

It means that at temperatures higher than 660 K the reaction will be spontaneous.

Best regards!

7 0
2 years ago
What mass of copper is deposited when a current of 10.0a is passed through a solution of copper(ii) nitrate for 30.6 seconds?
asambeis [7]
Data Given:

Time = t = 30.6 s

Current = I = 10 A

Faradays Constant = F = 96500

Chemical equivalent = e = 63.54/2 = 31.77 g

Amount Deposited = W = ?

Solution:
             According to Faraday's Law,

                                          W  =  I t e / F

Putting Values,
 
                            W  =  (10 A × 30.6 s × 31.77 g) ÷ 96500

                            W  =  0.100 g

Result:
           0.100 g of Cu
²⁺ is deposited.
3 0
3 years ago
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