hey there!:
H2S(aq) <=> H⁺(aq) + HS⁻(aq)
K'c = [H⁺][HS⁻]/[H₂S] = 9.5*10⁻⁸
HS⁻(aq) <=> H⁺(aq) + S²⁻(aq)
K"c = [H⁺][S²⁻]/[HS⁻] = 1.0*10⁻¹⁹
H₂S(aq) <=> 2 H⁺(aq) + S²⁻(aq)
Kc = [H⁺]²[S²⁻] / [H₂S]
= [H+][HS⁻] / [H₂S] * [H⁺][S²⁻]/[HS⁻]
= K'c *K"c
= ( 9.5*10⁻⁸ ) * ( 1.0 x 10⁻¹⁹ )
= 9.5*10⁻²⁷
Hope this helps!
The mass stays constant as a substance changes from a liquid to a gas.
The Law of Conservation of Mass states that, in ordinary chemical reactions, mass is neither destroyed nor created.
That is, the mass of the reactants must equal the mass of the products.
2H₂O(ℓ) ⟶ 2H₂O(g)
1 g 1 g
If the mass of liquid water is 1 g, the mass of the water vapour must be 1 g.
Even though the water vapour is a gas and you can’t see it, it still has a mass
of 1 g.
Answer:
C. at low temperature and low pressure.
Explanation:
- <em>Le Châtelier's principle </em><em>states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.</em>
<em />
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.</em>
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<em><u>Effect of pressure:</u></em>
- When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.
- The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.
<em>So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), </em><em>so the equilibrium partial pressure of CO (g) can be maximized at low pressure.</em>
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<u><em>Effect of temperature:</em></u>
- The reaction is exothermic because the sign of ΔH is (negative).
- So, we can write the reaction as:
<em>2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.</em>
- Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, <em>so the equilibrium partial pressure of CO (g) can be maximized at low temperature.</em>
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<em>C. at low temperature and low pressure.</em>
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