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VikaD [51]
3 years ago
15

H2S(g) is an endothermic process. A 6.1589−g sample of the solid is placed in an evacuated 4.000−L vessel at exactly 24°C. After

equilibrium has been established, the total pressure inside is 0.709 atm. Some solid NH4HS remains in the vessel. (a) What is the KP for the reaction? (b) What percentage of the solid has decomposed? % (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel? More than 50% would remain. Between 20% and 50% would remain. Between 10% and 20% would remain. Less than 10% would remain.
Chemistry
1 answer:
rewona [7]3 years ago
7 0

s an endothermic process. A 6.1589-g sample of the solid is placed in an evacuated 4.000-L vessel at exactly 24°C. After equilibrium has been established, the total pressure inside is 0.709 atm hope this helps

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Can you pls tell me the word equations for all these equations​
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Below

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5 0
3 years ago
The ph of a 0.0100 m solution of the sodium salt of a weak acid is 11.00. what is the ka of the acid?
Vitek1552 [10]
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
     pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
     [OH-] = 10^-pOH = 10^-3 = x

Considering the sodium salt NaA in water, we have the equation
     NaA → Na+ + A- 
hence, [A-] = 0.0100 M

Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
                             A- + H2O ⇌ HA + OH-
     Initial             0.0100            0       0
     Change        -x                    +x     +x
     Equilibrium    0.0100-x         x       x

We can now calculate the Kb for A-:
     Kb = [HA][OH-] / [A-] 
           = x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
     Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4

We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
     Ka = Kw / Kb 
           = 1.00x10^-14 / 1.00x10^-4
           = 1.00x10^-10
7 0
3 years ago
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