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PtichkaEL [24]
3 years ago
5

Given 6193 mL of a gas at 62.3 °C. What is its volume at 38.1 °C?

Chemistry
1 answer:
Luden [163]3 years ago
5 0

Answer:

5746.0 mL.

Explanation:

We can use the general law of ideal gas:<em> PV = nRT.</em>

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

<em>V₁T₂ = V₂T₁</em>

<em></em>

V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.

V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.

<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>

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Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of HNO_3 = \frac{15.5g}{63g/mol\times 9.5L}=0.026M

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Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c  

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K_c=3.72

Thus  the value of the equilibrium constant Kc for this reaction is 3.72

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