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djyliett [7]
3 years ago
9

2. What is the temperature (°C) of 1.50 moles of gas stored in a 10.0 L container at 1558 mm

Chemistry
1 answer:
nalin [4]3 years ago
6 0

Answer:

-106°C

Explanation:

1558 mmHg / 1 × 1 atm / 760. mmHg = 2.05 atm

PV = nRT

2.05 atm (10.0 L) = 1.50 mol (0.082 atmL/molK) T

20.5 atmL = (0.123 atmL/K) T

K/0.123 atm L × 20.5 atmL = (0.123 atmL/K) T × K/0.123 atmL

166.666 K = T

167 K = T (sig figs)

K = °C + 273

167 - 273 = °C - 273

-106 = °C

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The equilibrium concentrations of the reactants and products are [HA]=0.280 M, [H+]=4.00×10−4 M, and [A−]=4.00×10−4 M. Calculate
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6.24

Explanation:

The following data were obtained from the question:

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

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Next, we shall write the balanced equation for the reaction. This is given below:

HA <===> H+ + A-

Next, we shall determine the equilibrium constant Ka for the reaction. This can be obtained as follow:

Equilibrium constant for a reaction is simply the ratio of concentration of the product raised to their coefficient to the concentration of the reactant raised to their coefficient.

The equilibrium constant for the above equation is given below:

Ka = [H+] [A−] /[HA]

Concentration of HA, [HA] = 0.280 M,

Concentration of H+, [H+] = 4×10¯⁴ M

Concentration of A-, [A−] = 4×10¯⁴ M

Equilibrium constant (Ka) =

Ka = (4×10¯⁴ × 4×10¯⁴) / 0.280

Ka = 1.6×10¯⁷/ 0.280

Ka = 5.71×10¯⁷

Therefore, the equilibrium constant for the reaction is 5.71×10¯⁷

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Equilibrium constant, Ka = 5.71×10¯⁷

pKa =?

pKa = – Log Ka

pKa = – Log 5.71×10¯⁷

pKa = 6.24

Therefore, the pka for the reaction is 6.24.

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