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spin [16.1K]
3 years ago
6

What is the partial pressure of water vapor in an air sample when the total pressure is 1.00 atm, the partial pressure of nitrog

en is 0.79 atm, the partial pressure of oxygen is 0.19 atm, and the partial pressure of all other gases in air is 0.0044 atm?
Chemistry
1 answer:
Inga [223]3 years ago
7 0

Answer:

0.0156 atm

Explanation:

Let the partial pressure of water vapor be P_w.

Given:

Total pressure of air sample is, P_T=1.00\ atm

Partial pressure of nitrogen is, P_n=0.79\ atm

Partial pressure of oxygen is, P_o=0.19\ atm

Partial pressure of all other gases is, P_{other}=0.0044\ atm

From Dalton's law of partial pressure, we know that, the total pressure of a mixture of gases is equal to sum of partial pressure of each individual gas.

So, total pressure of air sample is equal to the sum of partial pressure of nitrogen, oxygen, other gases and water vapor.

Therefore, framing in equation form, we have:

P_n+P_o+P_{other}+P_w=P_T

Plug in the given values and solve for P_w. This gives,

0.79+0.19+0.0044+P_w=1.00\\\\0.9844+P_w=1.00\\\\P_w=1.00-0.9844\\\\P_w=0.0156\ atm

Therefore, the partial pressure of water vapor in an air sample is 0.0156 atm.

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3 years ago
A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati
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Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.  

\text{calculating the initial HI}= \frac{mol}{V}

                                       =\frac{9.3 \times 10 ^ -3}{2}

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\text{Similarly}\ \  I_2 \ \  \text{follows} \ \  H_2 = 0 }

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It is defined that:

I_2 = 6.29 \times 10^{-4}  \ M \\\\x = I_2 \\\\

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          =0.00465 -2 \times 6.29 \times 10^{-4} \\\\ =  0.00465 -\frac{25.16 }{10^4}  \\\\   = 0.003392\  M

Now, we calculate the position:  

For the reaction H 2(g) + I 2(g)\rightleftharpoons  2HI(g), you can calculate the value of Kc at 1000 K.  

data expression for Kc

2HI \rightleftharpoons  H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}

         = \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386

calculating the reverse reaction

H_2(g) + I_2(g)\rightleftharpoons  2HI(g)

Kc = \frac{1}{Kc} \\\\

     = \frac{1}{0.034386}\\ \\= 29.081\\

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Answer:

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