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olga_2 [115]
3 years ago
15

A 8.95 L container holds a mixture of two gases at 21 °C. The partial pressures of gas A and gas B, respectively, are 0.373 atm

and 0.650 atm. If 0.220 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
Chemistry
1 answer:
Amanda [17]3 years ago
6 0

Answer:

The total pressure is 1.616 atm

Explanation:

First of all we say:

In a closed system, sum of partial pressures is the value for the total pressure.

In our system, we have gas A and B

Total pressure without the third gas is = 0.373 atm + 0.650 atm = 1.023 atm

As we add a third gas, with no change in volume or T°, let's find out by the Ideal Gases Law, its pressure:

P . V = n . R . T

P = ( n . R . T) / V

P = (0.220 mol . 0.082 . (273 + 21°C)) / 8.95L = 0.593 atm

273 + 21°C → Absolute value of T°

Let's sum the partial pressures, then:

0.373 atm + 0.650 atm + 0.593 atm = 1.616 atm.

It's ok to say that the total pressure was increased, because we have more gas now.

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The dissociation of formic acid is:

HCOOH \rightleftharpoons HCOO^{-} + H^{+}

The acid dissociation constant of formic acid, k_a is:

k_a = \frac{[HCOO^{-}]  [H^{+}]}{HCOOH}

Rearranging the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

pH = 2.75

pH = -log[H^{+}]

[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}

pk_a = 3.75

k_a = 10^{-3.75} = 1.78\times 10^{-4}

Substituting the values in the equation:

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}

\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}

Hence, the ratio is \frac{1}{10}.

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How many grams of NaCL are required to prepare 50ml of a 2.0 molar solution​
Ksivusya [100]

Answer:

5.85 gm.

Explanation:

We know that,

Normality =<u> Molarity × Molecular </u><u>weight</u>

Equivalent weight

Since molecular weight of NaCl= equivalent weight = 23+35.5 =58.5

Normality of NaCl= molarity=2

Now,

Normality= <u>weight</u><u> </u><u>in</u><u> </u><u>gram</u><u> </u><u>×</u><u>1000</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

Volume ×equivalent weight

Weight in gram is given by,

<u>=</u><u>Normality × Volume × equivalent </u><u>weight</u>

1000

= <u>2× 50 × 58.</u><u>5</u>

1000

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3 years ago
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