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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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An acid with molar mass 84.48 g/mol is titrated with 0.650 M KOH. What volume of KOH solution is needed to titrate 1.70 grams of
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Answer:

V=0.0310L=3.10mL

Explanation:

Hello.

In this case, since the acid is monoprotic and the KOH has one hydroxyl ion only, we can see that at the equivalence point the moles of both of them are the same:

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