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3 years ago

10. What is the atmospheric pressure if the partial pressures of nitrogen, oxygen, and

Chemistry

1 answer:

sashaice [31]3 years ago

7 0

Answer:

1.009 atm is the total pressure for the mixture

Explanation:

To determine the total pressure in atm, of the three gases (N₂, O₂ and Ar) we have to sum all the values.

Sum of partial pressures in a mixture = Total pressure

First of all, we need to convert the values from mmHg to atm

604.5 mmHg . 1atm / 760 mmHg = 0.795 atm

162.8 mmHg . 1atm / 760 mmHg = 0.213 atm

0.500 mmHg . 1atm / 760 mmHg = 6.58×10⁻⁴ atm

Partial pressure N₂ + Partial pressure O₂ + Partial pressure Ar = Total P

0.795 atm + 0.213 atm + 6.58×10⁻⁴ atm = 1.009 atm

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2 years ago

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3 years ago

What pressure, in atm, would be exerted by 0.023 grams of oxygen (O2) if it occupies 31.6 mL at 91

Vikentia [17]

Answer: A pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

Explanation:

Given : Mass of oxygen = 0.023 g

Volume = 31.6 mL

Convert mL into L as follows.

$1 mL = 0.001 L\\31.6 mL = 31.6 mL \times \frac{0.001 L}{1 mL}\\= 0.0316 L$

Temperature = $91^{o}C = (91 + 273) K = 364 K$

As molar mass of $O_2$ is 32 g/mol. Hence, the number of moles of $O_2$ are calculated as follows.

$No. of moles = \frac{mass}{molar mass}\\= \frac{0.023 g}{32 g/mol}\\= 0.00072 mol$

Using the ideal gas equation calculate the pressure exerted by given gas as follows.

PV = nRT

where,

P = pressure

V = volume

n = number of moles

R = gas constant = 0.0821 L atm/mol K

T = temperature

Substitute the value into above formula as follows.

$PV = nRT\\P \times 0.0316 L = 0.00072 mol \times 0.0821 L atm/mol K \times 364 K\\P = \frac{0.00072 mol \times 0.0821 L atm/mol K \times 364 K}{0.0316 L}\\= 0.681 atm$

Thus, we can conclude that a pressure of 0.681 atm would be exerted by 0.023 grams of oxygen $(O_2)$ if it occupies 31.6 mL at $91^{o}C$ .

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2 years ago