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marshall27 [118]
3 years ago
8

A nitrox ii gas mixture for scuba diving contains oxygen gas at 53 atm and nitrogen gas at 94 atm . what is the total pressure,

in torrs, of the scuba gas mixture?
Chemistry
1 answer:
Alina [70]3 years ago
6 0

Answer: The total pressure will be 1,11,720 torr.

Explanation:

Partial pressure of nitrogen gas.p_{N_2}=53 atm=53\times 760 torr= 40,280 torr (1atm = 760 torr)

Partial pressure of oxygen gas,p_{O_2}=94 atm=94\times 760 torr=71,440 torr

Total pressure, in torrs, of the scuba gas mixture = 40280 + 71440 torr =111,720 torr

Hence , the total pressure in scuba gas mixture is 111,720 torr.

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Using the first volume and temperature reading on the table as V1 and T1, solve for the unknown values in the table below. Remem
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<u>Answer:</u> Th value of A is 3.8\times 10^2K, value of B is 0.84 L, value of C is 1.1 L and the value of D is 2.2\times 10^2K

<u>Explanation:</u>

To calculate the missing values of volume and temperature, we will use Charles' Law.

This law states that volume is directly proportional to the temperature of the gas at constant number of moles and pressure. Equation given by this law is:

\frac{V_1}{T_1}=\frac{V_2}{T_2}      ....(1)

where,

V_1\text{ and }T_1 are initial volume and temperature of the gas

V_2\text{ and }T_2 are final volume and temperature of the gas

  • <u>For A:</u>

V_1=1.0L\\T_1=295K\\V_2=1.3L\\T_2=AK

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{1.3L}{A}\\\\A=383.5K\approx 3.8\times 10^2K

Whenever multiplication and division are involved, the answer must not contain, more number of significant figures as there are in the least precise term.

Hence, the value of A is 3.8\times 10^2K

  • <u>For B:</u>

V_1=1.0L\\T_1=295K\\V_2=BL\\T_2=250K

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{B}{250K}\\\\B=0.84L

Hence, the value of B is 0.84L

  • <u>For C:</u>

V_1=1.0L\\T_1=295K\\V_2=CL\\T_2=325K

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{C}{325K}\\\\C=1.1L

Hence, the value of C is 1.1L

  • <u>For D:</u>

V_1=1.0L\\T_1=295K\\V_2=0.75L\\T_2=DK

Putting values in equation 1, we get:

\frac{1.0L}{295K}=\frac{0.75L}{D}\\\\D=221.25\approx 2.2\times 10^2K

Hence, the value of D is 2.2\times 10^2K

7 0
3 years ago
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