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svet-max [94.6K]
3 years ago
8

At 25 celsius, 1.00 ,ole of O2 was found to occupy a volume of 12.5 L at a pressure of 198 kPa. What value of the gas constant i

n L times kPa divided by mol Times K
Chemistry
1 answer:
mr Goodwill [35]3 years ago
4 0

Answer:

R=8.301\frac{L*kPa}{mol*K}

Explanation:

Hello,

In this case, assuming oxygen as an ideal gas, it is described by:

PV=nRT

Whereas the gas constant R is required, therefore, it is computed as shown below:

R=\frac{PV}{nT} =\frac{198kPa*12.5L}{1.00mol*(25+273.15)K} \\\\R=8.301\frac{L*kPa}{mol*K}

Best regards.

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adell [148]

Answer:

1.21x10^{-3} M

Explanation:

Henry's law relational the partial pressure and the concentration of a gas, which is its solubility. So, at the sea level, the total pressure of the air is 1 atm, and the partial pressure of O2 is 0.21 atm. So 21% of the air is O2.

Partial pressure = Henry's constant x molar concentration

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H = 152.17 atm/M

For a pressure of 665 torr, knowing that 1 atm = 760 torr, so 665 tor = 0.875 atm, the ar concentration is the same, so 21% is O2, and the partial pressure of O2 must be:

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0.1837 = 152.17x[O2]

[O2] = 0.1837/15.17

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