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sveticcg [70]
3 years ago
12

A 255 ml gas sample contains argon and nitrogen gas at a temperature of 65 oc. the total pressure of the sample is 725 mmhg, and

the partial pressure of argon is 231 mmhg. what mass of nigrogen gas is present in the sample?
Chemistry
1 answer:
Serhud [2]3 years ago
6 0
We calculate first for the number of moles of gases in the sample through the ideal gas equation.
                                       n = PV/RT
      n = (725 mmHg/760 mmHg/atm)(0.255 L) / (0.0821 L.atm/mol.K)(65 + 273.15)  
                                       n = 8.76 x 10^-3 mol
Then, we calculate for the mol N2 using the ratio of the pressure.
                          n N2 = (8.76 x 10^-3 mols)(231 mmHg/725 mmHg)
                           n N2 = 2.79 x 10^-3 moles
Then, multiply the value with the molar mass of N2 which is 28 grams per mol giving us the answer of 0.078 grams. 
                                   
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zavuch27 [327]

Answer:

K_2=\frac{[NOBr]^4_{eq}}{[NO]^4_{eq}[Br]^2_{eq}}

Explanation:

Hello,

In this case, for the equilibrium condition, the equilibrium constant is defined via the law of mass action, which states that the division between the concentrations of the products over the concentration of the reactants at equilibrium equals the equilibrium constant, for the given reaction:

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Be sure to convert T from Celsius to Kelvin by adding 273.

Also I prefer to deal with pressure in atm rather than mmHg, so divide the pressure by 760 to get it in atm.

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