betsylancer8522
05.05.2020 •
Chemistry
What is the pressure in mm of Hg , of a gas mixture that contains 1g of H2, and 8.0 g of Ar in a 3.0 L container at 27°C.
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Ответ:
4362.4 mmHg
Explanation:
Step 1:
Determination of the number of mole of H2 and the number of mole of Ar.
Molar Mass of H2 = 2x1 = 2g/mol
Mass of H2 = 1g
Number of mole of H2 =?
Number of mole = Mass/Molar Mass
Number of mole of H2 = 1/2 = 0.5 mole
Molar Mass of Ar = 40g/mol
Mass of Ar = 8g
Number of mole of Ar =?
Number of mole = Mass/Molar Mass
Number of mole of Ar = 8/40 = 0.2 mole
Step 2:
Data obtained from the question.
Volume (V) of the mixture = 3L
Temperature (T) = 27°C = 27°C + 273 = 300K
Number of mole (n) of the mixture = 0.5 + 0.2 = 0.7 mole
Gas constant (R) = 0.082atm.L/Kmol
Pressure (P) of the mixture =?
Step 3:
Determination of the pressure of the mixture.
The pressure of the mixture can be obtained as follow:
PV = nRT
P x 3 = 0.7 x 0.082 x 300
Divide both side by 3
P = (0.7 x 0.082 x 300) /3
P = 5.74 atm
Step 4:
Conversion of the pressure obtained from atm to mmHg.
1 atm = 760mmHg
Therefore, 5.74 atm = 5.74 x 760 = 4362.4 mmHg
Ответ:
that's a good question
Explanation:
very good question