Answer :
Answer:
[tex]m_{H_2}=6.2x10^{-3}gH_2[/tex]
Explanation:
Hello,
In this case, for the mentioned scheme, we can assume hydrogen could be ideal at the given temperature and 1 atm of pressure since its molecules will not present significant interactions taking it to non-idealities. Moreover, when we collect a gas in water, 1 atm is a common pressure under normal conditions. In such a way, we can compute the moles of hydrogen gas by using the ideal gas equation:
[tex]n=\frac{PV}{RT}=\frac{1atm*80mL*\frac{1L}{1000mL} }{0.082\frac{atm*L}{mol*K}*(40+273)K} =3.1x10^{-3}molH_2[/tex]
Finally, we compute the mass by using hydrogen's atomic mass:
[tex]m_{H_2}=3.1x10^{-3}molH_2*\frac{2gH_2}{1molH_2}\\ \\m_{H_2}=6.2x10^{-3}gH_2[/tex]
Best regards.
The mass of H₂ that is present in the collection tube is 6.22×10¯³ g
We'll begin by calculating the number of mole of the H₂ in the tube. This can be obtained by using the ideal gas equation as illustrated below:
Temperature (T) = 40 °C = 40 + 273 = 313 K
Volume (V) = 80 mL = 80 / 1000 = 0.08 L
Pressure (P) = 1 atm
Gas constant (R) = 0.0821 atm.L/Kmol
Number of mole (n) =?
PV = nRT
1 × 0.08 = n × 0.0821 × 313
0.08 = n × 25.6973
Divide both side by 25.6973
n = 0.08 / 25.6973
n = 0.00311 mole
- Finally, we shall determine the mass of the H₂ in the tube.
Mole of H₂ = 0.00311 mole
Molar mass of H₂ = 1 × 2 = 2 g/mol
Mass of H₂ =?
Mass = mole × molar mass
Mass of H₂ = 0.00311 × 2
Mass of H₂ = 6.22×10¯³ g
Thus, the mass of the H₂ in the tube is 6.22×10¯³ g
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