Based on the data given, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu
<h3>How can molar mass of a gas be obtained from density, temperature and pressure?</h3>
The molar mass of a gas can be obtained from density, temperature and pressure using the formula below:
- molar mass = density × molar gas constant × temperature/pressure
Molar gas constant, R = R = 0.082 L.atm/mol/K.
Temperature = 150 °C = 423 K
Pressure = 785 torr = 1.033 atm
density = 4.93 g/L
molar mass of gas = 4.93 × 0.082 × 423/1.033
molar mass of gas = 165.5 g/mol
Then, molecular weight of the gas = 165.5 amu
Therefore, the molar mass of the gas is 165.5 g/mol while the molecular weight of the gas is 165.5 amu
Learn more about molar mass of a gas at: brainly.com/question/26215522
Answer:
4g of NH₃
Explanation:
Hello,
To solve this question, we'll require the equation of reaction.
Equation of reaction
N₂ + 3H₂ → 2NH₃
From the equation of reaction, 1 mole of N₂ reacts with 3 moles of H₂ to produce 2 moles of NH₃.
Number moles = mass / molar mass
Molar mass of H = 1.0g/mol
Mass of H₂ = 2 ×1 = 2g
Molar mass of N₂ = 14.0g/mol
Mass = 2 × 14 = 28g
Molar mass of NH₃ = 17g/mol
Mass = 17g
Therefore, 28g of N₂ reacts with 6g of H₂ to produce 34g of NH₃
I.e 28g + 6g(N₂ + H₂) = 34g of NH₃
34g of (N2 + H2) = 34g of NH₃
4g of (N2 + H2) = x g of NH₃
x = (4 × 34) / 34
x = 4g of NH₃
4g of NH₃ will be produced from 1g of N2 and 3g of H2
When 2H+ + 2e- → H2(g)
when the total charge transfer = steady current * time
=1.83 * 34 min * 60s = 3733.2 C
and when each electron has 1.6022x10^-19 C
So the amount of charge = 3733.2 / 1.6022x10^-19
= 2.33x10^22 electrons
To get the moles of the electrons we divided by Avogadro's number
= 2.33x10^22 / 6.022x10^23 = 0.03869 moles
when 2mol e → 1 mol H2
So 0.03869→ ???
moles of H2 = 0.03869 /2 = 0.01935 moles
know we have n = 0.01935 & p = 1.05 atm & R= 0.0821 & T = 25+273=298 K
So by substitution in the ideal gas formula, we can get V:
PV = nRT
1.05 * V = 0.01935 * 0.0821 * 298
∴ V = 0.45 L
