A) what is the molar fraction of each gas in a mixture of 6g of hydrogen H2 and 7g of nitrogen N2 ?

B) what is the total pressurd exerted by the above mixture in a volume of 15 l at a temperature of 0ºc ?
C) what is the partial pressure of each gas in the above mixture?
R=8.314J.mol.k

a.

mols H2 = g/molar mass = ?
mols N2 = g/molar mass = ?
XH2 = mols H2/total mols
XN2 = mols N2/total mols

b.
PV = nRT where n is total mols. P will be in kPa if R is 8.314 and that should be J/mol.K

c.
pH2 = XH2*Ptotal
pN2 = XN2*Ptotal

A. Mole Fraction (X₁) = [ moles of substance(n₁) / Total moles of all substances (∑n)

Moles H₂ = (6 gms/ 2 gms/mol) = 3 moles
Moles N₂ = (7 gms/ 14 gms/mol) = 0.50 mole
∑moles = (3 + 0.50)moles = 3.5 moles
X(H₂) = [(moles H₂)/(moles H₂ + moles N₂)] = 3/(3.0 + 0.50) = 3/3.5 = 0.8571
X(N₂) = 1 – X(H₂) = 1 – 0.8571 = 0.1486

B.Given 15.0 Liters total Volume at 0⁰C (=273K); use the Ideal Gas Law to calculate the partial pressures of H₂ and N₂.
PV = nRT => P = nRT/V
For H₂: n = 3 moles
For N₂: n = 0.50 mole
R = Constant = 0.08206 L-Atm/mol-K
T = 0⁰C = 273K
V(Total) = 15 Liters

(The following answer C-part)
P(H₂) = (3 moles)(0.08206 l-atm/mol-K)(273K)/(15 L) = 4.48 Atm x 760 mm-Hg/Atm = 3405 mm-Hg

P(N₂) = (0.50 mole)(0.08206 l-atm/mol-K)(273K)/(15 L) = 0.747 Atm x 760 mm-Hg/Atm = 568 mm-Hg

(The following answer B-Part)
Total Pressure = P(H₂) + P(N₂) = 4.48 Atm + 0.747 Atm = 5.23 Atm x 760 mm-Hg/Atm =3973 mm-Hg

To answer these questions, we need to use the concept of molar mass, molar fraction, and the ideal gas law. Let's go step by step:

A) Molar Fraction:
The molar fraction of a gas in a mixture is the ratio of the number of moles of that gas to the total number of moles in the mixture.

1. Calculate the number of moles for hydrogen H2:
Given mass of hydrogen H2 = 6 g
Molar mass of hydrogen H2 = 2 g/mol (since 2 hydrogen atoms combine to form 1 molecule of H2)
Number of moles of hydrogen H2 = (Mass of hydrogen H2) / (Molar mass of hydrogen H2)

2. Calculate the number of moles for nitrogen N2:
Given mass of nitrogen N2 = 7 g
Molar mass of nitrogen N2 = 28 g/mol (since N2 molecule has 2 nitrogen atoms, each with atomic mass 14)
Number of moles of nitrogen N2 = (Mass of nitrogen N2) / (Molar mass of nitrogen N2)

3. Calculate the total number of moles in the mixture:
Total moles in the mixture = Moles of hydrogen H2 + Moles of nitrogen N2

4. Calculate the molar fraction for each gas:
Molar fraction of hydrogen H2 = (Moles of hydrogen H2) / (Total moles in the mixture)
Molar fraction of nitrogen N2 = (Moles of nitrogen N2) / (Total moles in the mixture)

B) Total Pressure:
To calculate the total pressure exerted by the gas mixture, we can use the ideal gas law equation:

PV = nRT

Where:
P is the pressure,
V is the volume,
n is the number of moles,
R is the ideal gas constant,
T is the temperature in Kelvin.

1. Convert the temperature from Celsius to Kelvin:
Temperature in Kelvin = Temperature in Celsius + 273.15

2. Plug in the values into the ideal gas law equation:
(Pressure in the mixture) * (Volume) = (Total moles in the mixture) * (Ideal gas constant R) * (Temperature in Kelvin)

3. Solve for the pressure in the mixture:
Pressure in the mixture = [(Total moles in the mixture) * (Ideal gas constant R) * (Temperature in Kelvin)] / (Volume)

C) Partial Pressure:
The partial pressure of each gas in a mixture is the pressure that the gas would exert if it occupied the entire volume alone at the same temperature.

To calculate the partial pressure of each gas, we can use Dalton's Law of Partial Pressures:

Partial Pressure of a gas = (Molar fraction of the gas) * (Total pressure of the mixture)

Plug in the molar fraction for each gas and the total pressure of the mixture calculated in part B to find the partial pressure of each gas.

Remember to use consistent units throughout the calculations.