Answer:

Explanation:
From the question we are told that:
Pressure 
Temperature 
Volume 
Heat Produced 
Generally the equation for ideal gas is mathematically given by



Therefore


Since
Heat of combustion of Methane=889 kJ/mol
Heat of combustion of Propane=2220 kJ/mol
Therefore

Comparing Equation 1 and 2 and solving simultaneously




Therefore
Mole fraction 0f Methane is mathematically given as



Answer:
east to west is the answer
1m = 100cm
so 10m = 100*10 = 1000cm or in scientific notation 1.00x10^3 cm
1g = 1/1000kg
1mL = 1/1000L
so 1g/mL = (1/1000)/(1/1000)kg/L
=1kg/L
37.5g/mL = 37.5kg/L or 3.75*10^1 kg/L