A0.02877 g sample of gas occupies 10.0-ml at 289.0 k and 1.10 atm. upon further analysis, the compound is found to be 38.734% c and 61.266% f. what is the molecular formula of the compound? (order: c, x)

Assuming ideal gas, we calculate the number of moles.

PV = nRT

n = PV / RT

n = (1.10 atm) (10 x 10^-3L) / [(0.08205746 L atm / mol K) (289 K)]

n = 4.64 x 10^-4 mol

SO the molar mass is:

molar mass = 0.02877 grams / 4.64 x 10^-4 mol

molar mass = 62.02 g/mol

The only possible combination is:

C2F2

C = 12 * 2 = 24

F = 19 * 2 = 38

24 + 38 = 62 g/mol

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

answerto predict;