At constant temperature, He effuses 6.04 times faster than a gas having an unknown molar mass. Determine the molar mass of the unknown gas.

Explanation:

Rate of effusion =

k is a constant and M is molecular mass .

Molecular mass of He = 4 .

rate of effusion of He = r₁

Molecular mass of unknown compound = M

rate of effusion of unknown gas  = r₂

r₁ = k / √4

r₂ = k / √M

r₁ / r₂ = √ ( M/4 )

6.04 = √ ( M/4 )

36.48 = M/4

M = 146 approx .

molecular mass of unknown gas = 146 .

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.