Ne(g) effuses at a rate that is times that of br2(g) under the same conditions.

Answer : The Ne(g) effuses at a rate that is 2.828 times that of under the same conditions.

Solution : Given,

Molar mass of neon = 20 g/mole

Molar mass of bromine = 160 g/mole

Rate of diffusion : It is defined as the rate of diffusion is inversely proportional to the square root of the molar mass of the gas.

Formula used :

where,

= Rate of diffusion of neon gas

= Rate of diffusion of bromine gas

= Molar mass of neon gas

= Molar mass of bromine gas

Now put all the given values in this formula, we get

Therefore, the Ne(g) effuses at a rate that is 2.828 times that of under the same conditions.

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 :

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