Please help with these two chemistry questions. Image attached.

The answers for the following questions is answered below.

Explanation:

1. Given:

no of molecules (n) = 6.023 ×

volume of gas at STP conditions (V) =22.4 liters

To solve:

Volume of gas (v)

We know;

1 × 22.4 = v

v =22.4 L

Therefore the volume of a gas is 22.4 L

2. Given:

volume of oxygen gas (v) = 53 L

Volume of gas at STP conditions (V) =22.4 L

To solve:

no of moles of oxygen gas

We know;

n =

n =2.37 moles

Therefore the number of moles of oxygen are 2.37 moles

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 :