How many grams of KOH are needed to make 0.500 L of a 0.10 M solution

2.8 g KOH

Explanation:

We can find the amount of moles we need by multiplying 0.500 L and the molarity given, 0.10 M (molarity is moles of solute divided by liters of solution). The liters cancel out, leaving us with moles; we can then take the moles and convert it to grams of KOH by using the molar mass of KOH.

0.500 L x 0.10 mol/1 L x 56.106 g /1 mol = 2.8 g KOH (two significant figures)

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 :