How many grams are there in 9.4 x1023 molecules of h2?

In order to solve this, it is recommended that you use stoichiometry. You need to use the mole hill and go from molecules to moles to grams. In order to go from molecules to moles, you must use Avogadro's number (6.02x10^23). Next, you need to go from moles to grams using atomic mass (I usually just go out to two decimal places). So, your atomic mass should be around 2.02g for H2. Your final answer should be around 3.1g H2 (don't forget  significant figures!). Hope this helped

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 :