How many grams of NH4OH are present in 37.0 grams of 6.00 M NH4OH which has a density of 0.960 g/mL?

The mass of NH₄OH  present in the solution is 8.085 g

Explanation:

Given;

molarity of NH₄OH = 6 M

density of NH₄OH = 0.960 g/mL

mass of NH₄OH  = 37.0 grams

volume of  NH₄OH = mass /density

                                = 37/0.96

                                = 38.54 mL

molarity = moles of solute / Liters of solution

moles of solute  = molarity  x Liters of solution

moles of solute  = 6 x  38.54 x 10⁻³

moles of solute  = 0.231 moles

Reacting mass = number of moles x molecular mass

Molecular mass of NH₄OH = 35 g/mol

Reacting mass = 0.231 mol x 35 g/mol

Reacting mass = 8.085 g

Therefore, the mass of NH₄OH  present in the solution is 8.085 g

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

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 :