How many liters of a 3.67 m k2so4 solution are needed to provide 57.3 g of k2so4(molar mass 174.01 g/mol)? recall that m is equivalent to mol/l.

Amount of volume needed in Litres is 0.09L

Explanation:

Given that the concentration in Mol/L =3.67M

Mass of K2SO4 = 57.3g

Molar mass of K2SO4 = 174.01g/mol

From stochiometry,

Mole = Mass/Molar mass

Mole = 57.3/174.01= 0.33mol.

Concentration in mol/L=

Mole/volume

3.67=0.33/volume

Volume= 0.33/3.67

Volume = 0.09L

Therefore the volume of K2SO4 needed is 0.09L

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 :

what are the statements