What is the mass of 1.7 × 1023 atoms of zinc

We are given with a compound, Zinc (Zn) having a 1.7 x 10 ^23 atoms. We are tasked to solve for it's corresponding mass in g. We need to find first the molecular weight of Zinc, that is

Zn= 65.38 g/mol

Not that 1 mol=6.022x10^{23} atoms, hence,

1.7 x 10 ^23 atoms x 1 mol/6.022x10^{23} atoms x65.38 g/ 1mol

=18.456 g of Zn

Therefore, the mass of Zinc 18.456 g

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 :

oxygen, selenium, tellurium, and polonium.

they're all from the same group as sulfur (group 16).