Question

M(CuSO1)=4 г Найти n=? N=?

Answer 1

n = m/M = 4г/112 г/моль = 0,036моль

N = n * Na = 0.036 моль * 6, 02 * 10^23 = 0,217 * 10^23 молекул

Answer 2

It looks like there may be a typo in the question as the subscript 1 after SO is not clear.

If the subscript 1 is meant to be a subscript for sulfur (S), the correct formula for copper(II) sulfate is CuSO4.

Assuming that M(CuSO4) = 4 g, we can use the molar mass of copper(II) sulfate to find the number of moles of the compound present:

Molar mass of CuSO4 = atomic mass of Cu + atomic mass of S + 4 x atomic mass of O = 63.55 g/mol + 32.06 g/mol + 4 x 15.99 g/mol = 159.61 g/mol

Number of moles of CuSO4 = Mass of CuSO4 / Molar mass of CuSO4 = 4 g / 159.61 g/mol = 0.0251 mol

To find the number of atoms of copper (Cu) present, we need to multiply the number of moles of CuSO4 by the Avogadro constant, which is approximately 6.022 x 10^23 mol^-1:

Number of atoms of Cu = Number of moles of CuSO4 x Avogadro constant = 0.0251 mol x 6.022 x 10^23 mol^-1 = 1.509 x 10^22 atoms

Therefore, the number of moles of CuSO4 is approximately 0.0251 mol and the number of atoms of copper (Cu) present is approximately 1.509 x 10^22 atoms.