Question

the ratio of oxygen atoms present in one mole of cupric nitrite and ferric sulphite

Answer 1

Answer:

Cupric Nitrite - Cu(NO2)2

Ferric Sulphite - Fe2 (SO3)3

The number of atoms of oxygen in Cupric Nitrite is 2×2×6×10^23 = 4×6×10^23

Ferric Sulphite ==> 3×3×6×10^23 atoms

Ratio is 4 : 9

Answer 2

Ratio of oxygen atoms in one mole of cupric nitrite and ferric sulfite is 4 : 9

Explanation:

We are given:

Two chemical compounds having chemical formula $Cu(NO_2)_2$ (cupric nitrite) and $Fe_2(SO_3)_3$ (ferric sulfite)

1 mole of cupric nitrite contains 4 moles of oxygen atoms.

1 mole of ferric sulfite contains 9 moles of oxygen atoms.

According to mole concept:

1 mole of a chemical substance contains $6.022\times 10^{23}$ number of particles.

So, 1 mole of cupric nitrite will contain $(4\times 6.022\times 10^{23})$ number of oxygen atoms

And, 1 mole of ferric sulfite will contain $(9\times 6.022\times 10^{23})$ number of oxygen atoms

Taking the ratio of oxygen atoms in two compounds:

$\frac{n_O(Cu(NO_2)_2)}{n_O(Fe_2(SO_3)_3)}=\frac{(4\times 6.022\times 10^{23})}{(9\times 6.022\times 10^{23})}\\\\\frac{n_O(Cu(NO_2)_2)}{n_O(Fe_2(SO_3)_3)}=\frac{4}{9}$

Learn more about mole concept:

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