Chemistry, asked by harshdeepsingh89, 8 months ago

How much does the mass of NH3 contain same number of atom as are present in 4 gram atom of Oxygen

Answers

Answered by Anonymous
2

Answer:

Let moles of NH3 = x

No. of atoms in 1 molecule of NH3 = (1+3) = 4

Therefore,

No. of atoms in x moles NH3 = x * N * 4 …..eq(i)

(where N is Avogadro no.)

No of moles in 4g O2 = 4/32 (Mol. Wt of O2 = 32)

No of atoms in 1 molecule of O2 = 2

Therefore,

No. of atoms in 4g O2 = (4/32) * N * 2 …..eq(ii)

Equating eq (i) & (ii),

x * N * 4 = (4/32) * N * 2

Therefore, x = 16

hope u like it!

Answered by Anonymous
8

Answer:

Let moles of NH3 = x

No. of atoms in 1 molecule of NH3 = (1+3) = 4

Therefore,

No. of atoms in x moles NH3 = x * N * 4 …..eq(i)

(where N is Avogadro no.)

No of moles in 4g O2 = 4/32 (Mol. Wt of O2 = 32)

No of atoms in 1 molecule of O2 = 2

Therefore,

No. of atoms in 4g O2 = (4/32) * N * 2 …..eq(ii)

Equating eq (i) & (ii),

x * N * 4 = (4/32) * N * 2

Therefore, x = 16

Explanation:

@MNF

Similar questions