Question

What weight of NH3 contains same number of atoms as in 64g of oxygen:​

Answer 1

Answer:

Let moles of NH3 = x

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,No. of atoms in x moles NH3 = x * N * 4 …..eq(i)

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,No. of atoms in x moles NH3 = x * N * 4 …..eq(i)(where N is Avogadro no.)

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,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)

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,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

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,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 = 2Therefore,

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,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 = 2Therefore,No. of atoms in 4g O2 = (4/32) * N * 2 …..eq(ii)

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,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 = 2Therefore,No. of atoms in 4g O2 = (4/32) * N * 2 …..eq(ii)Equating eq (i) & (ii),

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,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 = 2Therefore,No. of atoms in 4g O2 = (4/32) * N * 2 …..eq(ii)Equating eq (i) & (ii),x * N * 4 = (4/32) * N * 2

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,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 = 2Therefore,No. of atoms in 4g O2 = (4/32) * N * 2 …..eq(ii)Equating eq (i) & (ii),x * N * 4 = (4/32) * N * 2Therefore, x = 16

Explanation:

@MNF

Answer 2

Answer:

Let moles of NH3 = xNo. of atoms in 1 molecule of NH3 = (1+3) = 4Therefore,No.

Explanation:

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