Question

Moles of o atoms in 9.00g of mg(no3)2

Answer 1

Answer:

Explanation:relative molecular  mass og Mg(NO3)2=24+2(14+48)=124+124=248

248g of Mg(NO3)2 contain 6 atom of O

then 9g will contain 6*9/248=54/248=0.21 atom of O

Answer 2

Moles of O atoms in 9.00g of $(Mg(NO_3)_2$ is 0.402

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number $6.023\times 10^{23}$ of particles.

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{9.00g}{148.3g/mol}=0.067moles$

1 mole of magnesium nitrate $(Mg(NO_3)_2$ contains= 6 moles of oxygen atoms

0.067 moles of magnesium nitrate $(Mg(NO_3)_2$ contains= $\frac{6}{1}\times 0.067=0.402moles$ of oxygen atoms

Thus there are 0.402 moles of O atoms in 9.00g of $(Mg(NO_3)_2$

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