Question

Calculate the no. Of nitrate ions present in 462g of beryllium nitrate

Answer 1

Answer: 41.844×10^23 nitrate ions

Explanation:

Answer 2

The no. Of nitrate ions present in 462 g of beryllium nitrate are $41.82 \times 10^{23}$ atoms.

Explanation:

It is given that mass of beryllium nitrate is 462 g. And, according to mole concept, 1 mole of every substance contains $6.022 \times 10^{23}$ atoms.

As molar mass of beryllium nitrate is 133.022 g/mol. Therefore, first we will calculate the number of moles present in beryllium nitrate as follows.

            No. of moles = $\frac{mass}{\text{molar mass}}$

                                  = $\frac{462 g}{133.022 g/mol}$

                                  = 3.473 mol

Hence, number of ions or atoms present in 3.473 mol will be calculate as follows.

             No. of atoms or ions = $6.022 \times 10^{23} \times 3.473 mol$                                                                       = $20.91 \times 10^{23}$ atoms

As chemical formula of beryllium nitrate is $Be(NO_{3})_{2}$. Therefore, one molecule of beryllium nitrate will contain these many ions.

               $2 \times 20.91 \times 10^{23}$ atoms

                  = $41.82 \times 10^{23}$ atoms

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