Question

1. what is the mass of 1.28×10^24 atoms of zinc? grams 2. how many zinc atoms are there in 21.2 grams of zinc? atoms use the references to access important values if needed for this question. 2 item attempts remaining try another version submit answer

Explanation:

Step1: Recall molar - mass of zinc

The molar mass of zinc (Zn) is $M = 65.38\ g/mol$. One mole of any substance contains Avogadro's number of atoms, $N_A=6.022\times 10^{23}\ atoms/mol$.

Step2: Solve for the mass in the first question

We know that $n=\frac{N}{N_A}$, where $n$ is the number of moles, $N$ is the number of atoms, and $N_A$ is Avogadro's number. Given $N = 1.28\times 10^{24}$ atoms. First, find the number of moles: $n=\frac{1.28\times 10^{24}}{6.022\times 10^{23}}\ mol\approx2.13\ mol$. Then, use the formula $m = nM$, where $m$ is the mass. So $m=(2.13\ mol)\times(65.38\ g/mol)\approx139\ g$.

Step3: Solve for the number of atoms in the second question

First, find the number of moles of zinc in $21.2\ g$. Using $n=\frac{m}{M}$, with $m = 21.2\ g$ and $M = 65.38\ g/mol$, we get $n=\frac{21.2}{65.38}\ mol\approx0.324\ mol$. Then, find the number of atoms using $N=nN_A$. So $N=(0.324\ mol)\times(6.022\times 10^{23}\ atoms/mol)\approx1.95\times 10^{23}$ atoms.

Answer:

1. $139$
2. $1.95\times 10^{23}$