how many atoms of iron would you have in 4.506 g of beryllium?\na 3.01×10²³ atoms\nb 6.02×10²³ atoms\nc…

Explanation:

Step1: Find molar mass of beryllium

The molar mass of beryllium (Be) is approximately 9.012 g/mol.

Step2: Calculate number of moles of beryllium

The number of moles $n$ is calculated by the formula $n=\frac{m}{M}$, where $m = 4.506$ g and $M=9.012$ g/mol. So $n=\frac{4.506}{9.012}=0.5$ mol.

Step3: Use Avogadro's number

1 mole of any substance contains $6.02\times 10^{23}$ atoms (Avogadro's number $N_A$). For 0.5 mol of beryllium, the number of atoms $N=n\times N_A$. So $N = 0.5\times6.02\times 10^{23}=3.01\times 10^{23}$ atoms. But the question asks about atoms of iron which is a wrong - setup as the mass given is of beryllium. However, if we assume it's a mis - ask and we just calculate the number of atoms in the given mass of beryllium, the number of atoms is $3.01\times 10^{23}$ atoms.

Answer:

A. $3.01\times 10^{23}$ atoms