Question

1. what is the mass of 1.57×10^24 atoms of barium? grams 2. how many barium atoms are there in 22.9 grams of barium? atoms use the references to access important values if needed for this question. 3 item attempts remaining try another version submit answer

Explanation:

Step1: Recall molar - mass of barium

The molar mass of barium (Ba) is approximately 137.33 g/mol. One mole of any substance contains Avogadro's number ($N_A = 6.022\times10^{23}$) of atoms.

Step2: Solve for the mass of 1.57×10²⁴ atoms of barium

First, find the number of moles ($n$) of barium atoms. The formula for the number of moles is $n=\frac{N}{N_A}$, where $N$ is the number of atoms. So, $n=\frac{1.57\times 10^{24}}{6.022\times 10^{23}}$.
$n=\frac{1.57\times 10^{24}}{6.022\times 10^{23}}\approx2.61$ mol.
Then, use the formula $m = n\times M$, where $m$ is the mass, $n$ is the number of moles, and $M$ is the molar - mass. Substitute $n = 2.61$ mol and $M=137.33$ g/mol into the formula.
$m=2.61\times137.33\approx358$ g.

Step3: Solve for the number of barium atoms in 22.9 grams of barium

First, find the number of moles of barium in 22.9 grams. Using $n=\frac{m}{M}$, with $m = 22.9$ g and $M = 137.33$ g/mol.
$n=\frac{22.9}{137.33}\approx0.167$ mol.
Then, find the number of atoms using $N=n\times N_A$.
$N = 0.167\times6.022\times 10^{23}\approx1.006\times 10^{23}$ atoms.

Answer:

1. 358 grams
2. $1.01\times 10^{23}$ atoms