6. how many molecules are in a 189 g sample of carbon tetrabromide, cbr₄?

Explanation:

Step1: Calculate molar mass of $CBr_4$

The molar mass of $C$ is approximately $12.01\ g/mol$, and the molar mass of $Br$ is approximately $79.90\ g/mol$. For $CBr_4$, $M = 12.01+4\times79.90=12.01 + 319.6=331.61\ g/mol$.

Step2: Calculate number of moles

Use the formula $n=\frac{m}{M}$, where $m = 189\ g$ and $M = 331.61\ g/mol$. So $n=\frac{189}{331.61}\approx0.57\ mol$.

Step3: Calculate number of molecules

Use Avogadro's number $N_A = 6.022\times 10^{23}\ molecules/mol$. The number of molecules $N=n\times N_A$. So $N = 0.57\times6.022\times 10^{23}=3.43254\times 10^{23}\ molecules$.

Answer:

$3.43\times 10^{23}$ molecules