answer the following problem by converting moles to molecules, grams to molecules. write your answer in a 1…

Explanation:

Step1: Determine molar mass

Use molar - mass values of elements from the periodic table to find the molar mass of each compound. For $CCl_4$: The molar mass of $C$ is $12.01\ g/mol$, and the molar mass of $Cl$ is $35.45\ g/mol$. $M_{CCl_4}=12.01 + 4\times35.45=12.01+141.8 = 153.81\ g/mol$ For $(NO_2)2$: The molar mass of $N$ is $14.01\ g/mol$ and the molar mass of $O$ is $16.00\ g/mol$. $M{(NO_2)2}=2\times(14.01 + 2\times16.00)=2\times(14.01 + 32.00)=2\times46.01 = 92.02\ g/mol$ For $CO_2$: The molar mass of $C$ is $12.01\ g/mol$ and the molar mass of $O$ is $16.00\ g/mol$. $M{CO_2}=12.01+2\times16.00 = 44.01\ g/mol$ For $Na_2SO_4$: The molar mass of $Na$ is $22.99\ g/mol$, the molar mass of $S$ is $32.07\ g/mol$ and the molar mass of $O$ is $16.00\ g/mol$. $M_{Na_2SO_4}=2\times22.99+32.07 + 4\times16.00=45.98+32.07+64.00 = 142.05\ g/mol$ For $NH_4SO_4$ (should be $(NH_4)2SO_4$): The molar mass of $N$ is $14.01\ g/mol$, the molar mass of $H$ is $1.01\ g/mol$, the molar mass of $S$ is $32.07\ g/mol$ and the molar mass of $O$ is $16.00\ g/mol$. $M{(NH_4)2SO_4}=2\times(14.01 + 4\times1.01)+32.07+4\times16.00=2\times(14.01 + 4.04)+32.07 + 64.00=2\times18.05+32.07+64.00=36.10+32.07+64.00 = 132.17\ g/mol$ For $UF_4$: The molar mass of $U$ is $238.03\ g/mol$ and the molar mass of $F$ is $19.00\ g/mol$. $M{UF_4}=238.03+4\times19.00=238.03 + 76.00=314.03\ g/mol$

Step2: Convert grams to moles

Use the formula $n=\frac{m}{M}$, where $n$ is the number of moles, $m$ is the mass in grams, and $M$ is the molar mass in $g/mol$.

1. For $564\ g$ of $CCl_4$: $n_{CCl_4}=\frac{564\ g}{153.81\ g/mol}\approx3.67\ mol$ Then convert moles to molecules using Avogadro's number $N = n\times N_A$, where $N_A = 6.022\times10^{23}\ molecules/mol$ $N_{CCl_4}=3.67\ mol\times6.022\times10^{23}\ molecules/mol\approx2.21\times 10^{24}\ molecules$
2. For $5.65\times10^{26}$ molecules of $(NO_2)_2$: First, find the number of moles $n=\frac{N}{N_A}=\frac{5.65\times10^{26}\ molecules}{6.022\times10^{23}\ molecules/mol}\approx938\ mol$ Then find the mass $m = n\times M=938\ mol\times92.02\ g/mol\approx8.63\times10^{4}\ g$
3. For $7.53\ g$ of $CO_2$: $n_{CO_2}=\frac{7.53\ g}{44.01\ g/mol}\approx0.171\ mol$ $N_{CO_2}=0.171\ mol\times6.022\times10^{23}\ molecules/mol\approx1.03\times10^{23}\ molecules$
4. For $960\ g$ of $Na_2SO_4$: $n_{Na_2SO_4}=\frac{960\ g}{142.05\ g/mol}\approx6.76\ mol$ $N_{Na_2SO_4}=6.76\ mol\times6.022\times10^{23}\ molecules/mol\approx4.07\times10^{24}\ molecules$
5. For $8.6\ g$ of $(NH_4)2SO_4$: $n{(NH_4)2SO_4}=\frac{8.6\ g}{132.17\ g/mol}\approx0.0651\ mol$ $N{(NH_4)_2SO_4}=0.0651\ mol\times6.022\times10^{23}\ molecules/mol\approx3.92\times10^{22}\ molecules$
6. For $5.63\times10^{20}$ molecules of $UF_4$: First, find the number of moles $n=\frac{N}{N_A}=\frac{5.63\times10^{20}\ molecules}{6.022\times10^{23}\ molecules/mol}\approx9.35\times10^{-4}\ mol$ Then find the mass $m = n\times M=9.35\times10^{-4}\ mol\times314.03\ g/mol\approx0.294\ g$

Answer:

1. $2.21\times 10^{24}$ molecules
2. $8.63\times10^{4}\ g$
3. $1.03\times10^{23}$ molecules
4. $4.07\times10^{24}$ molecules
5. $3.92\times10^{22}$ molecules
6. $0.294\ g$