8. how many c atoms of are in 1.65 liters of butanoic acid (c4h8o2)?

Explanation:

Step1: Find the density of butanoic acid

The density of butanoic acid ($C_4H_8O_2$) is approximately $0.959\ g/mL$. First, convert the volume from liters to milliliters. Since $1\ L = 1000\ mL$, for a volume $V = 1.65\ L$, then $V=1.65\times1000 = 1650\ mL$.

Step2: Calculate the mass of butanoic acid

Use the density - mass - volume relationship $\rho=\frac{m}{V}$, so $m=\rho V$. Substituting $\rho = 0.959\ g/mL$ and $V = 1650\ mL$, we get $m=0.959\times1650=1582.35\ g$.

Step3: Calculate the molar mass of butanoic acid

The molar mass of $C_4H_8O_2$: $M=(4\times12.01)+(8\times1.01)+(2\times16.00)=48.04 + 8.08+32.00 = 88.12\ g/mol$.

Step4: Calculate the number of moles of butanoic acid

Use the formula $n=\frac{m}{M}$. Substituting $m = 1582.35\ g$ and $M = 88.12\ g/mol$, we have $n=\frac{1582.35}{88.12}\approx17.96\ mol$.

Step5: Calculate the number of moles of carbon atoms

In one molecule of $C_4H_8O_2$, there are 4 carbon atoms. So the number of moles of carbon atoms $n_{C}=4\times n$. Substituting $n = 17.96\ mol$, we get $n_{C}=4\times17.96 = 71.84\ mol$.

Step6: Calculate the number of carbon atoms

Use Avogadro's number $N_A = 6.022\times10^{23}\ atoms/mol$. The number of carbon atoms $N=n_{C}\times N_A$. Substituting $n_{C}=71.84\ mol$ and $N_A = 6.022\times10^{23}\ atoms/mol$, we have $N = 71.84\times6.022\times10^{23}\approx4.33\times10^{25}$ atoms.

Answer:

$4.33\times 10^{25}$ atoms