Question

10.9 kg k₂co₃
express your answer in formula units to three significant figures.

k₂co₃ formula units

part c

71.5 mg c₄h₁₀
express your answer in molecules to three significant figures.

c₄h₁₀ molecules

Explanation:

Part A:

Step1: Calculate molar mass of $K_2CO_3$

The molar mass of $K$ is approximately $39.10\ g/mol$, $C$ is $12.01\ g/mol$ and $O$ is $16.00\ g/mol$. So, $M_{K_2CO_3}=2\times39.10 + 12.01+3\times16.00=138.21\ g/mol$.

Step2: Convert mass to moles

The given mass $m = 10.9\ kg=10900\ g$. The number of moles $n=\frac{m}{M}=\frac{10900\ g}{138.21\ g/mol}\approx78.8655\ mol$.

Step3: Calculate formula - units

Using Avogadro's number $N_A = 6.022\times10^{23}\ formula - units/mol$. The number of formula - units $N=n\times N_A=78.8655\ mol\times6.022\times10^{23}\ formula - units/mol\approx4.75\times10^{25}\ formula - units$.

Step1: Calculate molar mass of $C_4H_{10}$

The molar mass of $C$ is $12.01\ g/mol$ and $H$ is $1.01\ g/mol$. So, $M_{C_4H_{10}}=4\times12.01 + 10\times1.01=58.14\ g/mol$.

Step2: Convert mass to moles

The given mass $m = 71.5\ mg = 0.0715\ g$. The number of moles $n=\frac{m}{M}=\frac{0.0715\ g}{58.14\ g/mol}\approx1.23\times10^{-3}\ mol$.

Step3: Calculate number of molecules

Using Avogadro's number $N_A = 6.022\times10^{23}\ molecules/mol$. The number of molecules $N=n\times N_A=1.23\times10^{-3}\ mol\times6.022\times10^{23}\ molecules/mol\approx7.40\times10^{20}\ molecules$.

Answer:

$4.75\times10^{25}\ K_2CO_3\ formula\ units$

Part B: