First we must determine the volume of the container:

$\displaystyle 3.2\hspace{1 mm}in\times 3.4\hspace{1 mm} in\times 3.3\hspace{1 mm} in= 35.9\hspace{1 mm}in^3$

The density is $\displaystyle \frac{0.370\hspace{1 mm}lbs}{35.9\hspace{1 mm}in^3}$.

Now we will convert this into grams per milliliter.

$\displaystyle \frac{0.370lbs}{35.9in^3}*(\frac{1in}{2.54cm})^3*\frac{1cm^3}{1mL}=6.29*10^{-4}\frac{lb}{mL}$

$\displaystyle 6.29*10^{-4}\frac{lb}{mL}*\frac{1kg}{2.2046lb}*\frac{1000g}{1kg}=0.285\frac{g}{mL}$

Each mole of methane to react releases 5kJ of heat. Since we have 18 moles of methane, we expect a potential of 90kJ of heat to be released.

$\displaystyle 18mol * \frac{5kJ}{mol} = 90kJ$

We also have to take into account that the process is only 60% efficient; therefore, we effectively only liberate 60% of the total expected amount.

$\displaystyle 18mol * 0.60 * \frac{5kJ}{mol} = 54kJ$

The net ionic equation that occurs is:

$\displaystyle Ag_{(aq)}^{+}\; +\;Cl _{(aq)}^{-}\; \rightarrow \; AgCl_{(s)}$

$\displaystyle NO_{3}^{-}$ and $\displaystyle Na^{+}$ are spectator ions. Hence, $\displaystyle NO_{3}^{-}$  concentration is not affected. The molecular mass of $\displaystyle NaCl$ is $\displaystyle 58.44 g$, then $\displaystyle 5.844 g$ of NaCl yields $\displaystyle 0.1000 \: mol$ of $\displaystyle Cl^{-}$. Since we have $\displaystyle 0.100\: L$ of a $\displaystyle 1.00 \: M$ solution of $\displaystyle AgNO_{3}$, we have $\displaystyle 0.100\: mol$ of $\displaystyle Ag^{+}$. There is no limiting reactant. Hence $\displaystyle 0.100\: mol$ of $\displaystyle AgCl$ are formed. Being the molecular mass of $\displaystyle AgCl \: 169.9 \: g/mol$, $\displaystyle 17.0 g \: of \: AgCl$ are formed. Result should be given with tree significant figures.

$\displaystyle 8.888 * 10^{9} atoms \ Fe * \frac{1 mol Fe}{6.022 * 10^{23}atoms\ Fe} * \frac{55.85 g Fe}{1 mol Fe} = 8.243 * 10^{-13} g Fe$

Remember: the given measurement has 4 significant figures, so the answer must also have 4 significant figures.

$\displaystyle 17.0 g\ S * \frac{1 mol S}{32.06 g\ S} * \frac{6.022^{23} atoms \ S}{1 mol\ S} = 3.19 * 10^{23} atoms \ S$

Remember, there really isn't a way to go straight from grams to atoms, but, it's possible to change grams into moles (using the molar mass), and it's possible to change moles into atoms (using Avogadro's number).

The first step is to determine the molecular weight of the molecule:

$\displaystyle MW=(atomic\ weight\ of\ C)+(atomic\ weight\ of\ H)+(atomic\ weight\ of\ Br)$

$\displaystyle MW_{CH_{3}CH_{2}Br}=2(12\frac{g}{mole})+5(1\frac{g}{mole})+(80\frac{g}{mole})=109 \frac{g}{mole}$

The molecular weight serves as a conversion factor to convert grams to moles as implied its units $\displaystyle \frac{g}{mole}$. So, given that we are dealing with 20.0 grams, the moles of 20.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$\displaystyle 20.0 g * \frac{mole}{109\ grams} =0.1835\ moles\ CH_{3}CH_{2}Br$

Based on the molecular formula, for every mole of the compound $\displaystyle CH_{3}CH_{2}Br$, there is 1 mole of bromine. Based on our calculations 20.0 grams of $\displaystyle CH_{3}CH_{2}Br$ is 0.1835 moles of the compound. Because there is a 1:1 molar ratio of bromine to every molecule, we have 0.1835 moles of bromine in 20 grams of our compound, $\displaystyle CH_{3}CH_{2}Br$.

The first step is to determine the molecular weight of the molecule ethanol, $\displaystyle CH_{3}CH_{2}OH$:

$\displaystyle MW=\text{respective equivalents *[(atomic weight of C)+(atomic weight of H)+(atomic weight of O)]}$

$\displaystyle MW_{CH_{3}CH_{2}OH}=2\left(\frac{12g}{mol}\right)+6\left(\frac{1g}{mol}\right)+\left(\frac{16g}{mol}\right)=46 \frac{g}{mol}$

The molecular weight serves as a conversion factor to convert grams to moles as implied by its units $\displaystyle \frac{grams}{mole}$. So, given that we are dealing with 50 grams of ethanol, the moles of 50 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$\displaystyle 50g * \frac{1mole}{46 g} =1.09 mol\ CH_{3}CH_{2}OH$

Based on the molecular formula, we can see that for every mole of the compound $\displaystyle CH_{3}CH_{2}OH$, there is 1 mole of oxygen. Based on our calculations, 50 grams of $\displaystyle CH_{3}CH_{2}OH$ is 1.09 moles of the compound. Because there is a 1:1 molar ratio of oxygen to every ethanol molecule, that is, 1 mole of oxygen in every mole of our molecule, we have 1.09 moles of oxygen in 50 grams of our compound $\displaystyle CH_{3}CH_{2}OH$. To convert 1.09 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which from the periodic table we know is $\displaystyle 16\frac{g}{mol}$:

$\displaystyle 1.09 mol * \frac{16g}{mol}= 17.4 g\ oxygen$

The first step is to determine the molecular weight of the molecule:

$\displaystyle MW=\text{respective equivalents *[(atomic weight of C)+(atomic weight of H)+(atomic weight of O)]}$

$\displaystyle MW_{C_{4}H_{9}O_{2}}=4(12\frac{g}{mole})+9(1\frac{g}{mole})+2(16\frac{g}{mole})= 89 \frac{g}{mole}$

The molecular weight serves as a conversion factor to convert grams to moles as implied its units $\displaystyle \frac{g}{mole}$. So, given that we are dealing with 45.0 grams, the moles of 45.0 grams can be calculated using the calculated molecular weight as seen below in which the grams cancel out leaving the units in moles:

$\displaystyle 45.0g * \frac{mole}{89.0\ grams} =0.506\ moles C_{4}H_{9}O_{2}$

Based on the molecular formula, we can see that for every mole of the compound $\displaystyle C_{4}H_{9}O_{2}$, there are 2 moles of oxygen. Based on our calculations 45.0 grams of $\displaystyle C_{4}H_{9}O_{2}$ is 0.506 moles of the compound. Because there is a 2:1 mole ratio of oxygen to every molecule, that is, 2 mole of oxygen in every mole of our molecule, we have double the moles of oxygen in our compound $\displaystyle C_{4}H_{9}O_{2}$ as calculated below:

$\displaystyle 0.506\ moles_{C_{4}H_{9}O_{2}} * \frac{2\ moles\ of\ oxygen}{1\ mole_{C_{4}H_{9}O_{2}}}=1.011\ moles\ of\ oxygen$

To convert 1.011 moles of oxygen to grams of oxygen, we perform the following equation based on the atomic weight of oxygen, which is $\displaystyle 16\frac{g}{mole}$:

$\displaystyle 1.012\ moles * \frac{16g}{mole}= 16.2 g\ oxygen$