First calculate the number of mols of each reactant:

$\displaystyle \frac{3.2 \:g \:Zn}{65.38\:\frac{g \:Zn}{mol \:Zn}} = 0.048945 \:mol \:Zn$

$\displaystyle \frac{8.1 \:g \:HCl}{36.45\:\frac{g \:HCl}{mol \:HCl}}=0.2222 \:mol \:HCl$ 

The ratio of $\displaystyle Zn$ to $\displaystyle HCl$ is 1:2. So, given that we have 0.048945 mol $\displaystyle Zn$, we only need 0.09789 mol $\displaystyle HCl$. Since we have more $\displaystyle HCl$ than is needed, $\displaystyle HCl$ is in excess. How much is in excess?

$\displaystyle 0.2222 \:mols \:HCl - 0.09789\:mols \:HCl \:needed = 0.12431 \:mols \:HCl \:excess$ 

Now convert mols of $\displaystyle HCl$ to grams:

$\displaystyle 0.12431\:mol \:HCl*36.45\:\frac{g \:HCl}{mol \:HCl} = 4.53 \:g \:HCl$ 

The system is open, so the pressure can be assumed atmospheric, e.g. 1 atm.

Using the ideal gas law,

$\displaystyle PV=nRT$

$\displaystyle V=\frac{nRT}{P}$

We need to convert $\displaystyle T$ to kelvin.

$\displaystyle T=515^{\circ}C+273.15 = 788.15 \:K$

The density of water at room temperature is approximately 1 g/mL, so we have 25 g of water. Convert to mols:

$\displaystyle n=\frac{25\:g}{18\:\frac{g}{mol}}=1.389 \:mol$

Plugging everything in:

$\displaystyle V=\frac{1.389\:mol*788.15\:K*0.08206\:\frac{L\cdot atm}{mol\cdot K}}{1\:atm}=89.83\:L \rightarrow90\:L$

First we need to calculate the mass of the wire. We will do this by calculating its volume then multiplying by the density.

$\displaystyle m=\rho V=\rho \pi r^{2}l =2.5\:\frac{g}{mL}(0.05\:cm)^{2}135\:cm =2.651\:g$

Next we need to find out how much energy the water gained from the metal wire.

$\displaystyle Q=mC_{p}\Delta T=(100\:g)(4.18\:\frac{J}{g^{\circ}C})(25^{\circ}C-35^{\circ}C)=-4180\:J$

The energy the water gained is equal to the energy the wire lost. This allows us to calculate the heat capacity for the metal: 

$\displaystyle 4180\:J = 2.6507\:g*C_{p,metal}(100^{\circ}C-35^{\circ}C)$

$\displaystyle C_{p,metal}=24.26\:\frac{J}{g^{\circ}C}\rightarrow 24\:\frac{J}{g^{\circ}C}$

We are given the pressure, temperature, and molar mass of the gas in the question stem and are asked to determine the density. To do this, we will need to make use of a manipulated form of the ideal gas equation.

$\displaystyle PV = nRT$

To find density, we will need a way to relate the variables given in the question stem to other parameters, namely mass and volume. Since we already have the volume, $\displaystyle V$, contained in the ideal gas equation, we just need a way to find the mass, $\displaystyle m$. We can do this by noting that the number of moles, $\displaystyle n$, can also be written as the mass divided by the molar mass, $\displaystyle \frac{m}{MM}$. Substituting this in gives us:

$\displaystyle PV = \left ( \frac{m}{MM} \right )RT$

And rearranging, we obtain the density:

$\displaystyle \frac{m}{V}=\frac{P\left ( MM \right )}{RT}=\rho$

But we're not done yet! Remember, we also have to convert the units given for pressure and temperature in the question stem into their proper form. Since we need pressure in terms of atmospheres, and there are $\displaystyle 760\: mmHg$ in $\displaystyle 1\: atm$ of gas, we can find that:

$\displaystyle 1140mmHg\left ( \frac{1atm}{760mmHg} \right )=1.5atm$

Furthermore, we need to convert degrees Celsius into Kelvins.

$\displaystyle 50^{o}C+273=323\: K$

Now that we have all the units in their correct form, we're finally ready to plug our values into the equation for density:

$\displaystyle \rho=\frac{P\left ( MM \right )}{RT}=\frac{\left ( 1.5atm \right )\left ( 120\frac{g}{mol} \right )}{\left ( 0.08206 \right \frac{L\cdot atm}{mol\cdot K})\left ( 323K \right )}=6.79\frac{g}{L}$ 

This is a typical dilution problem. The first step is to recognize that this is a dilution question. Whenever figuring out how to go from a higher concentrated solution to a lower concentrated solution, it is a dilution problem. The main equation to use here is:

$\displaystyle \small (Concentration_{1})$$\displaystyle \small (Volume_{1})$$\displaystyle \small = (Concentration_{2})(Volume_{2})$

or

$\displaystyle C_{1}V_{1} = C_{2}V_{2}$

or

$\displaystyle \small M_{1}V_{1}=M_{2}V_{2}$

When plugging in the numbers, make sure to assign the correct volumes and concentrations. Think of one side of the equation describing the 0.9% $\displaystyle \left(\frac{w}{v}\right)$ solution of sodium chloride and the other side describing the 0.45% solution of sodium chloride.

$\displaystyle 0.45\cdot500=0.9\cdot V_2$

$\displaystyle V_2= 250mL$

$\displaystyle \small V_{2}$ is the amount of the original solution needed to make the new 0.45% solution. However, we need to add water to the solution to dilute it. To find how much water is needed, subtract:

$\displaystyle \small 500mL - 250mL=250mL$ of water.  

42,000,000 has two significant figures because of the rule that states that trailing zeroes are never significant unless there is a decimal point. For example, if the number given was "42,000,000." it would be eight significant figures.

The correct answer is $\displaystyle 4.2*10^7$. This is found by moving the decimal place to the left 7 times in order to get one digit to the left of the decimal, and the rest of the digits to the right. The answer just has two significant figures because that is the number of significant figures in the original number.

To convert into standard notation we must move the decimal two places to the right because of the exponent of "2" above the 10. Then we re-write the number with the decimal in the new space. All of the digits that were in the original scientific notation must be accounted for. Thus, our answer must have six significant figures.

The answer is 500 because our rules of significant figures state that when two numbers are multiplied our answer can only have as many significant figures as our number with the fewest significant figures. In this case 5.02 has three significant figures, but 100 only has 1 significant figure. Therefore our answer can only have 1 significant figure.

The answer is 13.18 because the rules for adding and subtracting significant figures state that the answer can only be precise up to the least precise digit from the numbers that were added/subtracted. In this case 2.12 is only precise to the hundredth place, and therefore our answer can only go to the hundredth place.