When dealing with decimal numbers, zeros only count as long as there is a non-zero number somewhere before them. The first 4 zeros have no non-zero number before them, so they don't count as significant figures. The last 2 zeros have a 5 in front of them, so those zeros count as significant figures, and so does that 5; which gives us a total of 3 significant figures.

For scientific notation, the exponent next to the ten has nothing to do with significant figures; it's the decimal number before it that determines the number of significant figures. Based on those decimal numbers, $\displaystyle 2\cdot 10^3$  and $\displaystyle 4\cdot 10^3$ each have one significant figure, and $\displaystyle 8.11\cdot10^4$ has three significant figures. When dealing with non-decimal numbers, zeros only count if they are in between two non-zero numbers. That means that 1,340 only has three significant figures, and 1,001 has 4 significant figures.

Whenever multiplying, the final answer will have the same number of significant figures as the number with the least least number of significant figures in the question. 130.5 has 4 significant figures and 12 has 2 significant figures, that means that our answer must have 2 significant figures. $\displaystyle 130.5 \cdot 12 =1566$, but that's 4 significant figures instead of 2. So we need to round this to 2 significant figures. Our answer to the correct number of significant figures is 1600.

When adding numbers, the answer should use the the least number of decimal places of those in the question. 14.1 has one decimal place (one number behind the decimal), 12 has no decimal places, and 5.634 has 3 decimal places; this means that the answer must have no decimal places. $\displaystyle 14.1 + 12 + 5.634 =31.734$, this must be rounded to zero decimal places. Thus we round up to 32.

Notice that the most precise you can get with your volume measurement is $\displaystyle \pm0.1\,mL$ and your recorded value has 3 significant figures. You must take no more than 3 significant figures for your pressure measurement even if the device is more precise than this (which it is). Your recorded value must be

$\displaystyle 102\,kPa$

The formula for percent error is: 

$\displaystyle \frac{\left | \text{Accepted Value - Measured Value}\right |}{\text{Accepted Value}} \times100\%$

In this case, the measured value is 12.8g and the accepted value is 32.9g.

$\displaystyle \frac{\left | 32.9g-12.8g\right |}{32.9g}\times 100\%=61\%$

To find percent error we need to use the following equation:

$\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$

Plug in 42 for the theoretical yield and 32.73 for the actual yield and solve accordingly.

$\displaystyle \frac{42g-32.73g}{42g}*100\%=22.07\%$

To find the percent error we need to use the following equation:

$\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$

But in order to do this, we first have to convert moles of sodium oxide into grams:

$\displaystyle 4 mol\ Na_{2}O * \frac{62g}{1 mol}=248g\ NaO_2$

This gives us a theoretical yield of 248g, which we plug in with our 195g actual yield.

$\displaystyle \frac{248g-195g}{248g}*100\%=21.37\%\ \text{error}$

Our first step to complete this problem is to convert moles AgF into grams:

$\displaystyle 0.6 mol\ AgF*\frac{127g}{1 mol}=76.2g\ AgF$

This gives us a theoretical yield of 76.2 grams. We can find the percent error by plugging in 76.2g for our theoretical yield and 43g for the actual yield in the following equation:

$\displaystyle \frac{\text{theoretical yield}-\text{actual yield}}{\text{theoretical yield}}*100\%$

$\displaystyle \frac{76.2g-43g}{76.2g}*100\%=43.57\%\ \text{error}$

Use the following formula to find the percent error:

$\displaystyle \frac{|\text{accepted value-experimental value}|}{\text{accepted value}}\cdot100\%$

For this experiment, our accepted value is the same as the theoretical value.

$\displaystyle \text{Percent error}=\frac{|7.8-6.8|}{7.8}\cdot100\%$

$\displaystyle \textup{Percent error}=0.128\cdot100\%=12.8\%$