A 1.0mL vol.of seawater contains
about 4.0x10^{-12}g of gold. The total vol. of ocean water is
1.5x10^{21}L. Calculate the total amount of gold (in pounds) that
is present in seawater. There are 454g in 1 lb.

a.In order to solve this problem the first step that must be taken is to use dimensional analysis to change grams of gold to milligrams of gold. Also, in this process the amount of mg of gold per in seawater will be determined by using the given information that the total volume of ocean water is 1.5x10^{21}L.

1. First, I put 4.0 x 10 ^{-12}g over 1.0 mL because there are 4.0 x 10 ^{-12}g of gold in one mL of seawater.

2. Next, I multiplied 4.0 x 10 ^{-12}g by 1.5x10^{21}L because that is the total volume of ocean water, and it must be used in order to calculate the amount of mg of gold that is present in sea water.

3. In order to cancel out the units of liters and milliliters I used the following conversion factor.

1000mL = 1L

4. Finally, to change grams of gold to milligrams of gold I used the following conversion factor. The units of grams will then cancel out leaving the answer in mg.

1000mg = 1g

5. The answer is in two significant figures because in the problem the least amount of significant figures is two.

4.0 x 10 ^{-12}g x 1.5 x 10^{21}L x 1000mL x 1000mg = 6.0 x 10^{15}mg of gold

Thank you so much for explaining. I do not have that answer as an option. I have 6.0x10^{12}
lbs

Can you recheck and make sure.

Also are you overly good in Chemistry? I am looking to work with someone over the next 8 weeks if you are available. Some will be free questions and some I will pay for.

Are you still there? Sorry! So, the answer I gave was in milligrams. What are your options for units? And yes, I've tutored general chemistry and organic chemistry and am pursuing a doctorate right now. Let me know and sorry for the late reply!

Jun 10th, 2015

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