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A 1.0mL vol.of seawater contains about 4.0x10-12g of gold.  The total vol. of ocean water is 1.5x1021L.  Calculate the total amount of gold (in pounds) that is present in seawater.  There are 454g in 1 lb.

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Explanation & Answer

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 1.5x1021L.

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

2. Next, I multiplied 4.0 x 10 -12g by 1.5x1021L 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 -12g x 1.5 x 1021L x 1000mL x 1000mg = 6.0 x 1015mg of gold        

1.0 ml                                   1.0 L           1.0 g

xzyqk4 (159)
Carnegie Mellon University

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