SOLUTION: Use dimensional analysis and the given densities to make the following conversations.
A. 14.8 g of boron to cm^3 of boron. The density of boron is 2.34g/cm^3
B. 4.62g of me
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Question 210587: Use dimensional analysis and the given densities to make the following conversations.
A. 14.8 g of boron to cm^3 of boron. The density of boron is 2.34g/cm^3
B. 4.62g of mercury to cm^3 of mercury. The density of mercury is 13.5 g/cm^3
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Use dimensional analysis and the given densities to make the following conversations.
A. 14.8 g of boron to cm^3 of boron. The density of boron is 2.34g/cm^3
14.8 gm/2.34 gm/cc = 6.3248 cc of B
Notice how the units are correct if you get it correct. If you multiply it, the units would be gm*(gm/cc). Use the units to make sure you have the correct operation. cc = cubic centimeter, or cm^3, same as ml = millileter
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B. 4.62g of mercury to cm^3 of mercury. The density of mercury is 13.5 g/cm^3
4.62 gm/13.5 gm/cc = 0.34222... cc of Hg
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