Chemistry Math (Please help!)

  • Thread starter mahachit
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  • #1
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Homework Statement

1. Vanillin(used to flavour vanilla ice cream and other foods) is the substance whose aromathe human nose detects in the smallest amount. The threshold limit is 2.0 X 10^-11 g per liter of air. If the current price of 50g of vanilin is $112, determine the cost to suppply enough vanillin so that the aroma could be detected in a large aircraft hangar with a volume of 5.0 X 10^7 ft^3. (Leave our answer in dollars and cents)


The attempt at a solution

This is what I've done so far for #1 :

1ft^3 = 28.32 L
(5.0 X 10^7) X (2.832 X 10) = 14.16 X 10^8 L (volume of aircraft in L)
(2.0 X 10^-11) X (14.16 X 10^8) = 28.32 X 10^-3 (minimum amoun of vanillin needed)

This is the part where I got stuck; what do I have to do next?

Homework Statement

2. Magnesium (Mg) is a valuable metal use in alloys, in batteries, and in the manufacture of chemicals. It is obtained mostly from seawater, which contains about 1.3g of Mg for every kilogram of seawater. The total volume of seawater is 1.5 X 10^21 L. Its density is 1.03g/mL. Calculate the amount of seawater (in liters) needed to extract 8.0 X 10^4 tons of Mg, which is roughly the annual production in the United States. (1 ton= 2000 lb; 1lb = 453.6 g.)

The attempt at a solution

so...to find the mass i have to multiply the density and volume, right?

W=mg
1.5 X 10^21 L = 1500 X 10^21 mL
(1.5 X 10^24) mL X 1.03 g/mL = 1.545 X 10^24 g/mL^2 (mass of seawater)

So now I have to multiply the mass by gravity...but what is the formula gravity? And what do I have to do after that?
 
Last edited:

Answers and Replies

  • #2
Borek
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It is all about units - for example, you have calculated amount of vanilin to be 28.32 X 10^-3 of what? Pounds? Stones? Moles? Atoms? Boxes?
 
  • #3
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ok so the amount of vanillin I have calculated is in Litres....but what step should I do next?
 
  • #4
Borek
Mentor
28,702
3,190
No, it is not in litres. Pay attention to your units. Write them explicitely.
 

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