A gas has a solubility of .66g/L at 10.0 atm of pressure. What is the pressure on a 1.0 L sample that contains 1.5 grams of gas? Showing the work would be very helpful.
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There is an old math equation I remember from my school days called the product of the means is equal to the product of the extremes. In formula form it would look like this:
A/B = C/D
A and D would represent the extremes, while B and C would represent the means. One would simply set the values of your question up as an equality equation, like this:
10 atm/.66 g per L = x(unknown atm)/ 1.5 g per L
You would then cross multiply the extremes, which would be 10 times 1.5, which would give you 15. Then, cross multiply the means, which would be the x(unknown atm) times .66, which would give you .66x. This is what your results should look like:
15 = .66x
All we have to do now is divide both sides by .66. That will eleiminate the .66 on the right side, which will leave "x" equals the answer, the unknown pressure. That equation would look like this:
15/.66 = .66x/.66
22.73 = x
Rounding up, the correct answer for the new pressure would be 22.73 atm. You should remember this equation of the "product of the means is equal to the product of the extremes", I have used it numerous times since being out of school. It comes in quite handy!
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