Answer:
so you know the units in the problem and answer are the same, if they aren't you have to convert them to the same units.
Answer:
m(P4) = 46.175 (grams)
m (KClO3) = 149 (grams)
Explanation:
1) n(P4) = n(P4O10);
m(P4)/M(P4) = m(P4O10)/M(P4O10);
m(P4) = M(P4)*m(P4O10)/M(P4O10)
= 123.90*105.8/283.89
= 46.175 (grams)
2) Analogously, 10n(P4O10) = 3n(KClO3)
m (KClO3) = 10M(KClO3)*m(P4O10)/3M(P4O10)
= 10*122.55*105.8/283.89/3
= 149 (grams).
The answer is -17x-1. You have to combine like terms.
The answer is a bit confusing, so I did some research and fount the original question. This is the complete question:
"<span>If
measurements of a gas are 75L and 300 kilopascals and then the gas is
measured a second time and found to be 50L, describe what had to happen
to the pressure (if temperature remained constant). Include which law
supports this observation".
From that:
1) Data:
V1 = 75 liter
P1 = 300 kPa
V2 = 50 liter
P2 = x
T = constant.
2) Analysis (physical law)
Boyle's law states that the volume of a fixed amount of gas, at constant temperarature, is inversely related with the pressure.
In mathematical form that statement becomes into this equation:
PV = constant (at constant T)
=> P1V1 = P2V2.
Then, the decreasing of the volumen (compression) is accompanied by an increase of the pressure.
3) So, using Boyle's law with the data given:
300 kPa * 75 liter = x * 50 liter
=> x = 300 kPa * 75 liter / 50 liter = 450 kPa.
Answer: following Boyle's law, when the gas passed from 75 liter to 50 liter the pressure had to increase from 300 kPa to 450 kPa.
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