Doubling the mass doubles the energy, while doubling the velocity quadruples it. Your question is basically about order of operations; the exponent only applies to the variable it's immediately on. As you note, you'd have to put parentheses to make it cover the
m as well. We say that energy is linear in mass, but quadratic in velocity.
Answer:
given an element U of relative atomic mass 10.8 for example with isotopes U-10 and U-11 we can get their percentages by the following steps.
Explanation:
1:let the Percentage of U-10 equal to X% and percentage of U-11equal to Y%.
2:The addition of X% and Y% equal to 100%.
3:X%=100%-Y%
4:X% of U-10 +Y% of U-11=10.8
5: substitute 3 in 4.
6: i.e (100-Y)% of U-10+Y% of U-11=10.8.
7:(100-Y)/100 of 10 +Y/100 of 11=10.8.
8:100-Y/10 +11Y=10.8.
9:(1000-10Y+11Y)/100=10.8.
10:1000+Y=10.8*100
11:1000+Y=1080.
12:Y=1080-1000.
13:Y=80.
14: from 3,X=100%-Y%.
X=100%-80%=20%.
15: Therefore, the percentage of isotopes in the isotopic mixture are 20% U-10 and 80% U-11.
Answer:
a.)
To warm the liquid from 35°C to 78°C:
(2.3 J/g-K) x (42.0 g) x (78 - 35) = 4154 J
To vaporize the liquid at 78°C:
(38.56 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 35.154 kJ
Total:
4.154 kJ + 35.154 kJ = 39.3 kJ
b.)
To warm the solid from -155°C to -114°C:
(0.97 J/g-K) x (42.0 g) x (-114°C - (-155°C)) = 1670 J
To melt the solid at -114°C:
(5.02 kJ/mol) x (42.0 g C2H5OH / 46.06867 g C2H5OH/mol) = 4.5766 kJ
To warm the liquid from -114°C to 78°C:
(2.3 J/g-K) x (42.0 g) x (78 - (-114)) = 18547 J
To vaporize the liquid at 78°C:
35.154 kJ (as in part a.)
Total:
1.670 kJ + 4.5766 kJ + 18.547 kJ + 35.154 kJ = 59.9 kJ
Explanation:
Lin never forgot how it felt to be left out