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Brums [2.3K]
3 years ago
8

Which sample of ethanol has particles with the highest average kinetic energy?

Chemistry
2 answers:
ale4655 [162]3 years ago
6 0
The second one.
Remember, temperature is the AVERAGE kinetic energy. So, looking for highest average kinetic energy means looking for highest temperature, which is 55.
Note: the amount of the substance does not determine the average kinetic energy. 
pav-90 [236]3 years ago
5 0
The answer is two. All you are looking for with highest average kinetic energy is the one with the highest temperature. In this case it is 10mL ethanol at 55 degrees Celsius.
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Answer:

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70.44~g~H_2O\frac{1~mol~H_2O}{18~g~H_2O}=~3.9~mol~H_2O

Now, in 1 mol of CO2 we have 1  mol of C and in 1 mol of H_2O we have 1 mol of H. Additionally, if we want to calculate the moles of oxygen we need to <u>calculate the grams of C and O</u> and then do the <u>substraction</u> form the initial amount, so:

2.6~mol~CO_2\frac{1~mol~C}{1~mol~CO_2}\frac{12~g~C}{1~mol~C}=31.25~g~of~C

3.9~mol~H_2O\frac{2~mol~H}{1~mol~H_2O}\frac{1~g~H}{1~mol~H}=7.82~g~of~H

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grams~of~O=60.00~g-~39.08~g=20.92~g~of~O

Now we can <u>convert the grams</u> of O to moles, so:

20.92~g~of~O\frac{1~mol~O}{16~g~O}=1.30~mol~O

The next step is to divide all the mol values by the <u>smallest one</u>:

O=\frac{1.30~mol~O}{1.30~mol~O}=~1

C=\frac{2.6~mol~C}{1.30~mol~O}=~2

H=\frac{7.82~mol~H}{1.30~mol~O}=6

Therefore the formula is C_2H_6O

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