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Natasha_Volkova [10]
3 years ago
8

99% 1H 0.8% 2H 0.2% 3H Average Atomic Mass = [answer] amu PLEASE HELP ME

Chemistry
1 answer:
alukav5142 [94]3 years ago
5 0

Answer: 1.01 atomic mass units or amu

Explanation:

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How many moles of carbon, hydrogen, and oxygen are present in a 100-g sample of ascorbic acid?
Y_Kistochka [10]

There are:

3.41 moles of C

4.54 moles of H

3.40 moles of O.

Why?

To solve the problem, the first thing that we need to do is to write the chemical formula of the ascorbic acid.

C_{6}H_{8}O_{6}

Now, we know that there are 100 grams of the compound, so, the masses of each element will represent the percent in the compound.

We have that:

C_{6}=12.0107g*6=72.08g\\\\H_{8}=1.008g*8=8.064g\\\\O_{6}=15.999g*6=95.994g\\\\C_{6}H_{8}O_{6}=72.08g+8.064g+95.994g=176.138g

To know the percent of each element, we need to to the following:

C=\frac{72.08g}{176.138g}*100=0.409*100=40.92(percent)\\\\H=\frac{8.064g}{176.138g}*100=4.58(percent)\\\\O=\frac{95.994}{176.138g}*100=54.49(percent)

So, we know that for the 100 grams of the compound, there are:

40.92 grams of C

4.58 grams of H

54.49 grams of O

We know the molecular masses of each element:

C=12.0107\frac{g}{mol}\\\\H=1.008\frac{g}{mol}\\\\O=15.999\frac{g}{mol}{mol}

Now, to calculate the number of moles of each element, we need to divide the mass of each element by the molecular mass of each element:

C=\frac{40.92g}{12.010\frac{g}{mol}}=3.41mol\\\\H=\frac{4.58g}{1.008\frac{g}{mol}}=4.54mol\\\\O=\frac{54.49g}{15.999\frac{g}{mol}}=3.40mol

Hence, we have that there are 3.41 moles of C, 4.54 moles of H, and 3.40 moles of O.

Have a nice day!

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Why is governmental funding of research important?
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D. The government determines what is right & wrong for everyone

Explanation:

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If 7.0 mol sample of a gas has a volume of 12.2 L, what would the volume be if the amount of gas was increased to 16.8 mol
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Answer:

V_{2} = 29.28\,L

Explanation:

Let assume that gas behaves ideally and experiments an isobaric and isothermal processes. The following relationship is applied to determined the final volume:

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}

V_{2} = V_{1} \cdot \left(\frac{n_{2}}{n_{1}} \right)

V_{2} = (12.2\,L)\cdot \left(\frac{16.8\,moles}{7\,moles} \right)

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