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notsponge [240]
3 years ago
7

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

Chemistry
1 answer:
Y_Kistochka [10]3 years ago
5 0

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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Consider the titration of 82.0 mL of 0.140 M Ba(OH)2 by 0.560 M HCl. Calculate the pH of the resulting solution after the follow
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Answer:

a) pH = 13.447

b) pH = 13.176

Explanation:

Ba (OH)₂ is a strong base and ionizes completely in solution to give barium amend hydroxide ions.

The equation of the dissociation of Ba(OH)₂ is given below:

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a) Before the addition of HCl, i.e.,when 0.00 mL of HCl has been added;

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b) pH when 15.0 mL HCl has been added

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Volume of Ba(OH)₂ = 82.0 mL = 0.082 L, concentration = 0.140 M

moles of Ba(OH)₂ = 0.082 x 0.140 = 0.01148 moles

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0.0084 H+ willnreact with 0.0084 moles of OH-

moles OH- left after the reaction = 0.02296 - 0.0084 = 0.01456 moles

total volume of new solution = (82 +15) mL = 97 mL => 0.097 L

Concentration of OH- ions = moles / volume

[OH-] = 0.01456 / 0.097 = 0.1501

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