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12345 [234]
3 years ago
12

26.8 moles of argon contain how many atoms?

Chemistry
2 answers:
Maru [420]3 years ago
4 0
<span>1.61 × 1023 Multiply by 26.8 to get the answer.161.33 x 10 ^23   </span>
Alisiya [41]3 years ago
3 0

Answer : The number of atoms of argon are 1.61\times 10^{25}

Explanation : Given,

Moles of argon = 26.8 moles

As we know that,

1 mole of substance always contains 6.022\times 10^{23} number of atoms.

As, 1 mole of argon contains 6.022\times 10^{23} number of atoms

So, 26.8 mole of argon contains 26.8\times 6.022\times 10^{23}=1.61\times 10^{25} number of atoms.

Therefore, the number of atoms of argon are 1.61\times 10^{25}

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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!

5 0
3 years ago
A sample of potassium chlorate (15.0 g) is dissolved in 201 g of water at 70 °c with precautions taken to avoid evaporation of a
zlopas [31]
Answer is: unsaturated.
Solubility of potassium chlorate on 70°C is approximately 30 grams in 100 grams of water.
Solubility of potassium chlorate on 70°C is approximately 10 grams in 100 grams of water.
So if dissolve 15 g of potassium chlorate in 201 g of water, there is less salt than it solubility and solution is unsaturated.
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