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shutvik [7]
3 years ago
9

How many C atoms are there in 5.25 grams of a compound containing 92.26% C and 7.74% H?

Chemistry
1 answer:
Damm [24]3 years ago
7 0
Take 5.25 grams and give it proportionally to the carbon
0.9226 * 5.25 grams = 4.84 grams Carbon
Find the number of moles
(4.84 grams carbon)/(12.01 grams/mol) = 0.403 moles Carbon
The number of atoms in a mole of a substance is called Avogadro's number.
There are 6.022 * 10^23 atoms in one mole of anything
(0.403 moles Carbon) * (6.022 * 10^23 atoms/mole) = 2.43 * 10^23 atoms of Carbon
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Explanation:

The only flaw I can find is you squared 3 instead of cubing it and it will be 27X^4 instead of 9x^4.

This reduces the amount slightly, but the number is still incredibly high (about 10 ^ 5 L is what I've calculated). Your professor might want to point out that this will not be a effective experiment due to the large volume of saturated

The Ksp value of Ca(OH)2 on the site (I used 5.5E-6 [a far more soluble compound than Al(OH)3]) and estimated how much of it will be needed. My calculation was approximately 30 ml. If you were using that much in the experiment, it implies so our estimates for Al(OH)3 are right, that the high amount is unreasonably big and that Al(OH)3 will not be a suitable replacement unless the procedure was modified slightly.

6 0
3 years ago
What is infrared energy? simplify it
alex41 [277]
Infrared energy is the energy of light between microwave radiation and Ultraviolet radiation
7 0
2 years ago
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Equal masses of substance A and substance B are at 25°C. Substance A has a lower specific heat capacity than substance B. If 50
liraira [26]
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6 0
3 years ago
You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted
klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water =  600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = \frac{C_1}{C_2}

where; C_1= solubility of compound in the organic solvent and C_2 = solubility in aqueous water.

So; we can represent our data as:

K=(\frac{A_{(g)}}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL} )

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6=(\frac{0.6-B(mg)}{60mL} ) ÷ (\frac{B_{(mg)}}{100mL})

4.6 = \frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}

4.6 × (\frac{B_{(mg)}}{100mL}) = (\frac{0.6-B(mg)}{60mL} )

4.6 B *\frac{60}{100} = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = \frac{0.6}{3.76}

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0
3 years ago
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