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Agata [3.3K]
3 years ago
9

How many atoms are in 3 moles of carbon?

Chemistry
2 answers:
Finger [1]3 years ago
8 0
18........................
Leokris [45]3 years ago
7 0

Answer:

18.06 x 10^23

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Katena32 [7]

Answer:

Explanation:

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3 years ago
What specialized plant structures increase the probability of successful reproduction?
harkovskaia [24]

Answer:

The pistil and the stamen.

Explanation:

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Read 2 more answers
Methylation is a process that involves which of the following metals
Vesnalui [34]
Methylation is a process that involves Mercury metal.  Methylation is a process by which a methyl groups are added to the DNA molecule. when it occurs it can change the activity of a DNA segment without changing the sequence, additionally, when located in a gene promoter it may act to repress gene transcription. 
7 0
4 years ago
Balance the following equation on scrap paper:<br> AIF3+Li2O → Al2O3 + LiF
Oxana [17]

Answer:

2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF

Explanation:

AlF₃ + Li₂O —> Al₂O₃ + LiF

The above equation can be balanced as follow:

AlF₃ + Li₂O —> Al₂O₃ + LiF

There are 2 atoms of Al on the right side and 1 atom on the left side. It can be balance by writing 2 before AlF₃ as shown below:

2AlF₃ + Li₂O —> Al₂O₃ + LiF

There are 6 atoms of F on the left side and 1 atom on the right side. It can be balance by writing 6 before LiF as shown below:

2AlF₃ + Li₂O —> Al₂O₃ + 6LiF

There are 2 atoms of Li on the left side and 6 atoms on the right side. It can be balance by writing 3 before Li₂O as shown below:

2AlF₃ + 3Li₂O —> Al₂O₃ + 6LiF

Thus, the equation is balanced..!

8 0
3 years ago
A chemist makes 940. mL of sodium carbonate (Na_2CO_3) working solution by adding distilled water to 200. mL of a 0.171 m stock
Alexxandr [17]

Answer:

The answer is 0.0698 M

Explanation:

The concentration was prepared by a serial dilution method.

The formula for the preparation I M1V1 = M2V2

M1= the concentration of the stock solution = 0.171 M

V1= volume of the stock solution taken = 200 mL

M2 = the concentration produced

V2 = the volume of the solution produced = 940 mL

Substitute these values in the formula

0.171 × 200 = 490 × M2

34.2 = 490 × M2

Make M2 the subject of the formula

M2 = 34.2/490

M2 = 0.069795

M2 = 0.0698 M ( 3 s.f)

The concentration of the Chemist's working solution to 3 significant figures is 0.0698M

6 0
3 years ago
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