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erastova [34]
3 years ago
6

If 3.3167 moles of Al are produced, how many moles of AlCl3 were reacted? I

Chemistry
1 answer:
tresset_1 [31]3 years ago
8 0

Answer:

3.3167 moles Of AlCl3

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

3Ca + 2AlCl3 —> 3CaCl2 + 2Al

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Finally, we shall obtained the number of moles of AlCl3 that reacted to produce 3.3167 moles of Al as follow:

From the balanced equation above,

2 moles of AlCl3 reacted to produce 2 moles of Al.

Therefore, 3.3167 moles Of AlCl3 will also react to produce 3.3167 moles of Al.

Thus, 3.3167 moles Of AlCl3 is needed for the reaction.

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Answer:

A. The color change from blue to green happens in nature.

Explanation:

The color change is an indicator of a chemical reaction that has occurred. Although color changes are not necessarily good diagnostic tool to measure if a chemical reaction has occurred or not.

  • In this reaction, color changes is quite an effective tool
  • Chemical changes usually involves the formation of a new product from the chemical reaction.
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  • For this reaction, color change is the most appropriate signal for a chemical change.
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Which class of organic compound is least likely to be used as an organic solvent? aldehyde ketone alkyl halide ether
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Aldehyde

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3 years ago
If each NADHNADH generates 3 ATPATP molecules and each FADH2FADH2 generates 2 ATPATP molecules, calculate the number of ATPATP m
Nadusha1986 [10]

Answer:

128~ATP

Explanation:

The metabolic pathway by which energy can be obtained from a fatty acid is called <u>"beta-oxidation"</u>. In this route, acetyl-Coa is produced by removing <u>2 carbons</u> from the fatty acid for each acetyl-Coa produced. In other words, for each round, 1 acetyl Coa is produced and for each round 2 carbons are removed from the initial fatty acid. Therefore, the first step is to calculate the <u>number of rounds</u> that will take place for an <u>18-carbon fatty</u> acid using the following equation:

Number~of~Rounds=\frac{n}{2}-1

Where "n" is the <u>number of carbons</u>, in this case "18", so:

Number~of~Rounds=\frac{18}{2}-1~=~8

We also have to calculate the amount of Acetyl-Coa produced:

Number~of~Acetyl-Coa=\frac{18}{2}~=~9

Now, we have to keep in mind that in each round in the beta-oxidation we will have the <u>production of 1 FADH_2 and 1 NADH</u>. So, if we have 8 rounds we will have 8 FADH_2 and 8 NADH.

Finally, for the total calculation of ATP. We have to remember the <u>yield for each compound</u>:

-) 1~FADH_2~=~2~ATP

-) 1~NADH~=~3~ATP

-) Acetyl~CoA~=~10~ATP

Now we can do the total calculation:

(8*2)~+~(8*3)~+~(9*10)=130~ATP

We have to <u>subtract</u>  "2 ATP" molecules that correspond to the <u>activation</u> of the fatty acid, so:

130-2=128~ATP

In total, we will have 128 ATP.

I hope it helps!

6 0
2 years ago
A sample of gas contains 0.1700 mol of OF2(g) and 0.1700 mol of H2O(g) and occupies a volume of 19.5 L. The following reaction t
andreev551 [17]

Answer: The volume of the sample after the reaction takes place is 29.25 L.

Explanation:

The given reaction equation is as follows.

OF_{2}(g) + H_{2}O(g) \rightarrow O_{2}(g) + 2HF(g)

So, moles of product formed are calculated as follows.

\frac{3}{2} \times 0.17 mol \\= 0.255 mol

Hence, the given data is as follows.

n_{1} = 0.17 mol,      n_{2} = 0.255 mol

V_{1} = 19.5 L,         V_{2} = ?

As the temperature and pressure are constant. Hence, formula used to calculate the volume of sample after the reaction is as follows.

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

Substitute the values into above formula as follows.

\frac{V_{1}}{n_{1}} = \frac{V_{2}}{n_{2}}\\\frac{19.5 L}{0.17 mol} = \frac{V_{2}}{0.255 mol}\\V_{2} = \frac{19.5 L \times 0.255 mol}{0.17 mol}\\= \frac{4.9725}{0.17} L\\= 29.25 L

Thus, we can conclude that the volume of the sample after the reaction takes place is 29.25 L.

8 0
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