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3 years ago

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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3 years ago

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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 "beta-oxidation". In this route, acetyl-Coa is produced by removing 2 carbons 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 number of rounds that will take place for an 18-carbon fatty acid using the following equation:

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

Where "n" is the number of carbons, 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 production of 1 $FADH_2$ and 1 $NADH$ . 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 yield for each compound:

-) $1~FADH_2~=~2~ATP$

-) $1~NADH~=~3~ATP$

-) $Acetyl~CoA~=~10~ATP$

Now we can do the total calculation:

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

We have to subtract "2 ATP" molecules that correspond to the activation 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

2 years ago