Right now it is soluble. that's why it is dissolved in the water. which of the following can I add to make it soluble

Chemistry

1 answer:

Arisa [49]3 years ago

7 0

Answer:

Barium nitrate or silver nitrate based on the anion our solute contains

Explanation:

I assume the situation is that currently the solute is soluble in water and you wish to make it insoluble.

It really depends on the soluble material you have, however, let's look at some specific cases.

We have a salt in our solution. Addition of any of the three reagents will produce a double displacement reaction, that is, our cation will be replaced by another cation, either sodium, barium or silver cation.
According to the solubility rules, all sodium salts are soluble, so sodium nitrate won't precipitate our anion.
In case our solute contains sulfate, carbonate or phosphate, we may use barium nitrate to precipitate it, as barium sulfate, barium carbonate and barium phosphate are insoluble.
In case our solute contains chloride, then silver nitrate is the way to go to precipitate it in an insoluble form of AgCl. Similarly, silver would form precipitates with carbonate, phosphate, iodide, bromide and slightly soluble silver sulfate (barium is the choice for sulfate, however).

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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 $NADH$ .