26. If a saturated solution of AgNO3 at 20°C contains 216 g AgNO3 per 100.0 g of water, what mass of water could

Chemistry

1 answer:

givi [52]2 years ago

8 0

335.6g of water will hold 725 grams of silver nitrate.

<h3><u>Explanation:</u></h3>

The saturated solution of a solvent is the solution which can't hold more of the solvent at a particular temperature and pressure under normal conditions, if more solvent are added, it will precipitate out.

Silver nitrate solution at the particular temperature and pressure contains 216 grams of silver nitrate in 100g of water.

So, 1 gram of silver nitrate is held in 100/216 grams of water.

So, 725grams of silver Nitrate is held in 100*725/216 grams of water = 335.6 grams of water.

This gives the saturated solution of silver nitrate again.

You might be interested in

What is the difference between chemical change and physical change

Nadusha1986 [10]

In a physical change the appearance or form of the matter changes but the kind of matter in the substance does not. However in a chemical change, the kind of matter changes and at least one new substance with new properties is formed. Hope this helps!

7 0

3 years ago

The electron cloud model describes the __of electrons in an atom.

Alex Ar [27]

Answer:

it's location

Explanation:

4 0

2 years ago

8. How is acceleration dependent on the speed and direction of an object?

Sphinxa [80]

Answer:

if speed or direction changes then there is a change in velocity which is acceleration. Hope this helps!

Explanation:

3 0

3 years ago

Calculate the solubility (in g/L) of CaSO4(s) in 0.500 M Na2SO4(aq) at 25°C . The sp of CaSO4 is 4.93×10^−5 .

jeka94

When you calculate the first two numbers in 493×10

7 0

3 years ago

He structure of the product, C, of the following sequence of reactions would be:

NikAS [45]

Answer:

Explanation:

The complete question can be seen in the image attached.

We need to understand what is actually going on here. In the first step that yields product A, the sodamide in liquid ammonia attacks the alkyne and abstracts the acidic hydrogen of the alkyne. The second step is a nucleophilic attack of the C6H5C≡C^- on the alkyl halide to yield product B (C6H5C≡C-CH3CH2).

Partial reduction of B using the Lindlar catalyst leads to syn addition of hydrogen to yield structure I as the product C.

5 0

3 years ago