Answer’s pleas I don’t get this

Chemistry

2 answers:

Elenna [48]3 years ago

6 0

Allow the water to evaporate, it will leave the salt, it being a solid substance, behind.

Nostrana [21]3 years ago

3 0

C. Allow water to evaporate

You might be interested in

Can somebody plz help answer this science questions (only 1-2 sentences is good thx !

Mrac [35]

I believe it’s because they are trying to find other things that won’t pollute the environment as much, and they are also doing this because we are trying to reduce green house gas. Sorry if i’m wrong :(

8 0

3 years ago

What was the result of collisions between the early Earth and other, smaller protoplanets?

RoseWind [281]

The trajectory of their motion knocked the Earth into a different orbit.

3 0

4 years ago

An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?

umka2103 [35]

Answer: The boiling point of solution is 100.53

Explanation:

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

Or,

$\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

$K_b$ = molal boiling point elevation constant = 0.51°C/m

$m_{solute}$ = Given mass of solute (CsCl) = 8.00 g

$M_{solute}$ = Molar mass of solute (CsCl) = 168.4 g/mol

$W_{solvent}$ = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

$\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC$