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3 years ago

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1 answer:

leonid [27]3 years ago

8 0

Answer:

Which of the following

properties distinguishes a solution

oversaturated with a dilute?

The supersaturated solution is one in which the solvent has dissolved more solute than it can dissolve in the saturation equilibrium. The solute can be a solid, or a gas. The molecules of the solvent surround those of the solute and seek to open space between themselves to be able to harbor more amount of solute.

A dilute solution is a solution that has not reached the maximum concentration of solute dissolved in a solvent. The additional solute will dissolve when added in a dilute solution and will not appear in the aqueous phase. It is considered a state of dynamic equilibrium where the speeds in which the solvent dissolves the solute are greater than the recrystallization rate.

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kirill [66]

Well,

I guess the answer would be....

.

6 0

3 years ago

Read 2 more answers

2. How many moles of salt are present in 1.5L of a 5.OM salt water solution?

madam [21]

Answer:

There are 7.5 moles of salt

Explanation:

5.0M means that in every liter of solution, there are 5 moles of salt. So, 1.5L of solution times 5 moles per liter equals 7.5 moles

8 0

3 years ago

In the laboratory you are asked to make a 0.175 m barium iodide solution using 13.9 grams of barium iodide. How much water shoul

BARSIC [14]

<u>Answer:</u> The mass of water that should be added in 203.07 grams

<u>Explanation:</u>

To calculate the molality of solution, we use the equation:

$\text{Molality}=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

Where,

m = molality of barium iodide solution = 0.175 m

$m_{solute}$ = Given mass of solute (barium iodide) = 13.9 g

$M_{solute}$ = Molar mass of solute (barium iodide) = 391.14 g/mol

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

Putting values in above equation, we get:

$0.175=\frac{13.9\times 1000}{391.14\times W_{solvent}}\\\\W_{solvent}=\frac{13.9\times 1000}{391.14\times 0.175}=203.07g$

Hence, the mass of water that should be added in 203.07 grams

4 0

3 years ago

Black_prince [1.1K]

Answer:

efef2w

Explanation:

7 0

2 years ago

How many grams of Lead(II) nitrate are present in 150.0 mL of an Lead(II) nitrate solution that is 0.07268 M?

Lana71 [14]

Hey...
Use the molarity formula
M=moles/L and then convert to grams

0.07268*0.15=moles
<span>0.010902 mol
</span>Pb(NO3)2
1 mole=331.22g
0.010902 moles=

3.61 g

8 0

3 years ago