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Viefleur [7K]
3 years ago
12

What is unsaturated solution​

Chemistry
2 answers:
Natalija [7]3 years ago
6 0

Answer:

An unsaturated solution is a chemical solution in which the solute concentration is lower than its equilibrium solubility. All of the solute dissolves in the solvent. ... In an unsaturated solution, the rate of dissolution is much greater than the rate of crystallization.

topjm [15]3 years ago
3 0
It’s not saturated jwnsnenneen
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A sample of seawater weighs 158 g and has a volume of 156 mL. What is the density, in g/
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1.01 g/ml

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A 2 kg toy car and a 5 kg toy car are pushed with an equal amount of force. Which toy car will have the least acceleration?
pishuonlain [190]

Answer:

The 5 kg toy car will have least acceleration

Explanation:

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3 years ago
What is the molarity of a solution prepared by diluting 250 mL of a 40% H2SO4 solution to 1 Liter? The density of the stock solu
kiruha [24]

Answer:

1.195 M.

Explanation:

  • We can calculate the concentration of the stock solution using the relation:

<em>M = (10Pd)/(molar mass).</em>

Where, M is the molarity of H₂SO₄.

P is the percent of H₂SO₄ (P = 40%).

d is the density of H₂SO₄ (d = 1.17 g/mL).

molar mass of H₂SO₄ = 98 g/mol.

∴ M of stock H₂SO₄ = (10Pd)/(molar mass) = (10)(40%)(1.17 g/mL) / (98 g/mol) = 4.78 M.

  • We have the role that the no. of millimoles of a solution before dilution is equal to the no. of millimoles after dilution.

<em>∴ (MV) before dilution = (MV) after dilution</em>

M before dilution = 4.78 M, V before dilution = 250 mL.

M after dilution = ??? M, V after dilution = 1.0 L = 1000 mL.

∴ M after dilution = (MV) before dilution/(V after dilution) = (4.78 M)(250 mL)/(1000 mL) = 1.195 M.

3 0
3 years ago
Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as
nikitadnepr [17]

The question in incomplete, complete question is;

Determine the theoretical yield:

Excess aqueous copper(II) nitrate reacts with aqueous sodium sulfide to produce aqueous sodium nitrate and copper(II) sulfide as a precipitate. In this reaction 469 grams of copper(II) nitrate were combined with 156 grams of sodium sulfide to produce 272 grams of sodium nitrate.

Answer:

The theoretical yield of sodium nitrate is 340 grams.

Explanation:

Cu(NO_3)_2(aq)+Na_2S(aq)\rightarrow 2NaNO_3(aq)+CuS(s)

Moles of copper(II) nitrate = \frac{469 g}{187.5 g/mol}=2.5013 mol

Moles of sodium sulfide = \frac{156 g}{78 g/mol}=2 mol

According to reaction, 1 mole of copper (II) nitrate reacts with 1 mole of sodium sulfide.

Then 2 moles of sodium sulfide will react with:

\frac{1}{1}\times 2mol= 2 mol of copper (II) nitrate

As we can see from this sodium sulfide is present in limiting amount, so the amount of sodium nitrate will depend upon moles of sodium sulfide.

According to reaction, 1 mole of sodium sulfide gives 2 mole of sodium nitrate, then 2 mole of sodium sulfide will give:

\frac{2}{1}\times 2mol=4 mol sodium nitrate

Mass of 4 moles of sodium nitrate :

85 g/mol × 4 mol = 340 g

Theoretical yield of sodium nitrate = 340 g

The theoretical yield of sodium nitrate is 340 grams.

7 0
3 years ago
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