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2 years ago

8

0.65743 In scientific notation

1 answer:

Goryan [66]2 years ago

6 0

202827.0000 is the answer I think idrk

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In order for a solute to dissolve in a solvent, what must be true?

Svetradugi [14.3K]

In order for a solute to dissolve in a solvent, the attractive forces between solute particles and the solvent particles must be stronger than the attractive forces between solute-solute and solvent-solvent particles. This is important so that the solute will remain in solution.

7 0

2 years ago

PLEASE HURRY!!! I WILL GIVE BRAINLEIST!!!!

saul85 [17]

Answer:

A I did the exam and I saw A sorry if it is wrong dont have the best of memory

8 0

3 years ago

Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =

BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

Explanation :

As we know that CaCO₃ dissociates to give $Ca^{2+}$ ion and $CO_3^{2-}$ ion.

The solubility equilibrium reaction will be:

$CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=[Ca^{2+}][CO_3^{2-}]$

Let solubility of CaCO₃ be, 's'

$K_{sp}=(s)\times (s)$

$K_{sp}=s^2$

$8.7\times 10^{-9}=s^2$

$s=9.33\times 10^{-5}mol/L$

Therefore, the solubility of CaCO₃ is, $9.33\times 10^{-5}mol/L$

4 0

2 years ago

After reading the list of physical properties above, you realize that they are describing a substance that is a

Alecsey [184]

The answer is B

Hope this helps!

5 0

2 years ago

Titration of a 20.0-mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. If we assume that the acidit

Rasek [7]

Answer:

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

Explanation:

Volume of NaOH = 1.7 ml = 0.0017 L

Molarity of NaOH = 0.0811 M

Moles of NaOH = n

$0.0811 M=\frac{n}{0.0017 L}$

n = 0.0001378 mol

$H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O$

According to reaction, 2 mol of NaOH neutralize 1 mol of sulfuric acid.

Then 0.0001378 mol of NaOH will neutralize:

$\frac{1}{2}\times 0.0001378 mol=6.8935\times 10^{-5} mol$ of sulfuric acid.

Concentration of sulfuric acid in the acid rain sample: x

$x=\frac{6.8935\times 10^{-5}}{0.02 L}=0.0034467 mol/L$

Concentration of sulfuric acid in the acid rain sample is 0.0034467 mol/L.

7 0

3 years ago