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grigory [225]
3 years ago
6

One gram of salt in 100 liters of water could be considered a _______________________ solution. A) concentrated B) dilute C) sat

urated D) supersaturated
Chemistry
2 answers:
-Dominant- [34]3 years ago
6 0

The correct answer is B) Dilute

Explanation:

In a solution, the proportions of the solute vs the solvent determine the type of solution that includes dilute/concentrated or unsaturated, saturated and supersaturated. In the case of a dilute solution, this occurs if there is a small amount of the solute in comparison to the amount of solvent, which makes it possible to add more solute as this can be dissolved. This type of solution applies to the case presented because one gram of salt (solute or substance dissolved) is a small quantity of solute in comparison to the amount of solvent (the substance that dissolves) that in this case is 100 liters of water. Thus, in this case, there is a dilute solution.

vitfil [10]3 years ago
3 0

Answer:I believe your answer would be option B) dilute. Hope this helps.

Explanation:

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Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra
maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4  mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

 

The equation for the solubilization reaction of Mg(OH)2 can be given as:

 

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

 Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²  

<u>Step 2:</u> Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

 X = <u>2.412 * 10^-4 mol/L = solubility in water</u>

<u>Step 3</u>: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+  = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²  

5.61*10^-11 = X *(0.130)²  

5.61*10^-11 = X * (0.130)^2

X = <u>3.32*10^-9 = solubility in 0.130 M NaOH </u>

<u>Step 4:</u> Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0
3 years ago
Which is the best location for storing radioactive wastes?
vesna_86 [32]
<span>Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.

</span>Geological disposal<span> involves isolating radioactive waste </span>deep<span> inside a suitable rock volume to ensure that no harmful quantities of radioactivity ever reach the surface environment.
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Hope this helps :)
4 0
3 years ago
Read 2 more answers
How many block shells are the d block and the f block​
HACTEHA [7]

Answer:

1s, 3p, 5d, 7f

Explanation:

4 0
3 years ago
How did the work of Dmitri Mendeleev differ from that of John Newlands in the development of the periodic table?
alex41 [277]
When Newlands tried to create a periodic table, his tried to conform to the "Rule of Octaves" he had discovered. He had the right idea, in that if you arrange the elements by atomic weight there would be similarities every 7 elements (not 8 because noble gases hadn't been discovered yet) but he tried to push this rule so much that he would put multiple elements in the same box to try to keep the rule. Mendeleev, however, left gaps in this table for undiscovered elements, which paved the way for our modern periodic table.
3 0
3 years ago
A certain amount of H2S was added to a 2.0 L flask and allowed to come to equilibrium. At equilibrium, 0.072 mol of H2 was found
SIZIF [17.4K]

Answer:

0.098 moles H₂S

Explanation:

The reaction that takes place is

  • 2H₂(g) + S₂(g) ⇄ 2H₂S(g)  keq = 7.5

We can express the equilibrium constant as:

  • keq = [H₂S]² / [S₂] [H₂]² = 7.5

With the volume we can <u>calculate the equilibrium concentration of H₂</u>:

  • [H₂] = 0.072 mol / 2.0 L = 0.036 M

<em>The stoichiometric ratio</em> tells us that <u>the concentration of S₂ is half of the concentration of H₂</u>:

  • [S₂] = [H₂] / 2 = 0.036 M / 2 = 0.018 M

Now we <u>can calculate [H₂S]</u>:

  • 7.5 = [H₂S]² / (0.018*0.036²)
  • [H₂S] = 0.013 M

So 0.013 M is the concentration of H₂S <em>at equilibrium</em>.

  • This would amount to (0.013 M * 2.0 L) 0.026 moles of H₂S
  • The moles of H₂ at equilibrium are equal to the moles of H₂S that reacted.

Initial moles of H₂S - Moles of H₂S that reacted into H₂ = Moles of H₂S at equilibrium

Initial moles of H₂S - 0.072 mol = 0.026 mol

Initial moles of H₂S = 0.098 moles H₂S

8 0
3 years ago
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