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madreJ [45]
3 years ago
5

If the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in

the solution at that temperature, the solution is said to be
a.diluted.
b.unsaturated.
c.saturated.
d.supersaturated.
Chemistry
1 answer:
anygoal [31]3 years ago
5 0

Answer:

d. supersaturated.

Explanation:

A solution naturally contains a solute and a solvent. The solute is the solid substance that dissolves in the solvent, which is usually a liquid substance. A solution has a maximum amount of solute that can dissolve in its constituent solvent.

However, when the amount of dissolved solute in a solution at a given temperature is greater than the amount that can permanently remain in the solution at that temperature, the solution is said to be SUPERSATURATED. This means that the solution contains more than the maximum amount of solute.

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Assume that Aluminum and Silver Sulfide are the starting substances (reactants) in the reaction: a. Write a balanced chemical eq
Naddik [55]

Answer:

Check Explanation

Explanation:

a) The balanced chemical reaction between Aluminium and Silver Sulfide is represented below

2Al + 3Ag₂S → Al₂S₃ + 6Ag

Aluminium displaces Silver from Silver sulfide because it is higher than Silver in the electrochemical series.

b) Names of the products from this reaction

Ag is called Silver metal. Free Silver metal.

Al₂S₃ is called Aluminium Sulfide.

c) This reaction is a single-displacement reaction because an element directly displaces and replaces another element in a compound.

It is also a redox reaction (reduction-oxidation reaction) because there are species being oxidized and reduced simultaneously!

d) Yes, this reaction is an oxidation-reduction (redox) reaction because there are species being oxidized and reduced simultaneously!

e) The specie that is being oxidized is said to undergo oxidation. And oxidation is defined as the loss of electrons, thereby leading to an increase in oxidation number.

In this reaction, it is evident that Aluminium undergoes oxidation as its oxidation number increases from 0 in the free state to +3 when it displaces Silver and becomes Aluminium sulfide.

Al → Al³⁺

0 → +3 (Oxidation)

The specie that is being reduced is said to undergo reduction. And reduction is defined as the gain of electrons, thereby leading to a decrease in oxidation number.

In this reaction, it is evident that the silver ion undergoes reduction as its oxidation number decreases from +1 in the Silver Sulfide compound to 0 when it is displaced and becomes Silver in free state.

Ag⁺ → Ag

+1 → 0 (Reduction)

The reducing agent is the specie that brings about reduction. Since Silver ion in Silver sulfide is reduced, Aluminium is the reducing agent that initiates the reduction process.

The oxidation agent is the specie that brings about oxidation. Since, Aluminium is the specie that undergoes oxidation, the oxidizing agent is the Silver Sulfide that brings about the oxidation.

Hope this Helps!!!

6 0
2 years ago
2KI + Pb(NO3)2 → 2KNO3 + PbI2 Determine how many moles of KNO3 are created if 0.03 moles of KI are completely consumed.
ElenaW [278]
2:2 so the same proportion 
0,03

6 0
2 years ago
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What happens chemically when quick lime is added to water?
sineoko [7]

Answer:

CaO + H20 => Ca(OH)2

Explanation:

quick lime ia a oxyde and when it reacts with water it gives hydroxide

5 0
2 years ago
Read 2 more answers
How is Hydrogen in heavy water different from hydrogen in normal water​
soldier1979 [14.2K]

Answer: An oxygen atom in heavy water has an extra neutron. A hydrogen atom in heavy water has an extra proton.

Explanation:

7 0
2 years ago
For the aqueous solution containing 75 mg of compound C in 1.0 mL of water, what will be the total amount of the solute C that w
Sindrei [870]

Answer:

75 mg

Explanation:

We can write the extraction formula as

x = m/[1 + (1/K)(Vaq/Vo)], where

x = mass extracted

m = total mass of solute

K = distribution coefficient

Vo = volume of organic layer

Vaq = volume of aqueous layer

Data:

m = 75 mg

K = 1.8

Vo = 0.90 mL

Vaq = 1.00 mL

Calculations:

For each extraction,

1 + (1/K)(Vaq/Vo) = 1  + (1/1.8)(1.00/0.90) = 1 + 0.62 = 1.62  

x = m/1.62 = 0.618m

So, 61.8 % of the solute is extracted in each step.

In other words, 38.2 % of the solute remains.

Let r = the amount remaining after n extractions. Then  

r = m(0.382)^n.

If n = 7,

r = 75(0.382)^7 = 75 × 0.001 18 = 0.088 mg

m = 75 - 0.088 = 75 mg

After seven extractions, 75 mg (99.999 %) of the solute will be extracted.

5 0
3 years ago
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