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Rzqust [24]
3 years ago
13

Which of the following changes will likely increase the solubility of a solid solute but decrease the solubility of a gaseous so

lute? A. Increasing the molecular mass of the solute B. Increasing the temperature C. Decreasing the concentration D. Decreasing the atmospheric pressure
Chemistry
2 answers:
Readme [11.4K]3 years ago
6 0
A. Increasing the molecular mass of the solute, Because that would Increase the solubility of solid solute, but wouldn't increase the gaseous solute.
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MA_775_DIABLO [31]3 years ago
5 0

Answer: Option (B) is the correct answer.

Explanation:

When we increase the temperature then there is increase in solubility of a solid because of more number of collisions between its particles. As a result, there will be an increase in rate of reaction.

But when we increase the temperature for a gaseous solute then its molecules will gain more kinetic energy due to which they collide more rapidly. Hence, they will escape out of the reaction mixture so, there will be decrease in the solubility of a gaseous solute.

Thus, we can conclude that increasing the temperature will most likely increase the solubility of a solid solute but decrease the solubility of a gaseous solute.

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Which one of the following classes of organic compound does not contain the carbonyl (C=O) group?A) aldehydesB) carboxylic acids
vredina [299]

Answer: alcohols

Explanation:

The carbonyl group refers to C=O. It is contained in aldehyde, Ketones, carboxylic acids , esters, amides and acyl chlorides. They are not found in alcohols. The alchols are generally ROH. They do not contain any carbon-oxygen unsaturated bond in their structure hence the answer.

3 0
3 years ago
6. 100 ml of gaseous hydrocarbon consumes 300
mario62 [17]

Answer:

  • <u><em>a. C₂H₄</em></u>

Explanation:

At constant pressure and temperature, the mole ratio of the gases is equal to their volume ratio (a consequence of Avogadro's law).

Hence, the <em>complete combustion reaction</em> that has a ratio of 100 ml of gaseous hydrocarbon to 300 ml of oxygen, is that whose mole ratio is 1 mol hydrocarbon : 3 mol of oxygen.

Then, you must write the balanced chemical equations for the complete combustion of the four hydrocarbons in the list of choices, and conclude which has such mole ratio (1 mol hydrocarbon : 3 mol oxygen).

A complete combustion reaction of a hydrocarbon is the reaction with oxygen that produces CO₂ and H₂O, along with the release of heat and light.

<u>a. C₂H₄:</u>

  • C₂H₄ (g) + 3O₂ (g) → 2CO₂(g)  + 2H₂O (g)

Precisely, for this reaction the mole ratio is 1 mol C₂H₄: 2 mol O₂, hence, this is the right choice.

The following analysis just shows that the other options are not right.

<u>b. C₂H₂:</u>

  • 2C₂H₂ (g) + 5O₂ (g) → 4CO₂(g)  + 2H₂O (g)

The mole ratio for this reaction is 2 mol C₂H₂ :5 mol O₂.

<u>с. С₃Н₈</u>

  • C₃H₈ (g) + 5O₂ (g) → 3CO₂(g)  + 4H₂O (g)

The mole ratio is 1 mol C₃H₈ : 5 mol O₂

<u>d. C₂H₆</u>

  • 2C₂H₆ (g) +7 O₂ (g) → 4CO₂(g)  + 6H₂O (g)

The mole ratio is 2 mol C₂H₆ : 7 mol O₂

7 0
4 years ago
A mouse runs across a 4.56 m kitchen floor in 3.42 seconds . What is the velocity of the mouse
Lerok [7]
Velocity is the change in position divided by the time . In your problem, it should be expressed in meters per second. To solve, you must divide the distance, 4.56 meters, by the time it takes to travel that distance, which is 3.42 seconds ----->   4.56/3.42 = 1.333 meters per second is the velocity of the mouse
3 0
3 years ago
Which of the following is the weakest? A. hydrogen bond B. polar covalent bond C. dipole interaction D. ionic bond
aniked [119]

the real answer is c, i just took the test

6 0
4 years ago
Read 2 more answers
How to draw Hess' Cycle for this question ?
NISA [10]

Answer : The standard enthalpy of formation of ethylene is, 51.8 kJ/mole

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The formation reaction of C_2H_4 will be,

2C(s)+2H_2(g)\rightarrow C_2H_4(g)    \Delta H_{formation}=?

The intermediate balanced chemical reaction will be,

(1) C_2H_4(g)+3O_2(g)\rightarrow 2CO_2(g)+2H_2O(l)     \Delta H_1=-1411kJ/mole

(2) C(s)+O_2(g)\rightarrow CO_2(g)    \Delta H_2=-393.7kJ/mole

(3) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)    \Delta H_3=-285.9kJ/mole

Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equations, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ/mole

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.7kJ/mole)=-787.4kJ/mole

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.9kJ/mole)=-571.8kJ/mole

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

\Delta H=(+1411kJ/mole)+(-787.4kJ/mole)+(-571.8kJ/mole)

\Delta H=51.8kJ/mole

Therefore, the standard enthalpy of formation of ethylene is, 51.8 kJ/mole

7 0
3 years ago
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