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Ivan
1 year ago
14

The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C

represent individual particles.
A graph is shown with two inverted graph curves running close to each other. One of the curves labeled T1 is slightly more spread out than the other labeled T2. The x axis of the graph has the title Kinetic Energy. The y axis of the graph has the title Number of Particles. A vertical line perpendicular to the x axis is shown. This vertical line is labeled Activation Energy. A point labeled B is shown on the right hand side of the vertical line. A point labeled A is shown in the lower left side of the vertical line. A particle labeled C is shown in the upper left side of the vertical line.

Based on the graph, which of the following statements is likely to be true?
Particle A and C are more likely to participate in the reaction than particle B.
Most of the particles of the two gases have very high speeds.
A fewer number of particles of gas at T1 are likely to participate in the reaction than the gas at T2.
The average speed of gas particles at T2 is lower than the average speed of gas particles at T1.

Chemistry
1 answer:
coldgirl [10]1 year ago
7 0

Answer:

  • <u><em>The average speed of gas particles at T</em></u><em><u>₂</u></em><u><em> is lower than the average speed of gas particles at T</em></u><em><u>₁</u></em><u><em>.</em></u>

Explanation:

<em>Particles A and C</em> are shown as if they are on the same vertical line, which means with the same kinetic energy. Both particle A and C are to the lett of <em>particle B</em>, which means that the formers have a lower kinetic energy than the latter.

Since the likelyhood of a particle to participate in the reaction increases with the kinetic energy, particle B is more likely to participate in the reaction than particles A and C. Hence, the first choice is incorrect.

The graph, although not perfectly symmetrical, does show a bell shape, hence there are many particles will low kinetic energy and many particles with high kinetic energy. You cannot assert that most of the particles of the two gases have high high speeds. Hence, second statement is incorrect, too.

At high values of kinetic energy (toward the right of the curve), the line labeled T₁ is higher than the line labeled T₂, meaning that at T₁ more particles have an elevated kinetic energy than the number of particles that have an elevated kinetic energy at T₂.

On the other hand, at low values of kinetic energy (toward the left of the curve) the line T₂ is higher than the line T₁, meaning that at T₂ more particles have a low kinetic energy than the number of particles that have low kinetic energy at T₁.

Hence, the last two paragraphs are telling that the average kinetic energy of gas particles at T₂ is is lower than the average kinetic energy of gas particles at T₁.

Since the average speed is proportional the the square root of the temperature, the same trend for the average kinetic energy is true for the average speed, and you conclude that the last statement is true: "The average speed of gas particles at T₂ is lower than the average speed of gas particles at T₁".

Since more particles at T₁ have high kinetic energy than the number of particles at T₂ that have a high kinetic energy, more particles of gas at T₁ are likely to participate in the reaction  than the gas at T₂, and the third statement is incorrect.

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3 0
2 years ago
Assuming equal concentrations and complete dissociation, rank these aqueous solutions by their freezing points. NH4l, CoBr3, Na2
seraphim [82]
Answer:

NaI > Na2SO4 > Co Br3

meaning that NaI has the highest freezing point, and Co Br3 has the lowest freezing point.

Explanation:

The freezing point depression is a colligative property.

That means that it depends on the number of solute particles dissolved.

The formula to calculate the freezing point depression of a solution of a non volatile solute is:

ΔTf = i * Kf * m

Where kf is a constant, m is the molality and i is the van't Hoff factor.

Molality, which is number of moles per kg of solvent, counts for the number of moles dissolved and the van't Hoff factor multipllies according for molecules that dissociate.

The higher the number of molecules that dissociate, the higher the van't Hoff, the greater the freezing point depression and the lower the freezing point.

As the question states that you assume equal concentrations (molality) and complete dissociation you just must find the number of ions generated by each solute, in this way:

NH4 I → NH4(+) + I(-) => 2 ions

Co Br3 → Co(+) + 3 Br(-) => 4 ions

Na2SO4 → 2Na(+) + SO4(2-) => 3 ions.

So, Co Br3 is the solute that generate more particles and that solution will exhibit the lowest freezing point among the options given, Na2SO4 is next and the NaI is the third. Ordering the freezing point from higher to lower the rank is NaI > Na2SO4 > CoBr3, which is the answer given.
4 0
1 year ago
An unopened, cold 2.00 L bottle of soda contains 46.0 mL of gas confined at a pressure of 1.30 atm at a temperature of 5.0oC. If
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Answer:

20.8mL

Explanation:

7 0
1 year ago
For the reaction A + B − ⇀ ↽ − C + D A+B↽−−⇀C+D , assume that the standard change in free energy has a positive value. Changing
AURORKA [14]

Answer:

a. Not change the free energy value

b. Increase the free energy value

c. Decrease the free energy value

d. Decrease the free energy value

Explanation:

a. Adding a catalyst:

A catalyst is a substance that will reduce the activation energy of a reaction, it means that the reaction will occur fast. The values of enthalpy, entropy, and free energy are not affected by a catalyst, so ΔG remains the same.

b. Increasing [C] and [D]:

For a reversible reaction, the value of free energy can be calculated by:

ΔG = ΔG° + RT*lnK

Where ΔG° is the standard value for free energy, R is the gas constant, T is the temperature, and K is the constant of equilibrium, which in this case:

K = ([C]*[D])/([A]*[B])

When [C] and [D] increase, the value of K increases, and lnK also increases, then, the value of ΔG increases.

c. Coupling with ATP hydrolysis:

The free energy can be calculated by:

ΔG = ΔH - TΔS

Where ΔH is the change in enthalpy, and ΔS the change in entropy. The ATP hydrolysis is an exothermic reaction, so ΔH <0. When it is coupled, it will reduce the total value of ΔH, and because of that, the value of ΔG will decrease.

d. Increasing [A] and [B]:

As explained above, the increasing at [A] and [B] will decrease the value of K, so the value of lnK will decrease, and ΔG value will also decrease.

4 0
1 year ago
Read 2 more answers
ANSWER FAST PLEASE!!
Svetach [21]

Answer:D

Explanation:

Hope I helped

3 0
1 year ago
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