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Ivan
2 years 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]2 years 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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6 0
3 years ago
The student's lab manual says to mix some of his Na2CO3 solution with an aqueous solution of copper(II) sulfate (CuSO4)
lord [1]

Explanation:

When the student mixed the solution sodium carbonate with solution of copper(II) sulfate ; Copper Hydroxocarbonate , sodium sulfate and carbon dioxide gas was obtained as a products.

The balanced chemical reaction

2Na_2CO_3+2CuSO_4\rightarrow Cu_2(OH)_2CO_3+2Na_2SO_4+CO_2

Where:

Cu_2(OH)_2CO_3 = Copper(II) Hydroxocarbonate

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3 0
3 years ago
What is the molarity of a solution with 23 moles of solute, and 100 ml of a solvent?
Marina CMI [18]
Molarity is calculated by using following formula,

                            Molarity  =  Moles / Volume

Data Given:
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                  Volume  =  100 ml ÷ 1000  =  0.1 L

Putting values in eq. 1,

                              Molarity  =  23 mol / 0.1 L

                              Molarity  =  230 mol/dm³

Result:
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3 0
3 years ago
(The radioisotope 224Ra decays by alpha emission via two paths to the ground state of its daughter 94% probability of alpha deca
pav-90 [236]

Answer:

a) ²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b) the Q-value of this reaction is 5.789 MeV

c) the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

       

Explanation:

a)

The decay equation for the alpha decay is expressed as;

²²⁴Ra₈₈ ---> α + ²²⁰Rn₈₆ + Q

b)

Calculate the Q-value (in MeV) of this reaction.

Q = Mparent - Mdaughter -Mg

Q = MRa - MRn -Mg

= 224.020202 - 220.011384 - 4.00260305

= 0.00621495 amu

= 5.789 MeV

therefore the Q-value of this reaction is 5.789 MeV

c)

Energy of alpha particle is expressed as;

E∝ = MQ / ( m + M)

now this is the maximum energy available for the daughter, ²²⁰Rn going to the ground state;

The energy of the alpha particle gives;

E∝  = 220(5.789) / ( 4 + 220) = 5.69 MeV

as given in the question,The other less frequent alpha occurring 5.5% of the time leaves the daughter nucleus in an excited state of 0.241 MeV above the ground state.

Therefore the energy of this alpha is

E∝ = 5.69 - 0.241 = 5.449 Mev

Therefore the energies (in MeV) of the two associated alpha particles are; 5.69 MeV and 5.449 Mev

d)

Sketch of the nuclear decay scheme have been uploaded along side this answer.

7 0
3 years ago
PLEASE SHOW YOUR WORK!!
Rom4ik [11]

Answer:

6. 355.1 g of Na₂SO₄ can be formed.

7. 313 g of LiNO₃ were needed

Explanation:

<u>Excersise 6</u>.

The reaction is:  2 NaOH + H₂SO₄ --> 2 H₂O + Na₂SO₄

2 moles of sodium hydroxide react with 1 mol of sulfuric acid to produce 2 moles of water and 1 mol of sodium sulfate.

If we were noticed that the acid is in excess, we assume the NaOH as the limiting reactant. Let's convert the mass to moles (mass / molar mass)

200 g / 40 g/mol = 5 moles.

Now we apply a rule of three with the ratio in the reaction, 2:1

2 moles of NaOH produce 1 mol of sodium sulfate.

5 moles of NaOH would produce (5 .1)/2 = 2.5 moles

Let's convert these moles to mass (mol . molar mass)

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<u>Excersise 7.</u>

The reaction is:

Pb(SO₄)₂+ 4 LiNO₃ → Pb(NO₃)₄  +  2Li₂SO₄

As we assume that we have an adequate amount of lead (IV) sulfate, the limiting reactant is the lithium nitrate.

Let's convert the mass to moles (mass / molar mass)

250 g / 109.94 g/mol = 2.27 moles

Let's make a rule of three. Ratio is 2:4.

2 moles of lithium sulfate were produced by 4 moles of lithium nitrate

2.27 moles of Li₂SO₄ would have been produced by ( 2.27 .4) / 2 = 4.54 moles.

Let's convert these moles to mass (mol . molar mass)

4.54 mol . 68.94 g/mol = 313 g

5 0
2 years ago
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