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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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Two weak acids, A and B, have pKa values of 4 and 6, respectively. Which statement is true?A) Acid A dissociates to a greater ex
zalisa [80]

Answer:

A) Acid A dissociates to a greater extent in water than acid B

Explanation:

A) Acid A dissociates to a greater extent in water than acid B

We are given the pKa for both acids, and we know

pKa = - log Ka

Taking antilog to both sides of the equation we can solve for Ka

⇒ -pKa = log Ka

-antilog pKa = Ka

10 ^-pka = Ka

So ka for acid A = 10⁻⁴

and

ka for acid B = 10⁻⁶

True the equilibrium constant for acid A is greater, so it dissociates more.

B) For solutions of equal concentration, acid B will have a lower pH.

We know the stronger acid is A, and it dissociates more. Since pH is the negative log of H₃O⁺ concentration, it follows that at equal concentrations the acid A will have at equilibrium a greater [H₃O⁺] and hence a lower pH

C) B is the conjugate base of A

False:

If B were the conjugate base of A, its  Kb would have been given by:

Ka x Kb = Kw

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Kb = 10⁻¹⁴/10⁻⁴ = 10⁻¹⁰ for the conjugate base of acid A

which are not equal.

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False

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E) The equivalence point of acid A is higher than that of acid B

False

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The equivalence point is reached in the titration when the number of equivalents of base equals the number of equivalents of acid:

# equivalents acid = # equivalents of base  @ end point

and

# equivalents acid = Molarity of acid x Volume of acid

4 0
2 years ago
In an experiment, the density of an unknown liquid was calculated to be 0.78 g/ml. if the accepted value is 0.75 g/ml, calculate
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Percentage error is the relative error your measured value is from the true or accepted value. The formula for percentage error is written below:

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6 0
1 year ago
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Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

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volume of HA, V_A  = 25.0cm³, = 0.025dm³

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volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

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nika2105 [10]
With reference to present question, following things may be noted
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