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nikitadnepr [17]
3 years ago
8

 A reaction takes place between two gases at 298 K. There are 2.0 moles of gas A and 3.0 moles of gas B. Which scenario would lo

wer the collision frequency between gases as well as the rate of reaction between gases A and B?A)Increase the moles of gas B.B)Lower the temperature of the reaction to 250 K.C)Compress the gases into a smaller volume.D)Introduce a catalyst to the container.
Chemistry
2 answers:
Free_Kalibri [48]3 years ago
7 0

Answer:

B)Lower the temperature of the reaction to 250 K.

Explanation:

Option B is correct. Lowering the temperature will decrease the kinetic energy of the gas molecules decreasing the frequency of collision and the rate of the reaction.

Option A is incorrect. Increasing the moles or the number of gas molecules of A and B will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option C is incorrect. Compressing the gas into a small volume will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option D is incorrect. A catalyst works by lowering the activation energy and thus increasing the frequency and orientation of the collision between gas molecules and thus increasing the rate of the reaction.  

Soloha48 [4]3 years ago
3 0
Choice B I think, because if the temperature decreases the particles move slower due to less kinetic energy, and thus won't collide as frequently.
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The half-life for the radioactive decay of C-14 is 5730 years and is independent of the initial concentration. How long does it
DedPeter [7]

Answer:

It will take 2378 years for 25% of the C-14 atoms in a sample of the C-14 to decay.

Explanation:

Radioactive decays/reactions always follow a first order reaction dynamic.

Let the initial amount of C-14 atoms be A₀ and the amount of atoms at any time be A

The general expression for rate of reaction for a first order reaction is

(dA/dt) = -kA (Minus sign because it's a rate of reduction)

k = rate constant

(dA/dt) = -kA

(dA/A) = -kdt

 ∫ (dA/A) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from A₀ to A and the right hand side from 0 to t.

We get

In (A/A₀) = -kt

(A/A₀) = e⁻ᵏᵗ

A(t) = A₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus

T(1/2) = (In2)/k

T(1/2) = 5730 years

k = (In 2)/5730 = 0.000120968 = 0.000121 /year.

So, the amount of C-14 atoms left at any time is given as

A(t) = A₀ e⁻⁰•⁰⁰⁰¹²¹ᵗ

How long does it take for 25% of the C-14 atoms in a sample of the C-14 to decay?

When 25% of C-14 atoms in a sample decay, 75% of C-14 atoms in the sample remain.

Hence,

A(t) = 75%

A₀ = 100%

100 = 75 e⁻⁰•⁰⁰⁰¹²¹ᵗ

e⁻⁰•⁰⁰⁰¹²¹ᵗ = (75/100) = 0.75

In e⁻⁰•⁰⁰⁰¹²¹ᵗ = In 0.75 = - 0.28768

-0.000121t = -0.28768

t = (0.28768/0.000121) = 2,377.54 = 2378 years

Hope this Helps!!!

3 0
3 years ago
For the reaction C6H14(g) ------> C6H6(g) + 4H2(g), the rate of formation of hydrogen gas, H2 was found to be 2.5 x 10-2 atm/
riadik2000 [5.3K]

Answer : The rate of consumption of hexane is, 6.25\times 10^{-3}atm/s

Explanation :

The general rate of reaction is,

aA+bB\rightarrow cC+dD

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.

The expression for rate of reaction will be :

\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}

\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}

\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}

\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}

Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}

From this we conclude that,

In the rate of reaction, A and B are the reactants and C and D are the products.

a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.

The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.

The given rate of reaction is,

C_6H_{14}(g)\rightarrow C_6H_6(g)+4H_2(g)

The expression for rate of reaction :

\text{Rate of disappearance of }C_6H_{14}=-\frac{d[C_6H_{14}]}{dt}

\text{Rate of formation of }C_6H_6=+\frac{d[C_6H_6]}{dt}

\text{Rate of formation of }H_2=+\frac{1}{4}\frac{d[H_2]}{dt}

As we know that,  the partial pressure is directly proportional to the concentration. So,

\text{Rate of reaction}=-\frac{dP_{C_6H_{14}}}{dt}=+\frac{dP_{C_6H_6}}{dt}=+\frac{1}{4}\frac{dP_{H_2}}{dt}

Given:

+\frac{dP_{H_2}}{dt}=2.5\times 10^{-2}atm/s

As,  

-\frac{dP_{C_6H_{14}}}{dt}=+\frac{1}{4}\frac{dP_{H_2}}{dt}=2.5\times 10^{-2}atm/s

and,

\frac{dP_{C_6H_{14}}}{dt}=\frac{1}{4}\times 2.5\times 10^{-2}atm/s

\frac{dP_{C_6H_{14}}}{dt}=6.25\times 10^{-3}atm/s

Thus, the rate of consumption of hexane is, 6.25\times 10^{-3}atm/s

4 0
3 years ago
Someone people help me with this chemistry question
o-na [289]

Answer:

6 moles

Explanation:

From the equation you can see that twice as many HCL moles are used as H2 produced

2 x 3 = 6 moles  of HCL required

6 0
2 years ago
What are the number if protons, neutrons, and electrons
marishachu [46]

protons and electrons are both always the atomic number which is 9 in this case.

For neutrons you subtract the atomic number (9) from the weight of the atom (18.998) some teachers will want you to round to the nearest whole (19). We do this because the number of protons is the atomic number so if you subtract the protons from the whole weight of the atom you would have the electrons and neutrons left. Since electrons weigh so little we don't have to subtract them. Weighing neutrons and electrons would be like weighing an elephant (neutrons) and then putting one marshmallow on the scale (electron).  

5 0
3 years ago
Can some one plz help
blsea [12.9K]

The reaction uses B) 9.0 g Br₂.

 iron + bromine ⟶ product

2.0 g +     <em>x</em> g     ⟶   11.0 g

According to the <em>Law of Conservation of Mass</em>, the total mass of the reactants must equal the total mass of the products.

∴2.0 + <em>x</em> = 11.0

<em>x</em> = 11.0 – 2.0 = 9.0

The reaction uses 9.0 g Br₂.

5 0
3 years ago
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