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2 years ago

(pls help !!) question is in image

Chemistry

1 answer:

Anastaziya [24]2 years ago

6 0

Answer:

the second option is correct

Explanation:

maybe I think b is correct

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What is Supersaturation? (I think its D. But i dont know)

trapecia [35]

Explanation:

Supersaturation occurs with a chemical solution when the concentration of a solute exceeds the concentration specified by the value equilibrium solubility. Most commonly the term is applied to a solution of a solid in a liquid.

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3 years ago

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Reaction Rates

inn [45]

Answer:

PART A 1st order in A and 0th order in B

Part B The reaction rate increases

Explanation:

PART A

The rate law of the arbitrary chemical reaction is given by

$-r_A=k\times\left[A\right]^\alpha\times\left[B\right]^\beta\bigm$

Replacing for the data

Expression 1 $0.00639=k\times{0.12}^\alpha\times{0.22}^\beta$

Expression 2 $0.01280=k\times{0.24}^\alpha\times{0.22}^\beta$

Expression 3 $0.00639=k\times{0.12}^\alpha\times{0.11}^\beta$

Making the quotient between the fist two expressions

$\frac{0.00639}{0.01280}=\left(\frac{0.12}{0.24}\right)^\alpha$

Then the expression for $\alpha$

$\alpha=\frac{ln\frac{0.00639}{0.01280}}{ln\frac{0.12}{0.24}}=1\bigm$

Doing the same between the expressions 1 and 3

$\frac{0.00639}{0.00639}=\left(\frac{0.22}{0.11}\right)^\beta$

Then

$\beta=\frac{ln\frac{0.00639}{0.00639}}{ln\frac{0.22}{0.11}}=0\bigm$

This means that the reaction is 1st order respect to A and 0th order respect to B .

PART B

By the molecular kinetics theory, if an increment in the temperature occurs, the molecules will have greater kinetic energy and, consequently, will move faster. Thus, the possibility of colliding with another molecule increases. These collisions are necessary for the reaction. Therefore, an increase in temperature necessarily produces an increase in the reaction rate.

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3 years ago

Four people make observations recorded in the table.

77julia77 [94]

Answer:

D. They involve changes in energy.

The last one

Explanation:

have a good day

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2 years ago

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