Pls I need help with these ques. help plsssssssssssssss

Chemistry

1 answer:

snow_lady [41]4 years ago

6 0

Answer:

q1..no.2 and 4 are aromatic

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The burning of gasoline in an automobile engine is an example of a(n) _____.

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Carrie went with her mother to the rifle range to practice shooting. Firing the rifle created the force needed to send the bulle

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The reaction A+B-> C+D rate=k[A][B]^2 has an initial rate of 0.0810 M/s. What will the initial rate be if [A] is halved and [

makkiz [27]

The first to do is to isolate the constant, k, to one side. The rate expression would then be:

k = rate/ [A][B]^2

This constant, k, will be the same for the given reaction at a certain temperature. Then, we can relate a second rate to this reaction.

rate1/ [A1][B1]^2 = rate2/ [A2][B2]^2

First question:
0.0810 / [A1][B1]^2 = rate2/ [A2/2][3B2]^2
rate2 = (0.0810 [A2/2][3B2]^2 )/ [A1][B1]^2
rate2 = 0.3645 M/s

Second Equation:
0.0810 / [A1][B1]^2 = rate2/ [A2/2][3B2]^2
rate2 = (0.0810 [3A2][B2/2]^2 )/ [A1][B1]^2
rate2 = 0.0608 M/s

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3 years ago

4. How many weeks are equal to 2.94 x 10 to the power of 6 minutes?

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Answer:

292 weeks

Explanation:

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RATE LAW QUESTION !

vivado [14]

In general, we have this rate law express.:

$\mathrm{Rate} = k \cdot [A]^x [B]^y$
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation $\Delta \mathrm{Rate} = \Delta [B]^y$

$3.09 = (3)^y \implies y \approx 1$

To this point, $\mathrm{Rate} = k \cdot [A]^x [B]^1$

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for $\Delta \mathrm{Rate} = \Delta [A]^x$

$4= (2)^x \implies x = 2$

the rate law is

Rate = k·[A]²[B]

6 0

3 years ago