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erastova [34]
3 years ago
8

HURRY What is true for the energy involved in an explosion? A There is more energy at the end of the explosion than there is at

the beginning. B There is less energy at the end of the explosion than there is at the beginning. C There is the same amount of energy at the end of the explosion than there is at the beginning. D The amount of energy before and after the explosion depends on the type of reaction.
Chemistry
1 answer:
andreyandreev [35.5K]3 years ago
5 0
B .........................................


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Cr 6+ & SO4 2-

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What happens when chlorine forms an ion?
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Predict whether the bond between each pair of atoms will be no polar convalent, polar covalent or ionic
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What is the value of R in the ideal gas law?
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3 years ago
For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
Alik [6]

Answer : The rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

6.0\times 10^{-4}=k(0.20)^2(0.20)^0(0.20)^1

k=7.5\times 10^{-2}M^{-2}s^{-1}

Thus, the value of the rate constant 'k' for this reaction is 7.5\times 10^{-2}M^{-2}s^{-1}

Now we have to calculate the rate for trial 5 that starts with 0.90 M of reagent A, 0.60 M of reagents B and 0.70 M of reagent C.

\text{Rate}=k[A]^2[B]^0[C]^1

\text{Rate}=(7.5\times 10^{-2})\times (0.90)^2(0.60)^0(0.70)^1

\text{Rate}=4.25\times 10^{-2}Ms^{-1}

Therefore, the rate for trial 5 will be 4.25\times 10^{-2}Ms^{-1}

3 0
3 years ago
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