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Step2247 [10]
1 year ago
9

b → p experiment number [a] (m) [b] (m) initial rate (m/s) 1 0.273 0.763 2.83 2 0.273 1.526 2.83 3 0.819 0.763 25.47 22) the rat

e law for this reaction is rate
Chemistry
1 answer:
Mazyrski [523]1 year ago
8 0

The rate law for this reaction is [A]².

Balanced chemical reaction used in this experiment: A + B → P

The reaction rate is the speed at which reactants are converted into products.

Comparing first and second experiment, there is no change in initial rate. The concentration of reactant B is increased by double. Initial rate does not depands on concentration of reactant B.

Comparing first and third experiment, initial rate is nine times greater, while concentration of reactant A is three times greater. Conclusion is that concentration of reactant A is squared and the rate is [A]².

More info about rate law: brainly.com/question/16981791

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The decomposition of nitrogen dioxide is described by the following chemical equation: Suppose a two-step mechanism is proposed
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Answer:

<u>first step </u>

NO2(g)  ------------------------------------> NO(g) + O(g)

<u>second step</u>

NO2(g) + O(g) -----------------------------> NO(g) + O2(g)

Explanation:

<u>first step </u>

NO2(g)  ------------------------------------> NO(g) + O(g)

<u>second step</u>

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2 years ago
Which of the following is NOT a type of joint?
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3 years ago
For the reaction: 3 H2(g) + N2(g) &lt;--&gt; 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

Read more on equilibrium constant;

brainly.com/question/1619133

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5 0
2 years ago
To what volume should 5.0 g of KCl be diluted in order to prepare a 0.15 M solution? L
Nesterboy [21]

Answer:0.45L

Explanation:

molarity=0.15M

Mass=5g

No of moles=mass ➗ molecular mass

Molecular mass of KCL=39.0983x1+35.453x1

Molecular mass of KCL=74.5513

No of moles=5 ➗ 74.5513

No of moles=0.067

Volume in liters=No of moles ➗ molarity

Volume in liters=0.067 ➗ 0.15

Volume in liters=0.45L

7 0
3 years ago
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