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djverab [1.8K]
3 years ago
6

In the reaction a b c, doubling the concentration of a doubles the reaction rate and doubling the concentration of b does not af

fect the reaction rate. what is the rate law for this reaction?
Chemistry
1 answer:
PolarNik [594]3 years ago
3 0
The reaction must be a + b --> c

Then you can predict a reaction rate, r o the type r = k * a^n * b^m

Given that the reaction rate is not affected by the concentration of b you can state that m = 0 and r = k * a^n.

Now given, that there is a proportional relation between the reaction rate and a (double a gives double rate), then n = 1 and r = k*a. You can verify that if you dobule a r also doubles.

Answer: r = k*a
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Newton's third law

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10.8 days (3 sig.figs.)

Explanation:

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An empty water bottle is full of air at 15°C and standard pressure. The volume of the bottle is 0. 500 liter. How many moles of
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The gi<em>ve</em>n gas has standard pressure, P=1\rm atm

The volume of the gas has been, V= 0. 500 \;\rm  L

The temperature of the gas has been, T=15^\circ \text C\\&#10;T=288\;\rm K

Substituting the values for the moles of gas, <em>n:</em>

<em />

<em />\rm 1\;\times\; 0. 500 =\textit n\;\times\;0.08214\;atm.L/mol.K\;\times\;288\;K\\\\&#10;\textit n=\dfrac{0. 500}{0.08214\;\times\;288} \;mol\\\\&#10;\textit n=0.021\;mol

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Learn more about ideal gas, here:
brainly.com/question/8711877

5 0
3 years ago
Given the following equilibrium constants: Kb B(aq) + H2O(l) ⇌ HB+(aq) + OH−(aq) 1/Kw H+(aq) + OH−(aq) ⇌ H2O(l) What is the equi
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<u>Answer:</u> The value of K_c for the net reaction is \frac{K_b}{K_w}

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  B(aq.)+H_2O(l)\rightleftharpoons HB^+(aq.)+OH^-(aq.);K_b

<u>Equation 2:</u>  H^+(aq.)+OH^-(aq.)\rightleftharpoons H_2O(l);\frac{1}{K_w}

The net equation follows:

B(aq.)+H^+(aq.)\rightleftharpoons HB^+(aq.);K_c

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The value of equilibrium constant for net reaction is:

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We are given:  

K_1=K_b

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Putting values in above equation, we get:

K_c=K_b\times \frac{1}{K_w}=\frac{K_b}{K_w}

Hence, the value of K_c for the net reaction is \frac{K_b}{K_w}

7 0
3 years ago
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