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2 years ago

Consider these two entries from a fictional table of standard reduction potentials.

Chemistry

1 answer:

Vika [28.1K]2 years ago

8 0

Answer:

1.89 V

Explanation:

To calculate the standard cell potential, subtract the reduction potential of the anode from the reduction potential of the cathode.

So for your calculation,

-0.22 V - (-2.11 V)= 1.89 V

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What data should be plotted to show that experimental concentration data fits a first-order reaction? A) 1/[reactant] vs. time B

natita [175]

Answer:

C) In[reactant] vs. time

Explanation:

For a first order reaction the integrated rate law equation is:

$A = A_{0}e^{-kt}$

where A(0) = initial concentration of the reactant

A = concentration after time 't'

k = rate constant

Taking ln on both sides gives:

$ln[A] = ln[A]_{0}-kt$

Therefore a plot of ln[A] vs t should give a straight line with a slope = -k

Hence, ln[reactant] vs time should be plotted for a first order reaction.

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3 years ago

For the simple decomposition reactionAB(g)→ A(g) + B(g)Rate =k[AB]2 and k=0.2 L/mol*s . How long will it takefor [AB] to reach 1

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Answer:

6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.

Explanation:

$Rate = k[AB]^2$

The order of the reaction is 2.

Integrated rate law for second order kinetic is:

$\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt$

Where, $[A_0]$ is the initial concentration = 1.50 mol/L

$[A_t]$ is the final concentration = 1/3 of initial concentration = $\frac{1}{3}\times 1.50\ mol/L$ = 0.5 mol/L

Rate constant, k = 0.2 L/mol*s

Applying in the above equation as:-

$\frac{1}{0.5} = \frac{1}{1.50}+0.2t$

$\frac{1}{1.5}+0.2x=\frac{1}{0.5}$

$t = 6.66\ s$

<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>

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2 years ago

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