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cricket20 [7]
3 years ago
8

The rate constant k for a certain reaction is measured at two different temperatures:

Chemistry
1 answer:
bogdanovich [222]3 years ago
8 0

Answer:

Ea=5.5 Kcal/mole

Explanation:

Let rate constant are K_1  and K_2  at temperature T_1  and T_2

By using Arrhenius equation at two different two different temperature,

Log K_1/K_2 =E_a/2.303R*(1/T_2 -1/T_2 );T_1=273+376=649K  ;K_1=4.8*10^8;T_2=273+280=553K  ;K_2=2.3*10^8;R=2 cal/(mole.K);Log (4.8*10^8)/(2.3*10^8 )=E_a/2.303R*(1/553K-1/649); Log 4.8/2.3=E_a/2.303R*96/358897 ;0.32=E_a/2.303R*96/358897;E_a=(0.32*2.303R*358897)/96;  

By putting value of R=2 cal/mole.K

E_a=5510.265cal/mole;

By rounding off upto 2 significant figure;

E_a=5.5Kcal/mole;

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Answer:

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Explanation:

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Taking negative logarithm on both sides:

-\log[1.01\times 10^-{14}]=(-\log [H^+])+(-\log [OH^-])

The pH is the negative logarithm of hydrogen ion concentration in solution.

The pOH is the negative logarithm of hydroxide ion concentration in solution.

13.99=pH+pOH

a) 1.39\times 10^{-2} M of NaOH.

Concentration of hydroxide ions:

NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)

So, [OH^-]=1\times [NaOH]=1\times 1.39\times 10^{-2} M=1.39\times 10^{-2} M

pOH=-\log[1.39\times 10^{-2} M]=1.86

13.99=pH+pOH

13.99=pH+1.86

pH=13.99-1.86=12.13

b) 0.0051 M of NaOH.

Concentration of hydroxide ions:

Al(OH)_3(aq)\rightarrow Al^{3+}(aq)+3OH^-(aq)

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13.99=pH+pOH

13.99=pH+1.82

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6 0
3 years ago
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vladimir2022 [97]

Answer:

4.7485 g

Explanation:

4.50 x 10^22 Cu atoms * (1 mol Cu / 6.022 x 10^23 Cu atoms) * 63.546 g Cu/(mol Cu) = 4.7485 g

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TEA [102]
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Answer:

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