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Sauron [17]
3 years ago
8

Which coefficient before potassium (K) balances this chemical equation?

Chemistry
2 answers:
miskamm [114]3 years ago
5 0

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

Zinaida [17]3 years ago
4 0

Answer:

The coefficient before potassium (K) balances this chemical equation is 2.

Explanation:

_K +Cl₂ → 2KCl

K =1 ; Cl =2

K=1 × 2 = 2

Cl = 1 × 2 = 2

2 K +Cl₂ = 2 KCl

Explanation:

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Answer:

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Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

Then;

the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

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Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

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The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

8 0
3 years ago
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