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shepuryov [24]
3 years ago
5

Given the rate equation, when the concentration of A is doubled while that of B is halved, the rate will Given the rate equation

, when the concentration of A is doubled while that of B is halved, the rate will blank the original rate. the original rate.
Chemistry
1 answer:
melisa1 [442]3 years ago
6 0

Answer:

I believe the question is incomplete as the rate equation was not included in the question, but no need to be worried I can be of help. Here is the complete question:

Given: A + 3B  2C + D

This reaction is first order with respect to reactant A and second order with respect to reactant B. If the concentration of A is doubled and the concentration of B is halved, the rate of the reaction would _____ by a factor of _____.

(a) increase, 2

(b) decrease, 2

(c) increase, 4

(d) decrease, 4

(e) not change

Explanation:

Solving the above question, the rate of the reaction will decrease and by factor of 2. How?

K1 = [A] [B]^2 i.e, reactant A is first order, reactant B is second order.

k2 = 2[A] [B/2]^2 that is, concentration of reactant A is doubled, concentration of reactant B is halved.

Comparing both rate of reactions:

k1/k2 = [A] [B/2] / 2[A] [B/2]^2

k1/k2 = 4 [A] [B]^2 /2 [A] [B]^2

k1 /k2 = 4 [B]^2 / 2[B]^2

k1/k2 = 2

The rate of the reaction is therefore decreased and by 2.

With this ideas of how to solve this kind of question, you can use it to solve for the actual question if the question is solved is not the actual one.

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3 0
3 years ago
Two equilibrium reactions of nitrogen with oxygen, with their corresponding equilibrium constants (Kc) at a certain temperature,
7nadin3 [17]

Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

     N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)   Kc = 1/2.16e - 24  

    now adding 1 and reversed equation (2)

       N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)

      N_{2}O(gas)\rightarrow N_{2}(gas) + 1/2O_{2}(gas)

   we get ,

                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

3 0
3 years ago
which nonmetals could form an ionic compound with the formula MgX^2 ( where X represents the nonmetal)? (possible, not possible)
djverab [1.8K]

Answer:

The answer to your question is:

Explanation:

Ionic compound is formed when a metal and a nonmetal are attached.

If we have MgX₂, that means that the nonmetal must have a valence of -1.

From the list the nonmetals with a valence of -1 are:

Bromine(Br) and  fluorine(F).

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Explanation:

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Which of these pairs indicates an incorrect coupling of reversible reactions?dehydration synthesis and hydrolysisanabolic and ca
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Answer:

option C= hydrolysis and break down

Explanation:

All other three pairs are correct coupling of each others.

Option A= dehydration synthesis and hydrolysis

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In dehydration synthesis monomers combine through the covalent bonds and form large molecules. The large molecules are called polymers. The water as a byproduct also released when monomers joints together.

Hydrolysis:

In hydrolysis the polymers are break down into monomers by using water molecules. The catalysts are also required in this process.

Option B= Catabolic and Anabolic

Anabolic:

In this process smaller molecules combine to gather to form large complex molecules by using energy.

For example simple glucose molecules join together to form large disaccharides.

Catabolic:

It is the break down of large complex molecules to the smaller molecules.

For example during cellular respiration sugar molecules break down and generate energy.

Option D=  Break down and synthesis

The break down and synthesis are also reverse pair of each others. The  synthesis involve the formation of molecules form smaller component while the break down involve destruction of molecules into smaller units.

8 0
3 years ago
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