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Soloha48 [4]
2 years ago
7

Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.

Chemistry
1 answer:
mars1129 [50]2 years ago
7 0

Answer: acid dissociation constant Ka= 2.00×10^-7

Explanation:

For the reaction

HA + H20. ----> H3O+ A-

Initially: C. 0. 0

After : C-Cx. Cx. Cx

Ka= [H3O+][A-]/[HA]

Ka= Cx × Cx/C-Cx

Ka= C²X²/C(1-x)

Ka= Cx²/1-x

Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2

Ka= 0.2(0.001²)/(1-0.001)

Ka= 2.00×10^-7

Therefore the dissociation constant is

2.00×10^-7

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Answer:

An external stimuli is a stimulus that comes from outside an organism and causes a reaction.An internal stimuli is a stimulus that comes from inside an organism.

3 0
2 years ago
What is the value for (delta)G at 1000 K if (delta)H = -220 kJ/mol and (delta)S = -0.05 kJ/(molK)?
tankabanditka [31]
The system is isothermal, so we use the formula:
(delta)G = (delta)H - T (delta) S

Plugging in the given values:
(delta)G = -220 kJ/ mol - (1000K) (-0.05 kJ/mol K)
(delta)G = -170 kJ/mol

If we take a basis of 1 mol, the answer is
D. -170 kJ 
6 0
3 years ago
Read 2 more answers
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andrey2020 [161]

Answer:

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Explanation:

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4 0
2 years ago
Consider the following system at equilibrium:
Vika [28.1K]

Answer:

1) Rightward shift

2) Rightward shift

3) Leftward shift

4) Leftward shift

5) Leftward shift

6) Rightward shift

7) No shift

8) No shift                                                              

   

Explanation:

To evaluate each case we need to consider Le Chatelier's Principle, which states that the adding of additional reactants or products to a system will shift the equilibrium in the opposite direction, to maintain the equilibrium of the system. On the contrary, if we remove a reactant or a product in the system, the equilibrium will be shifted in the direction of the reactant or product reduced, to produce more of it (and thus maintain balance).        

Taking into account the above, let's see each statement, in the following equation:

A(aq) + B(aq)  ⇄  2C(aq)    (1)

1) Increase A. This will cause a rightward shift in equation 1 in order to consume the reactant added.

2) Increase B. Same as 1), this will cause a rightward in equation 1.

3) Increase C. This will cause a leftward shift in order to consume the excess of product in the system.  

4) Decrease A. This will produce a leftward shift to produce the reactant that is being reduced.

5) Decrease B. Same as 4), a leftward shift.

6) Decrease C. This will produce a rightward shift to produce the product that is being reduced.

7) Double A, half B. The double A will cause a rightward shift and the half B will produce a leftward shift, which results in no shift.

8) Double both B and C. Double B will produce a rightward shift and double C will produce the contrary, a leftward shift, so the final result is no shift.

               

I hope it helps you!

4 0
3 years ago
In uncompetitive inhibition, the inhibitor can bind to the enzyme only after the substrate binds first. (T/F)
Veronika [31]

Answer:

True

Explanation:

In an uncompetitive inhibition, initially the substrate [S] binds to the active site of the enzyme [E] and forms an enzyme-substrate activated complex [ES].

The inhibitor molecule then binds to the enzyme- substrate complex [ES], resulting in the formation of [ESI] complex, thereby inhibiting the reaction.

This inhibition is called uncompetitive because the inhibitor does not compete with the substrate to bind on the active site of the enzyme.

Therefore, in an uncompetitive inhibition, the inhibitor molecule can not bind on the active site of the enzyme directly. The inhibitor can only bind to the enzyme-substrate complex formed.

       

4 0
2 years ago
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