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.
95.6 cal
are needed.
Explanation:
Use the following equation:
q
=
m
c
Δ
T
,
where:
q
is heat energy,
m
is mass,
c
is specific heat capacity, and
Δ
T
is the change in temperature.
Δ
T
=
T
final
−
T
initial
Known
m
=
125 g
c
Pb
=
0.130
J
g
⋅
∘
C
T
initial
=
17.5
∘
C
T
final
=
42.1
∘
C
Δ
T
=
42.1
∘
C
−
17.5
∘
C
=
24.6
∘
C
Unknown
q
Solution
Plug the known values into the equation and solve.
q
=
(
125
g
)
×
(
0.130
J
g
⋅
∘
C
)
×
(
24.6
∘
C
)
=
400. J
(rounded to three significant figures)
Convert Joules to calories
1 J
=
0.2389 cal
to four significant figures.
400
.
J
×
0.2389
cal
1
J
=
95.6 cal
(rounded to three significant figures)
95.6 cal
are needed.
When Ka =1.37 x 10^-4
∴Pka = - ㏒Ka
= -㏒ 1.37 x 10^-4
= 3.86
According to H- H equation:
when PH = PKA + ㏒[conjugate base / weak acid]
when the lactate is the conjugate base and lactic acid is the weak acid
∴ PH = Pka +㏒[lactate] / [lactic acid]
so, by substitution:
4.49 = 3.86 + ㏒[lactate]/[lactic acid]
∴[lactate]/[lactic acid] = 4.3
∴ [lactic acid] / [lactate] = 1/4.3
= 0.23
The answer is D) <span>The Ka is small, and the equilibrium favors reactants.
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