Your answer would be, I believe, The same number of decimal places as the least precise values!!!!
Hope that helps!!!
Answer: This is a typical acid/base equilibrium problem, that involves the use of logarithms.
Explanation:We assume that both nitric acid and hydrochloric acid dissociate to give stoichiometric
H
3
O
+
.
Moles of nitric acid:
26.0
×
10
−
3
⋅
L
×
8.00
⋅
m
o
l
⋅
L
−
1
=
0.208
⋅
m
o
l
H
N
O
3
(
a
q
)
.
And, moles of hydrochloric acid:
88.0
×
10
−
3
⋅
L
×
5.00
⋅
m
o
l
⋅
L
−
1
=
0.440
⋅
m
o
l
H
C
l
(
a
q
)
.
This molar quantity is diluted to
1.00
L
. Concentration in moles/Litre =
(
0.208
+
0.440
)
⋅
m
o
l
1
L
=
0.648
⋅
m
o
l
⋅
L
−
1
.
Now we know that water undergoes autoprotolysis:
H
2
O
(
l
)
⇌
H
+
+
O
H
−
. This is another equilibrium reaction, and the ion product
[
H
+
]
[
O
H
−
]
=
K
w
. This constant,
K
w
=
10
−
14
at
298
K
.
So
[
H
+
]
=
0.648
⋅
m
o
l
⋅
L
−
1
;
[
O
H
−
]
=
K
w
[
H
+
]
=
10
−
14
0.648
=
?
?
p
H
=
−
log
10
[
H
+
]
=
−
log
10
(
0.648
)
=
?
?
Alternatively, we know further that
p
H
+
p
O
H
=
14
. Once you have
p
H
,
p
O
H
is easy to find. Take the antilogarithm of this to get
[
O
H
−
]
.
Answer link
Answer:
See explaination
Explanation:
The Cys3-cys97 and cys21-cys142 disulfides restrict the unfolded state of lysozyme enzyme to a class of more compact structures with a less exposed hydrophobic surface, compared to the unfolded states of reduced/non-crosslinked lysozyme. there are 2 major factors which lead to the stabilization of lysozyme due to disulfide bonds-
1- increase in the loop size due to the formation of disulfide bonds that leads to an increase in the even entropic effect.
2- the region formed should be flexible. the strain energy due to the formation of the disulfide bond is lower.
cys21-cys142 has a higher Tm than the cys3-cys97 because it involves flexible parts of the molecule. 21 and 142 residues are located on opposite sides of the active-site cleft where significant hinge-bending motion is seen. this introduces minimal strain in the protein.
Answer:
Explanation:
Molar heat capacity at constant volume Cv of a gas = n x .5 R where n is degree of freedom of the gas molecules
CO₂ is a linear molecule , so number of degree of freedom = 3 + 2 = 5
3 is translational and 2 is rotational degree of freedom . There is no vibrational degree of freedom given .
So Cv = 5 / 2 R
= 2.5 R .
Answer: try to understand coz the question is not valid
Explanation: Explain the relationship between forward and reverse reactions at equilibrium and predict how changing the amount of a reactant or product (creating a stress) will affect that relationship.For example (select one from each underlined section)If the amount of (reactant or product) increases, the rate of the (forward or reverse)reaction will (increase or decrease)to reach a new equilibrium. If the amount of (reactant or product) decreases, the rate of the (forward or reverse)reaction will (increase or decrease)to reach a new equilibrium. Procedure: Access the virtual lab and complete the inquiry experiment
