Answer:
The correct option is: salinity increases
Explanation:
Ocean is a large water body that is a division of the World Ocean. An ocean is divided into various vertical zones based upon the physical and biological conditions.
The halocline zone is characterized by a <u>rapid increase in the salinity with the increase in the depth.</u> Thus resulting in decrease in the temperature of maximum density with the increase in salinity.
With the Bronsted-Lowry theory of acids and bases, acids donate protons to other substances. A monoprotic acid is an an acid that can only supply one proton (hydrogen). A triprotic acid is an an acid that can supply up to three protons (hydrogens). A triprotic acid get get its hydrogen ions by receiving them from a base.
<h3>
Answer: <em>
pH=2.25 </em></h3>
Explanation:
monochloroacetic acid also means: chloroacetic acid
pKa of monochloroacetic acid= 1.4 x 10^-3 (I believe this should have been given in the problem or perhaps in the textbook)
Formula: pH= pKa + log ( some number in M)
pH= -log (1.4 x 10^-3) + log (0.25M)= 2.85 + -0.602= 2.25
pH= 2.25
Answer:
Cdiasuu alprguta bonrradse??
Answer:
0.0159m
Explanation:
9 M
Explanation:
Lead(II) chloride,
PbCl
2
, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.
Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,
K
sp
, will be established between the solid lead(II) chloride and the dissolved ions.
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
Now, the molar solubility of the compound,
s
, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.
Notice that every mole of lead(II) chloride will produce
1
mole of lead(II) cations and
2
moles of chloride anions. Use an ICE table to find the molar solubility of the solid
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
I
−
0
0
C
x
−
(+s)
(
+
2
s
)
E
x
−
s
2
s
By definition, the solubility product constant will be equal to
K
sp
=
[
Pb
2
+
]
⋅
[
Cl
−
]
2
K
sp
=
s
⋅
(
2
s
)
2
=
s
3
This means that the molar solubility of lead(II) chloride will be
4
s
3
=
1.6
⋅
10
−
5
⇒
s
= √
1.6
4
⋅
10
−
5 =
0.0159 M
