<em>FeCl2</em><em> </em><em>+</em><em> </em><em>Na2CO3</em><em> </em><em>=</em><em> </em><em>FeCO3</em><em> </em><em>+</em><em> </em><em>2</em><em>N</em><em>a</em><em>C</em><em>l</em>
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
Using the significant figure it would be 27.3
Answer:
C. move left
Explanation:
The object will move towards the left direction due to the unbalanced forces that are acting on it.
The resultant force on the object will be 1N in the left direction
- The resultant force on a body is that singular force that will have the same effect as the different forces that acts on a body
- When forces acts in opposite directions, they are subtracted
- The object will move in the direction of the one with the greater force
So;
Resultant force = 26N - 25N = 1N
The body moves 1N to the left
You need the solubility of the LiCl in water at 20°C.
The solubilities are shown in tables. Many books contain those tables. You have to make sure that the solubility is shown in the same solvent and at the same temperature that you are going to prepare the solution.
In this case the solubility of LiCl in water at 20°C is 83.05 g of LiCl in 100 g of H2O. Check if your book and your teacher work with the same value.
Using 8.05 g LiCl / 100 g of water you get:
300 g of water * 83.05 g LiCl / 100 g of water = 249.15 g of LiCl.
Answer: 249.15 g of LiCl