Answer:
The equillibrium constant Kc = 11.2233
Explanation:
Step 1:
aA + bB ⇔ cC + dD
with a, b, c and d = coefficients
Kc = equillibrium constant =( [C]^c [D]^d ) / ( [A]^a [B]^b)
Concentration at time t
[A] = 0.300 M
[B] =1.10 M
[C] = 0.450 M
Change :
A: -x
B: -2x
C: -x
The following reaction occurs and equillibrium is established
A + 2B ⇔ C
[A] = 0.110M
[B] = ?
[C] = 0.640 M
For A we see that after change: 0.3 -x = 0.11
Then for B we have 1.1 - 2x = ? ⇒ 1.1 -2 *0.19 = 0.72
This gives us for the equillibrium constant Kc = [C] / [A][B] ²
Kc = 0.64 / (0.11) * (0.72)² = 11.2233
Answer:
Aquatic plants have aur filled tissues to make them easy to float.....
Explanation:
It provides buoyancy
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
