Answer:
ΔH3 = -110.5 kJ.
Explanation:
Hello!
In this case, by using the Hess Law, we can manipulate the given equation to obtain the combustion of C to CO as shown below:
C(s) + 1/2O2(g) --> CO(g)
Thus, by letting the first reaction to be unchanged:
C(s) + O2(g)--> CO2 (g) ; ΔH1 = -393.5 kJ
And the second one inverted:
CO2(g) --> CO(g) + 1/2O2(g) ; ΔH2= 283.0kJ
If we add them, we obtain:
C(s) + O2(g) + CO2(g) --> CO(g) + CO2 (g) + 1/2O2(g)
Whereas CO2 can be cancelled out and O2 subtracted:
C(s) + 1/2O2(g) --> CO(g)
Therefore, the required enthalpy of reaction is:
ΔH3 = -393.5 kJ + 283.0kJ
ΔH3 = -110.5 kJ
Best regards!
Answer:
What is the charge of CA?
2+
Table of Common Element Charges
Number Element Charge
17 chlorine 1-
18 argon 0
19 potassium 1+
20 calcium 2+
88 more rows•Dec 24, 2018
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
Answer:
The solubility of the mineral compound X in the water sample is 0.0189 g/mL.
Explanation:
Step 1: Given data
The volume of water sample = 46.0 mL.
The weight of the mineral compound X after evaporation, drying, and washing = 0.87 g.
Step 2: Calculate the solubility of X in water
46.00 mL of water sample contains 0.87 g of the mineral compound X.
To calulate how many grams of the mineral compound 1.0 mL of water sample contains:
0.87 g/46.0 mL = 0.0189 g.
This means the solubility of the mineral compound X in the water sample is 0.0189 g/mL.
