The value of delta H° for aniline = 6415 Kj/mol
<u><em>calculation</em></u>
Step 1: find heat
Q (heat) = C (specific heat capacity) x ΔT (change in temperature)
C= 4.25 kj/c°
ΔT = 69.8-29.5 = 40.3 c°
Q= 4.25 kj/c x 40.3 c = 171.28 kj
Step 2: find the moles of aniline
moles = mass/molar mass
= 2.49 g/ 93.13 g/mol =0.0267 moles
Step 3: find delta H
171.28 kj/0.0267 mol = 6415 kj/mol
since the reaction is exothermic delta H = 6415 Kj/mol
No it is not they are molecules connected making the leaf but they are made out of elements.
Answer:
Solid liquids , plasma gas
Explanation:
It retains its shape regardless of the shape of the container
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
5CO2 should be in the blank spot.
the reaction would be 7