The elements that fill the d orbitals on the periodic table are called transition metals.
Answer:
The given reaction is a combustion reaction of benzene,
C
6
H
6
. From its balanced chemical equation,
2
C
6
H
6
+
15
O
2
→
12
C
O
2
+
6
H
2
O
,
the mass of carbon dioxide
(
C
O
2
)
produced from 20 grams (g) of
C
6
H
6
is determined through the molar mass of the two compounds, given by,
M
M
C
O
2
=
44.01
g
/
m
o
l
M
M
C
6
H
6
=
78.11
g
/
m
o
l
and their mole ratio:
12
m
o
l
C
O
2
2
m
o
l
C
6
H
6
→
6
m
o
l
C
O
2
1
m
o
l
C
6
H
6
With this,
m
a
s
s
o
f
C
O
2
=
(
20
g
C
6
H
6
)
(
1
m
o
l
C
6
H
6
78.11
g
C
6
H
6
)
(
6
m
o
l
C
O
2
1
m
o
l
C
6
H
6
)
(
44.01
g
C
O
2
1
m
o
l
C
O
2
)
=
(
20
)
(
6
)
(
44.01
)
g
C
O
2
78.11
=
5281.2
g
C
O
2
78.11
m
a
s
s
o
f
C
O
2
=
67.6
g
C
O
2
Therefore, the mass in grams of
C
O
2
formed from 20 grams of
C
6
H
6
is
67.6
g
C
O
2
.
it is a problem of app
Answer : The 
for this reaction is, -88780 J/mole.
Solution :
The balanced cell reaction will be,
Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half oxidation-reduction reaction will be :
Oxidation : 
Reduction : 
Now we have to calculate the Gibbs free energy.
Formula used :
where,
= Gibbs free energy = ?
n = number of electrons to balance the reaction = 2
F = Faraday constant = 96500 C/mole
= standard e.m.f of cell = 0.46 V
Now put all the given values in this formula, we get the Gibbs free energy.
Therefore, the for this reaction is, -88780 J/mole.
Answer:
increase
Explanation:
Let's suppose we have a sample of air in a closed container. We heat the container and we want to predict what would happen to the pressure.
According to Gay-Lussac's law, the pressure of a gas is directly proportional to its absolute temperature.
Thus, if we increased the temperature of the air by heating it, its pressure would increase.
If a sample of air in a closed container was heated, the total pressure of the air would increase.
E. co and n2Effusion is the process where gas escapes through a hole. Gases with a lower molecular mass effuse more speedy than gases with a higher molecular mass. R<span>elative rates of effusion is related to the molecular mass.
a) M(N</span>₂)/M(O₂) = 28/32 = 0,875
b) M(N₂O)/M(NO₂) = 44/46 = 0,956
c) M(CO)/M(CO₂) = 28/44 = 0,636
d) M(NO₂)/M(N₂O₂) = 44/58= 0,758
e) M(CO)/M(N₂) = 28/28 = 1, <span>CO and N</span>₂ <span>have iexact molecular masses and will effuse at nearly identical rates.</span>
