Answer:
4. 1.18 mol·L⁻¹
14. See below.
Explanation:
4. Dilution calculation
V₁c₁ = V₂c₂
Data:
V₁ = 200 mL; c₁ = 5.6 mol·L⁻¹
V₂ = 950 mL; c₂ = ?
Calculation:
c₂ = c₁ × V₁/V₂
c₂ = 5.6 mol·L⁻¹ × (200/950) = 1.18 mol·L⁻¹
The new concentration is 1.18 mol·L⁻¹
.
14. Boyle's Law graphs
We can write Boyle's Law as
pV = k or p = k/V or V= k/p
p and V are inversely related.
(a) As pressure increases, volume decreases. Thus, a graph of V vs p is a hyperbola.
(b) p = k/V =k(1/V)
1/V = (1/k)p
y = m x + 0
A graph of 1/V vs p is a straight line.
Answer:
You have to apply the base to an acid burn and vice versa.
Explanation:
Answer:
yes I am the best person on earth
Explanation:
im awesome
E
θ
Cell
=
+
2.115
l
V
Cathode
Mg
2
+
/
Mg
Anode
Ni
2
+
/
Ni
Explanation:
Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]
Mg
2
+
(
a
q
)
+
2
l
e
−
→
Mg
(
s
)
−
E
θ
=
−
2.372
l
V
Ni
2
+
(
a
q
)
+
2
l
e
−
→
Ni
(
s
)
−
E
θ
=
−
0.257
l
V
The standard reduction potential
E
θ
resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (
298
l
K
,
1.00
l
kPa
) is defined as
0
l
V
for reference. [2]
A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.
Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher
E
θ
and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower
E
θ
will experience oxidation and act the anode.
E
θ
(
Ni
2
+
/
Ni
)
>
E
θ
(
Mg
2
+
/
Mg
)
Therefore in this galvanic cell, the
Ni
2
+
/
Ni
half-cell will experience reduction and act as the cathode and the
Mg
2
+
/
Mg
the anode.
The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:
E
θ
cell
=
E
θ
(
Cathode
)
−
E
θ
(
Anode
)
E
θ
cell
=
−
0.257
−
(
−
2.372
)
E
θ
cell
=
+
2.115
Indicating that connecting the two cells will generate a potential difference of
+
2.115
l
V
across the two cells.
