Answer:
V₂ = 106.5 mL
Explanation:
Given data:
Initial volume =200 mL
Initial pressure = 2 atm
Initial temperature = 35 °C (35 +273 = 308 K)
Final temperature = 55°C (55+273 = 328 K)
Final volume = ?
Final pressure = 4 atm
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
Solution:
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2 atm ×200 mL × 328 K / 308 K ×4 atm
V₂ = 131200 atm .mL. K / 1232 K.atm
V₂ = 106.5 mL
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.
a. Hydrogen
Hydrogen has a line at 410 nm.
Mercury has a line at 405 nm.
Sodium and neon have no lines near 412 nm.
Answer:
(a) 
(b)
(c) 
Explanation:
Hello,
(a) In this case, with the given formula we easily compute the mass of gold contained in the sovereign as shown below:
(b) Now, by knowing the density of gold and copper, 19.32 and 8.94 g/cm³ respectively, we compute each volume, by also knowing that the rest of the coin contains copper:
(c) Finally, the volume is computed by dividing the total mass over the total volume containing both gold and copper:
Best regards.
