The class can break or if you put it in cold water it can cool down down fast.
Answer:
234.35 °C
Explanation:
Given data:
Volume of balloon = 125000 mL
Moles of oxygen = 3 mol
Pressure = 1 atm
Temperature = ?
Solution:
Formula:
PV = nRT
P = Pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Volume of balloon = 125000 mL × 1 L /1000 mL
Volume of balloon = 125 L
Now we will put the values:
Ideal gas constant = R = 0.0821 atm.L/mol.K
PV = nRT
T = PV/nR
T = 1 atm × 125 L/ 0.0821 atm.L/mol.K × 3 mol
T= 125 /0.2463 /K
T = 507.5 K
K to °C
507.5 K - 273.15 = 234.35 °C
Answer:
From point, 1 mole of water = molar mass of water =18g 20 moles of water = 18 g x 20 = 360g (iv) From point, 6.022 x 1023 molecules of water = 1 mole = 18g of water 1.2044 x 1025 molecules of water Therefore, points (ii) and (iv) represent 360 g of water.
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.