H+ ions
That's the answer to your question
<h2><u>Answer:</u></h2>
The correct answer is A) 1.04 mol Cu
{65.8 g / 63.55 g/mol}
= 1.04 mol Cu
Explanation:
In 63.55 g of copper metal there are 1 m
o
l of C
u atoms. By dividing the mass of Cu and molar mass, we can easily get the number of moles.
Answer:
The specific heat of water is 4.18 J/g C.
Explanation:
q
=
m
C
s
Δ
T
Never forget that!
2200
=
m
⋅
4.18
J
g
⋅
°
C
⋅
66
°
C
∴
m
≈
8.0
g
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.
Answer: The mass of the sample will be 1417.7 grams.
Explanation:
We are given:
This means that 1 mole of NaCl has an enthalpy of fusion of 30.2 kJ
1 mole of NaCl has a mass of 58.44 grams.
So, 30.2 kJ of heat is require for a mass 58.44 grams of NaCl
So, 732.6 kJ of heat will be required for = 
= 1417.65 grams of NaCl.
Hence, the mass of NaCl sample will be 1417.7 grams.
