Answer:
None of the above.
Explanation:
The limiting factor of this is the chlorine. You only have 2 moles of chlorine.
2Na + Cl_2 ==> 2*NaCl
The equation tells you that for every mol of Cl2 that you have, you require 2 moles of Na and you get 2 moles of NaCl
So what that means is that if you have 2 mols of Cl2, you need just 4 moles of Na, and you will get 4 mols of NaCl
Since 4 is not one of your choices, the answer is none of the above
0.02375 moles
Explanation:
Fe2O3+CO :. Fe+CO2
We know that
56*2+16*3=160 gm of Fe2O3 can give 56 gm of Fe
So
3.8 gm can give 1.33 gm of Fe
It is equal to 1.33/56 moles of Fe which is equal to 0.02375 moles
Ionic bonds are metals and non metals and covalent are non metals so C.
Answer:
Here's what I get
Explanation:
(i) Voltaic cell
A voltaic cell is a device that uses a chemical reaction to produce electrical energy.
(ii) Overall Cell Potential
The standard reduction potentials for the half-reactions are
<u>ℰ°/V
</u>
Cu²⁺ + 2e⁻ ⇌ Cu 0.34
Zn²⁺ + 2e⁻ ⇌ Zn -0.76
The half-reaction with the more positive potential is the reduction half-reaction. It is the reaction that occurs at the cathode.
The half-reaction with the more negative potential is the oxidation half-reaction. It is the reaction that occurs at the anode.
We reverse that half-reaction and subtract the voltages to get the cell reaction.
<u> ℰ°/V
</u>
Cathode: Cu²⁺ + 2e⁻ ⇌ Cu 0.34
<u>Anode: Zn ⇌ Zn²⁺ + 2e⁻ -0.76</u>
Cell: Zn + Cu²⁺ ⇌ Zn²⁺ + Cu 1.10

(iii) Diagram
The specific labels will depend on your textbook.
They are often as follows.
a. Electron flow
b. Voltmeter or lightbulb
c. Electron flow
d. Cathode or Cu
e. Cu²⁺(aq) and NO₃⁻(aq)
f. Salt bridge
g. Zn²⁺(aq) and NO₃⁻(aq)
h. Anode or Zn
The salt bridge enables ions to flow in the internal circuit and to maintain electrical neutrality in the two compartments.
It often consists of a saturated solution of KCl.
As Zn²⁺ ions form in the anode compartment, Cl⁻ ions move in to provide partners for them.
As Cu²⁺ ions are removed from the cathode compartment, K⁺ ions move in to replace them.