The balanced half-cell equation for the reaction occurring at the anode is H2 ---> 2H(+) + 2e(-)
E<u>xplanation:</u>
- The balanced half-cell equation taking place at the anode is explained below
- The product produced in the reaction in the fuel cell is water.
- H2 ---> 2H(+) + 2e(-)
- In the above reaction, the oxidation state of hydrogen switches from 0 to +1.
- It is becoming oxidized by delivering two electrons at the anode.
- In the fuel cell, hydrogen molecules get oxidized to hydronium ions.Thus half-reaction is the oxidation reaction.
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Answer:
1.096g
Explanation:
You must know the atomic mass of Hydrogen, Fluorine, and Sodium before you can start:
Hydrogen: 1.008g/mol
Fluorine: 18.99g/mol
Sodium: 22.98g/mol
Next, find the composition percentage of NaF
22.98 + 18.99 = 41.97
Fluorine is 18.99/41.97 =45.25%
Sodium is 100-45.25 = 54.75%
Ultimately we want to know about HF so find how much F is in 2.3g: 2.3 * 0.4525 = 1.041g
Find comp. percentage of HF
18.99+1.008 = 19.998; H/total F/total
Hydrogen 5.041%
Fluorine 94.959%
Laws of conservation of say we have 1.041g of fluorine in our HF. We know 1.041 is 94.959% of the mass of HF so do some simple math to find the remaining: 1.041/0.94959 = 1.096g
Answer:
D) 21.62 B
Explanation:
I took a quiz and it says that D is correct. Sorry I couldn't be of more assistance.
Answer:
2 NO (g) → N2 (g) + O2 (g)
2 NOCl (g) → 2 NO (g) + Cl2 (g)
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2NOCl (g) ⟶ N2 (g) + O2 (g) + Cl2 (g)
ΔH = [90.3 kJ x 2 x -1] + [-38.6 kJ x -1 x 2] = -103.4 kJ
The ΔH for the reaction is -103.4 kJ
It should be a chemical change
