Answer:
(a) The anode electrode which comprises the zinc electrode being placed in a water solution with low oxygen concentration.
(b) Cathodic reaction is: 
⇒ 
Anodic reaction is: 
⇒
Explanation:
In the given problem, we have an oxygen-concentration cell consisting of two zinc electrodes. One is immersed in a water solution with a low oxygen concentration and the other in a water solution with a high oxygen concentration. The zinc electrodes are connected by an external copper wire.
(a) Which electrode will corrode?
The electrode that will corrode is the anode electrode which comprises the zinc electrode being placed in a water solution with low oxygen concentration.
(b) Write half-cell reactions for the anodic reaction and the cathodic reaction.
Cathodic reaction is: ⇒
Anodic reaction is: ⇒
Answer:
you can correct it by adding an acid to make it neutral or less basic
The one you have selected is correct. CO is a compound because it contains more than one element.
The expectancy of how long a product will last divided by two!
Answer is: C₃H₃N₃O₃.
Chemical reaction: CₓHₓNₓOₓ + O₂ → aCO₂ + x/2H₂ + x/2N₂.
m(CₐHₓNₓ) = 5,214 g.
m(CO₂) = 5,34 g.
m(H₂) = 1,09 g.
m(N₂) = 1,70 g.
n(CO₂) = n(C) = 5,34 g ÷ 44 g/mol = 0,121 mol.
n(H₂O) = 1,09 g ÷18 g/mol = 0,06 mol.
n(H) = 2 · 0,0605 mol = 0,121 mol.
n(N₂) = 1,7 g ÷ 28 g/mol = 0,0607 mol.
n(N) = 0,0607 mol · 2 = 0,121 mol.
n(C) : n(H) : n(N) = 0,121 mol : 0,121 mol : 0,121 mol /: 0,121
n(C) : n(H) : n(N) = 1 : 1 : 1.
M(CHN) = 27 g/mol.
m(O₂) = 8,13 g - 5,214 g = 2,914 g.
n(O₂) = 2,914 g ÷ 32 g/mol = 0,09 mol.
n(CₓHₓNₓOₓ) = 5,214 g ÷ 129,1 g/mol = 0,0404 mol.
n(CₓHₓNₓOₓ) : n(CO₂) = 1 : 3.
