Answer & Explanation:
(a)

reducing agent = Fe²⁺
Oxidizing agent = NO₃⁻
oxidation
Fe²⁺ ⇒ Fe(OH)₃
reduction
NO₃⁻ ⇒ N₂
Oxidation Half Reaction
(<em>redox reactions are balanced by adding appropriate H⁺ and H₂O atoms)</em>
Fe²⁺ ⇒ Fe(OH)₃
Balance O atoms
Fe²⁺ + 3H₂O ⇒ Fe(OH)₃
Balance H atoms
Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺
balance Charge
Fe² + 3H₂0 ⇒ Fe(OH)₃ + 3H⁺ + e⁻..............(1)
reduction Half Reaction
NO₃⁻ ⇒ N₂
Balance N atoms
2NO₃⁻ ⇒N₂
Balance O atoms by adding appropriate H₂O
2NO₃⁻ ⇒ N₂ + 6H₂O
Balance H atoms
2NO₃⁻ + 12H⁺ ⇒ N₂ + 6H₂O
Balance Charge
2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O.................(2)
Combine Equation (1) and (2)
(1) × 10: 10Fe² + 30H₂0 ⇒ 10Fe(OH)₃ + 30H⁺ + 10e⁻
(2) × 1: 2NO₃⁻ + 12H⁺ + 10e⁻⇒ N₂ + 6H₂O
(1) + (2): 10Fe² + <u><em>30H₂0</em></u> + 2NO₃⁻ + <u><em>12H⁺</em></u> + <u><em>10e⁻</em></u> ⇒10Fe(OH)₃ + <u><em>30H⁺</em></u><u><em> </em></u>+ <em><u>10e⁻</u></em> +
N₂ + <u><em>6H₂O</em></u>
10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂
this is the balanced reaction
REDUCTION POTENTIAL
10Fe²⁺(aq) + 10e⁻ ⇒ 10Fe(OH)₃(aq) E°ox = 10(-0.44) = -4.4V
2NO₃⁻(aq) - 2e⁻ ⇌ N₂(g) + 18H⁺ E°red = 2(+0.80) = +1.6
10Fe² + 24H₂0 + 2NO₃⁻ ⇒ 10Fe(OH)₃ + 18H⁺ + N₂ E°cell = -2.8V
E°cell = E°red + E°ox
There is 12 inches in 1 foot.
Divide 16 with 12 to get feet
16/12 = ~1.33 (rounded)
1.33 is your answer
hope this helps
Answer: The rate of the loss of
is 0.52M/s
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate of disappearance of
=![-\frac{1d[O_3]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_3%5D%7D%7B2dt%7D)
Rate of formation of
=![+\frac{1d[O_2]}{3dt}](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BO_2%5D%7D%7B3dt%7D)
![-\frac{1d[O_3]}{2dt}=+\frac{1d[O_2]}{3dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_3%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BO_2%5D%7D%7B3dt%7D)
Rate of formation of
= 
Thus Rate of disappearance of
=![\frac{2d[O_2]}{3dt}=\frac{2}{3}\times 7.78\times 10^{-1}M/s=0.52M/s](https://tex.z-dn.net/?f=%5Cfrac%7B2d%5BO_2%5D%7D%7B3dt%7D%3D%5Cfrac%7B2%7D%7B3%7D%5Ctimes%207.78%5Ctimes%2010%5E%7B-1%7DM%2Fs%3D0.52M%2Fs)
Answer:
55.4324℃
Explanation:
Q=cm△t
50J = 0.902J/(g·℃) x 1g x △t
△t =55.4324℃