Answer:
The second one - rich/poor
Explanation:
However it would be in reverse order so when filling in the blank, poor would come first then rich
but if you are not able to change it then you can just leave it like that.
I hope this helped☺
Answer:
Explanation:
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
destroyerplayz20062
13 hours ago
Chemistry
High School
Predict the Eº values for all (6) combinations of the following: Cu(s) and Cu(NO3)2(aq) Fe(s) and Fe(NO3)3(aq) Zn(s) and Zn(NO3)2(aq) Pb(s) and Pb(NO3)2(aq)
Answer:
C. 0.35cm
Explanation:
The length indicated by the arrow along the ruler should recorded be recorded as "0.35cm".
This is correct because when counting the measurement on the ruler, the first line on the ruler is 0.1cm, the second line is 0.2cm, and so on. The spaces between each line is 0.05cm. So, the arrow is pointing on the space between 0.3cm and 0.4cm.
Therefore, 0.3cm + 0.05cm = 0.35cm.(answer).
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