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Chemistry
Chemical Reactions
Redox Reactions
Fe3+ and CN-
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What about them?
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Consider a cell at 255 K:
line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe Given the standard reduction potentials calculate the cell
Top answer:
To calculate the cell potential after the reaction, we can use the Nernst equation, which takes into
Read more.
Consider a cell at 255 K:
line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe Given the standard reduction potentials calculate the cell
Top answer:
To calculate the cell potential after the reaction, you need to use the Nernst equation, which takes
Read more.
Consider a cell at 255 K:
line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe Given the standard reduction potentials calculate the cell
Top answer:
To calculate the cell potential in this scenario, you need to use the Nernst equation, which relates
Read more.
Consider a cell at 255 K:
line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe Given the standard reduction potentials calculate the cell
Top answer:
To calculate the cell potential after the reaction, you need to use the Nernst equation, which
Read more.
Consider a cell at 255 K:
line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe Given the standard reduction potentials calculate the cell
Top answer:
To calculate the cell potential after the reaction, we need to use the Nernst equation: Ecell =
Read more.
Consider a cell at 255 K:
line notation Pb-Pb2+(1.27M)--Fe3+(2.29M)-Fe Given the standard reduction potentials calculate the cell
Top answer:
To calculate the cell potential, you need to use the Nernst equation, which takes into account the
Read more.
Hi I'm stuck on these questions can you please help me
A soil sample contains 0.2234 g of Fe3+ per 100g. Fe3+ is precipitated
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Sarah, I read this last night. Truth be know, I really don't understand the problem.
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Kc is 1.67 × 1020 at 25°C for the formation of iron(III) oxalate complex ion:
Fe3+(aq) + 3 C2O42-(aq) [Fe(C2O4)3]3-(aq). If
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867687
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balance the net ionic equation for the overall reaction.
Pt | Fe2+(aq), Fe3+(aq) || I-(aq) | AgI(s) | Ag(s) a) Fe3+(aq) + Ag(s) +
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no
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Fe2+ + MnO-4 --> Fe3+ + Mn2+
Fe2+ = +2 Mn = +7 O-4 = 4(-2) = -8 Fe3+ = 3+ Mn2+ = 2+ ok so next I gave 8H to MnO-4 and 4 H20 to
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The simple answer is that Mn changed frm +7 to +2 so it must have gained 5e. MnO4^- + 8H^+ + 5e ==>
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