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ValentinkaMS [17]
3 years ago
15

What are the subscripts for the molecular formula of C2H3

Chemistry
1 answer:
jarptica [38.1K]3 years ago
7 0
The subscript are the numbers under
C(2)H(3)

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A graduated cylinder

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What is true about an atom in an excited state? Select all that apply:
AnnyKZ [126]

Explanation:

What is true about an atom in an excited state? Select all that apply:

A. It is less likely to enter chemical reactions.

B. It is more likely to enter chemical reactions.

C. It has less energy than a ground state atom.

D. It has more energy than a ground state atom.

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Why is carbon dioxide used in fire extinguisher?
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3 years ago
Is the electrochemical cell spontaneous or not spontaneous as written at 25 ∘C ? spontaneous not spontaneous Calculate the poten
givi [52]

Answer:

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

Explanation:

Let's consider the oxidation and reduction half-reactions and the global reaction.

Anode (oxidation): Sn²⁺(0.0023 M) ⇒ Sn⁴⁺(0.13 M) + 2 e⁻

Cathode (reduction): 2 Fe³⁺(0.11 M) + 2 e⁻ ⇒ 2 Fe²⁺(0.0037 M)

Global reaction: Sn²⁺(0.0023 M) + 2 Fe³⁺(0.11 M) ⇒ Sn⁴⁺(0.13 M) + 2 Fe²⁺(0.0037 M)

The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red,cat - E°red,an

E° = 0.771 V - 0.154 V = 0.617 V

The Nernst equation allows us to calculate the cell potential (E) under the given conditions.

E=E\° -\frac{0.05916}{n} logQ\\E=E\° -\frac{0.05916}{n} log\frac{[Sn^{+4}].[Fe^{2+}]}{[Sn^{2+}].[Fe^{3+} ]} \\E=0.617V-\frac{0.05916}{2} log\frac{(0.13).(0.0037)}{(0.0023).(0.11)} \\E=0.609V

The cell potential is 0.609 V. Given E > 0 the electrochemical cell is spontaneous as written.

4 0
3 years ago
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