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BARSIC [14]
4 years ago
13

For each pair below, select the sample that contains the largest number of moles .

Chemistry
2 answers:
aliya0001 [1]4 years ago
6 0

Answer:

Pair A: B

Pair B: A

Pair C: B

LenKa [72]4 years ago
5 0

If you need the answer to this problem you need to post more information about the problem.


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Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v​
bogdanovich [222]

Answer:

Depending on the E^\circ value of \rm Ag^{+} + e^{-} \to Ag\; (s), the cell potential would be:

  • 0.55\; \rm V, using data from this particular question; or
  • approximately 0.46\; \rm V, using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

  • The conversion between \rm Ag\; (s) and \rm Ag^{+} ions, \rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s), and
  • The conversion between \rm Cu\; (s) and \rm Cu^{2+} ions, \rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s).

Note that the standard reduction potential of \rm Ag^{+} ions to \rm Ag\; (s) is higher than that of \rm Cu^{2+} ions to \rm Cu\; (s). Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if \rm Ag^{+} ions are reduced while \rm Cu\; (s) is oxidized.

Therefore:

  • The reduction reaction at the cathode will be: \rm Ag^{+} + e^{-} \to Ag\; (s). The standard cell potential of this reaction (according to this question) is E(\text{cathode}) = 0.89\; \rm V. According to the 2012 CRC handbook, that value will be approximately 0.79\; \rm V.
  • The oxidation at the anode will be: \rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}. According to this question, this reaction in the opposite direction (\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)) has an electrode potential of 0.34\; \rm V. When that reaction is inverted, the electrode potential will also be inverted. Therefore, E(\text{anode}) = -0.34\; \rm V.

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}.

Using data from the 1985 and 2012 CRC Handbook:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}.

5 0
4 years ago
Please help... will give brainiest
Dvinal [7]

Answer:

The answer will be :

In year 1869

Russian scientist named Dmitri Mendeleev

Chart called Mendeeleves periodic table

increasing atomic masses

physical and chemical properties.

Hope this Helps

Best of luck !!

7 0
3 years ago
Did Adam Jackson win by
devlian [24]

Answer:

by a number

Explanation:

4 0
3 years ago
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Calculate the number of grams of carbon found in a 5 mole sample of carbon.
Ksenya-84 [330]

Answer:

60g

Explanation:

1mol of carbon has 12g

3 0
3 years ago
What volume of a 2.5 M NaOH solution is required to make 1 liter of a 0.75 M NaOH
defon
V1M1 = V2M2 

<span>V1 × 2.5 = 1 × 0.75,
so     V1 = 0.75/2.5
              = 0.3 </span>
6 0
3 years ago
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