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Tema [17]
3 years ago
14

A TLC plate showed 2 spots with Rf values of 0.25 and 0.26. The plate was removed from the developing chamber, the residual solv

ent was allowed to evaporate from the plate, and then the plate was returned to the developing chamber. What would you expect to see after the second development was complete
Chemistry
1 answer:
Katena32 [7]3 years ago
4 0

Answer:

See explanation

Explanation:

TLC is a chromatographic method in which the solute is spotted on a plate and the plate is placed in an air tight chamber containing a solvent. The solvent is maintained below the level of the spot. The capillary movement of the solvent through the plate achieves the required separation.

If two spots have Rf values of 0.25 and 0.26 respectively and then the plate was removed from the developing chamber, subsequently, the residual solvent was allowed to evaporate from the plate, and then the plate was returned to the developing chamber.

It will be observed after the second development is complete that the new Rf values will be 0.50 and 0.52 respectively. It will just be as though the second chromatogram picked up from where the first chromatogram stopped.

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The complete question is:

An aqueous solution of ammonium sulfate is allowed to react with an aqueous solution of calcium nitrate.

The net ionic equation contains which of the following species (when balanced in standard form)?

a. 2NO3-(aq)

b. Ca2+(aq)

Answer:

b. Ca2+(aq)

Ca2+ (aq) + SO4^2-(aq) --------------> CaSO4(s)

Explanation:

The overall ionic equation is:

Ca2+(aq) + 2NO3-(aq) + 2NH4+(aq) + SO4^2-(aq) ---------------> CaSO4(s) + 2NH4NO3(aq)

The NO3- and NH4+ are spectator ions as they do not participate in the formation of the precipitate CaSO4.

The net ionic equation is:

Ca2+ (aq) + SO4^2-(aq) --------------> CaSO4(s)

The spectator ions form the soluble ammonium trioxonitrate V

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3 years ago
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235 92 U >144 56 Ba + 3 1 on
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Answer:?

Explanation:

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2 years ago
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For the galvanic (voltaic) cell Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s) (E°= 0.77 V at 25°C), what is [Fe2+] if [Mn2+] = 0.040 M and
avanturin [10]

Answer:

0.01836 M

Explanation:

Again the reaction equation is;

Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)

E°cell= 0.77 V

Ecell= 0.78 V

[Mn2+] = 0.040 M

[Fe2+] = the unknown

n=2

From Nernst's equation;

Ecell= E°cell- 0.0592/n log Q

0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]

0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]

0.01/ -0.0296= log [Fe2+] /[0.040]

-0.3378= log [Fe2+] /[0.040]

Antilog(-0.3378) = [Fe2+] /[0.040]

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