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LekaFEV [45]
4 years ago
8

In Becquerel’s initial experiment, in which he exposed uranium salt crystals to sunlight, the photographic paper became exposed

even though it was shaded from the sun by a dark cloth. At first, how did Becquerel explain this result?
Chemistry
1 answer:
marshall27 [118]4 years ago
3 0
<span>In Becquerel’s initial experiment, in which he exposed uranium salt crystals to sunlight, the photographic paper became exposed even though it was shaded from the sun by a dark cloth. The explanation that he gave was that the crystals were emitting light without any external light source to stimulate or excite them. He called this phenomenon as "radioactivity'". Further, more and more experiments were done to analyse actually what happened between the crystals and the light</span>
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Calculate E o , E, and ΔG for the following cell reactions (a) Mg(s) + Sn2+(aq) ⇌ Mg2+(aq) + Sn(s) where [Mg2+] = 0.025 M and [S
Rudik [331]

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

<u>Explanation:</u>

Mg(s) + Sn²⁺(aq) ⇌ Mg²⁺(aq) + Sn(s)

[Mg2+] = 0.025 M

[Sn2+] = 0.040 M

First we need the standard reduction potentials:

. . . . . . . . . . . . . . . . . E°(V)

Mg²⁺ + 2 e⁻ ⇌ Mg(s). . .−2.372

Sn²⁺ + 2 e⁻ ⇌ Sn(s) . . . −0.13

Take the more negative (or less positive in other cases) one, and write it as an oxidation:

Mg(s) ⇌ Mg²⁺ + 2 e⁻. . .+2.372 V

Combine them,

Mg(s) + Sn²⁺ ⇌ Mg²⁺ + Sn(s)

E°(cell) = +2.372 – 0.13 V = 2.24 V

To get the cell potential under the conditions given, use the Nernst Equation:

E(cell) = E°(cell) – [(0.059)/n]•logQ = 2.24 V – 0.0295 V • log [Mg²⁺]/[Sn²⁺]

Note that the solids don't appear in Q, only the concs. of the dissolved ions.

E(cell) = 2.24 V – 0.0295 V X log (0.025)/(0.040)

          = 2.24 + 0.006 V ≈ 2.246 V

The concentration ratio in Q (Sn²⁺ and Mg²⁺) is too close to 1 to shift E(cell) significantly from E°(cell) given the precision I have for the Sn reduction potential.

∆G = –nFE(cell) = –2(96.485 kJ/mol•V)(2.246 V) = –433 kJ/mol

E⁰(cell) = 2.24V

E(cell) = 2.246V

∆G = -433 KJ/mol

4 0
4 years ago
How are electrons excited and what happens when the electrons relax
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3 years ago
The vapor pressure of a substance describes how readily molecules at the surface of the substance enter the gaseous phase. At th
geniusboy [140]

Answer:

84.75°C is the boiling point of water at an elevation of 7000 meter.

Explanation:

Rate of change of pressure = 19.8 mmHg/1000 ft

1 foot = \frac{1}{3.28} meter

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Elevation      Pressure

     0 m       760 mmHg

1000 m       695 mmHg

2000 m       630 mmHg

Pressure drop at the elevation of 7000 m: 7000\times 0.065 mmHg=455 mmHg

Pressure at 7000 m = 760 mmHg - 455mmHg = 305 mmHg

The boiling point of water decreases 0.05°C for every 1 mmHg drop in atmospheric pressure.

At 7000 meter elevation the boiling of water will be :

0.05^oC\times 305=15.25^oC

Boiling point of water at 7000 meter elevation :

100.0^oC-15.25^oC =84.75^oC

84.75°C is the boiling point of water at an elevation of 7000 meter.

8 0
4 years ago
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