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Semmy [17]
3 years ago
10

Which of the following accurately describes semiconductor diodes? A. Unlike junction diodes, point-contact diodes are enclosed i

n a suitable casing and have terminals for connecting them to a circuit. B. Unlike point-contact diodes, junction diodes utilize a point of metal wire in contact with a single wafer of P-type or N-type material. C. Junction diodes are preferred over point-contact diodes for most purposes. D. Point-contact diodes are more likely to be used as rectifiers than junction diodes.
Chemistry
2 answers:
Anna007 [38]3 years ago
5 0
The correct answer is letter A: <span>Unlike junction diodes, point-contact diodes are enclosed in a suitable casing and have terminals for connecting them to a circuit.</span>
monitta3 years ago
4 0
I think the answer is D
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Answer:

Redox type

Explanation:

The reaction is:

2Cr +  3Fe(NO₃)₂ → 2Fe + 2Cr(NO₃)₃

2 moles of chromium can react to 3 moles of iron (II) nitrate in order to produce 2 moles of iron and 2 moles of chromium nitrate.

If we see oxidation state, we see that chromium changes from 0 to +3

Iron changed the oxidation state from +2 to 0

Remember that elements at ground state has 0, as oxidation state.

Iron is being reduced while chromium is oxidized. Then, the half reactions are:

Fe²⁺  +  2e⁻ ⇄  Fe    (Reduction)

Cr ⇄ Cr³⁺  +  3e⁻    (Oxidation)

When an element is being  reduced, while another is being oxidized, we are in prescence of a redox reaction.

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2 years ago
The standard cell potential (E°cell) for the reaction below is +0.63 V. The cell potential for this reaction is ________ V when
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Answer : The cell potential for this reaction is 0.50 V

Explanation :

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The half-cell reactions are:

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Reduction half reaction (cathode):  Pb^{2+}+2e^-\rightarrow Pb

First we have to calculate the cell potential for this reaction.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}

where,

F = Faraday constant = 96500 C

R = gas constant = 8.314 J/mol.K

T = room temperature = 25^oC=273+25=298K

n = number of electrons in oxidation-reduction reaction = 2

E^o_{cell} = standard electrode potential of the cell = +0.63 V

E_{cell} = cell potential for the reaction = ?

[Zn^{2+}] = 3.5 M

[Pb^{2+}] = 2.0\times 10^{-4}M

Now put all the given values in the above equation, we get:

E_{cell}=(+0.63)-\frac{2.303\times (8.314)\times (298)}{2\times 96500}\log \frac{3.5}{2.0\times 10^{-4}}

E_{cell}=0.50V

Therefore, the cell potential for this reaction is 0.50 V

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