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liberstina [14]
3 years ago
10

If the p-side has a higher doping concentration, explain how to keep tuning to the same radio channel?

Chemistry
1 answer:
zzz [600]3 years ago
6 0

Answer:

increase in temperature of the intrinsic semiconductor

Explanation:

  • If the p-side has a higher doping concentration, it implies that number of holes (positive ion) increased which is greater than number of electron (negative ion) in the n-side
  • in order to balance the intrinsic concentration, that is to balance the number of holes and electrons which depends on temperature.
  • an increase in the temperature of the intrinsic semiconductor (p-side), increases the number of electron but number of holes remains constant.

A balance in the intrinsic concentration helps in tuning to the same radio channel.

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What happens when nitrogen fills its valence shell?
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8 0
3 years ago
50.0 g of NaNO3 are dissolved into enough water to make 250 mL of solution. What is the molarity of this solution?
d1i1m1o1n [39]

Answer:

2.35 M

Explanation:

Molarity is mol/L of solution. We have to convert the g to mol and the mL to L. G to mol uses the molar mass of the compound. The molar mass of NaNO₃ is 85.00g/mol.

50.0gNaNO3*\frac{1molNaNO3}{85.00gNaNO3} = 0.588molNaNO3

Then you have to convert mL to L.

250mL*\frac{1L}{1000mL} = 0.250L

Now divide the mol by the L.

\frac{0.588mol NaNO3}{0.250L} = 2.352 M

Round to the smallest number of significant figures = 2.35M

7 0
3 years ago
Balance the following skeleton reactions and identify the oxidizing and reducing agents:(b) P₄(s) → HPO₃²⁻(aq) + PH₃(g) [acidic]
m_a_m_a [10]

The balanced chemical equation is :

5P₄ + 36OH → 12HPO₃⁻² (aq) + 8PH₃ (acidic)

Here the oxidation number of P changed from 0 to -3 in PH₃ and increases  from 0 to +3 in HPO₃⁻². When P₄ changes to PH₃ reduction reaction is taking place as there is addition of hydrogen and when P₄ changes to HPO₃⁻² oxidation takes place as there is addition of oxygen.

Thus clearly both reduction and oxidation are taking place.

Thus, we can infer that here P₄ is both oxidizing as well as reducing agent.

To know more about oxidation number here:

brainly.com/question/13182308

#SPJ4

5 0
2 years ago
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