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S_A_V [24]
1 year ago
10

Predict the effect (if any) of an increase in temperature on the electrical conductivity of (b) a semiconductor;

Chemistry
1 answer:
attashe74 [19]1 year ago
8 0

If we increase the temperature then the more no of electrons get the energy to jump from conduction band to valence band , thereby increase the conduction of the semiconductor .

Semiconductors :

In the semiconductor the energy gap between the valence band and the conduction band is small . this small energy gap allows some electrons to move from the valence band to the conduction band . thus the semiconductor conduct electricity and conductivity increases with increase in temperature .

There are two type of semiconductor : intrinsic semiconductor and extrinsic semiconductor .

Pure semiconducting substances are called intrinsic semiconductor .

The semiconductor obtained by doping are called extrinsic semiconductors.

Extrinsic semiconductor are of two types : n-type semiconductor and     p-type semiconductor .

n-type semiconductor is due to metal excess defect and p-type semiconductor is due to metal deficiency defect .

Semiconductors are used in transistors , photo-electric devices and rectifiers .

Learn more about semiconductors here :

brainly.com/question/15844033

#SPJ4

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Someone please help me I will give BRAINLIST
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Is MgCO3 organic or inorganic
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g Ammonia has been studied as an alternative "clean" fuel for internal combustion engines, since its reaction with oxygen produc
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\text{Ammonia has been studied as an alternative "clean" fuel for internal combustion}

\text{engines, since its reaction with oxygen produces only nitrogen and water vapor,}

\text{and in the liquid form it is easily transported. An industrial chemist studying this}

\text{reaction fills a} \ \mathbf{100 \  L }\ \text{tank with} \ \mathbf{8.6 \ mol} \ \text{of ammonia gas and} \ \mathbf{28 \ mol} \ \  \text{of oxygen gas, }

\text{to be} \  \mathbf{2.6\  mol} \ .\ \text{Calculate the concentration equilibrium constant for the combustion of}

\text{ammonia at the final temperature of the mixture. Round your answer to  2 significant digits.}

Answer:

Explanation:

From the correct question above:

The reaction can be represented as:

\mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

From the above reaction; the ICE table can be represented as:

                    \mathbf{4 NH_3_{(g)}+ 3O_{2(g)} \iff 2N_{2(g)}+ 6H_2O_{(g)} }

I (mol/L)     0.086            0.28                 0              0

C                   -4x                -3x               +2x           +6x

E                 0.086 - 4x     0.28 - 3x      +2x             +6x

At equilibrium;

The water vapor = \dfrac{2.6 \ mol}{100 \ L} = 6x

x = \dfrac{2.6}{100} \times \dfrac{1}{6}

x = 0.00433

\text{equilibrium constant}  ({k_c}) =  \dfrac{ [N_2]^2 [H_2O]^6 }{ [[NH_3]^4] [O_2]^3 }

\implies \dfrac{(2x)^2 (6x)^6}{(0.086-4x)^4\times (0.28-3x)^3} \\ \\

Replacing the value of x, we have:

K_c = \dfrac{4 \times 46,656 \times x^8}{(0.086-4x)^4\times (0.28 -3x)^3} \\ \\ K_c = \dfrac{4 \times 46656 \times (0.00433)^8}{(0.06868)^4(0.26701)^3} \\ \\ K_c = \mathbf{5.4446 \times 10^{-8}}

K_c = \mathbf{5.5 \times 10^{-8} \ to  \ 2 \ significant \ figures}

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