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Basile [38]
3 years ago
9

Atoms of arsenic (As) are often added to silicon (Si) in a process called doping to change the conductivity of the silicon. How

does the addition of arsenic change the conductivity of silicon?
a) An atom of arsenic has one fewer valence electron and more electron shells than an atom of silicon, so the conductivity decreases because the arsenic atom gains an electron.

b) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity decreases because the arsenic atom loses the electron.

c) An atom of arsenic has one fewer valence electron and fewer electron shells than an atom of silicon, so the conductivity increases because the arsenic atom gains an electron.

d) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity increases because the arsenic atom loses the electron.


HAD TO POST THIS THIS BC THE OTHER POSTED IS WRONG

Chemistry
2 answers:
astra-53 [7]3 years ago
8 0

Answer:

D is the answer, the other one is wrong.

Explanation:

IRINA_888 [86]3 years ago
5 0

Answer:

d) An atom of arsenic has one more valence electron and more electron shells than an atom of silicon, so the conductivity increases because the arsenic atom loses the electron.

Explanation:

This is an example of a n-type semiconductor. The additional electron introduced to the 'grid' of silicon atoms causes an increase in the conductivity of the silicon. This additional electron is introduced as arsenic loses its extra electron.

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At constant pressure, which of these systems do work on the surroundings? A ( s ) + B ( s ) ⟶ C ( g ) A(s)+B(s)⟶C(g) 2 A ( g ) +
Tju [1.3M]

Correct question:

At constant pressure, which of these systems do work on the surroundings?

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

(c) A ( g ) + B ( g ) ⟶ C ( g )

(d) 2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

Answer:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

Explanation:

Work done by a system on the surroundings at a constant pressure is given as;

W = -PΔV

Where;

ΔV is gas expansion, that is final volume of the gas minus initial volume of the gas must be greater than zero.

Part (a)

A ( s ) + B ( s ) ⟶ C ( g )

ΔV = 1 - (0) = 1 (expansion)

Part (b)

2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

ΔV = 5 - ( 2+ 2) = 1 (expansion)

Part (c)

A ( g ) + B ( g ) ⟶ C ( g )

ΔV = 1 - ( 1 + 1) = -1 (compression)

Part (d)

2 A ( g ) + 2 B ( g ) ⟶ 3 C ( g )

ΔV = 3 - ( 4) = -1 (compression)

Thus, systems where there is gas expansion are in part (a) and part (b). The correct answers are:

(a) A ( s ) + B ( s ) ⟶ C ( g )

(b) 2 A ( g ) + 2 B ( g ) ⟶ 5 C ( g )

4 0
3 years ago
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Answer:

see explaination

Explanation:

Molecular equation;

2Li3PO4(aq) + 3CaCl2(aq) >>>> Ca3(PO4)2(s) + 6LiCl(aq)

Total ionic equation; . Includes all ions ;

6Li^+(aq) + 2PO4^-3(aq) + 3Ca^+2(aq) + 6Cl^-(aq) >>>> Ca3(PO4)2(s) + 6Li^+(aq) + 6Cl^-(aq)

Net ionic equation; remove common ions from total ionic;

2PO4^-3(aq) + 3Ca^+2(aq) >>>> Ca3(PO4)2(s)

6 0
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Pls help question 2 ill attach a picture​
Finger [1]

Answer:

X 154

Check solution in explanation

Explanation:

Average atomic mass = ( mass 1× abudance) + ( mass 2× abudance)+ ( mass 3× abudance) / 100

(149×13.8)+(152×44.9) +(154×41.3)/100

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