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asambeis [7]
3 years ago
7

Assuming dopant atoms are uniformly distributed in a silicon crystal, how far apart are these atoms when the doping concentratio

ns are 5.0x1015 cm-3 and 5.0x1020 cm-3?
Chemistry
1 answer:
enyata [817]3 years ago
8 0

Answer:

d =~ 5.8μm

d =~ 0.13 μm

Explanation:

when the doping concentrations are 5 × 10^15 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^15

d = 1/ 170997.5

d = 5.85 × 10 ^ -6

d =~ 5.8μm

when the doping concentrations are 5 × 10^20 cm^-3

d = v^-1/3  ; where d represent the distance between the atoms , and v  represent the volume

d =1/ ∛v

d = 1/ ∛5 × 10^20

using the principle of surds and standard forms, we have

d = 1/ ∛0.5 × 10^21  

d = 1/7937005.26

d = 1.26 × 10 ^ -7

d = 0.126 × 10 ^ -6

d =~ 0.13 μm

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The name given to the Earth's climate when there was no ice at the poles
Citrus2011 [14]

Answer:

i think snowball, it sounds weird but its true (i think im sorry if its wrong)

Explanation:

7 0
3 years ago
Which of the following electron configurations are written incorrectly?
Lynna [10]

Answer:

The electronic configuration that are incorrectly written is 1s²2s³2p⁶, 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴.

Explanation:

The electronic configuration of the elements corresponds to how all the electrons of an element are arranged in energy levels and sub-levels.

There are 7 energy levels —from 1 to 7— whose sublevels are described as s, p, d and f.

All electronic configurations begin with the term "1s" —corresponding to the sublevel s of level 1— so 4s²3d¹⁰4p⁷, 3s¹ and 2s²2p⁴ are incorrectly written. In addition, 4s²3d¹⁰4p⁷ is written incorrectly because is impossible to jump from the sublevel "s" to the sublevel "d" —which is found from level 3 and up— without passing through the sublevel "p".

In the case of 1s²2s³2p⁶, the wrong thing is that the sublevel "s" can only hold two electrons, not three.

The other options are correctly written.

3 0
2 years ago
In a solution of a carbonated beverage, what is the dissolved carbon dioxide?
insens350 [35]

Answer:

D.

Explanation:

D is the correct answer because, in aqueous solution, solvent is water and solute (in this example carbon dioxide CO₂) is a substance dissolved in water. The amount of solute that can be dissolved in a solvent depends of chemical composition, temperature and pressure

4 0
3 years ago
Read 2 more answers
The solution you identified in question (1) acts as a buffer due to reactions that occur within the solution when an acid or a b
weeeeeb [17]

Answer:

The answer is "\bold{CH_3COO^{-} \ (aq) + H^{+}\ (aq) \longrightarrow CH_3COOH \ (aq)}"

Explanation:

When HCI is added in the chemical equation it reacts with sodium acetate so, it will give the following chemical equation:

CH_3COONa\ (aq) + HCl\ (aq)\longrightarrow CH_3COOH\ (aq) + NaCl\ (aq)\\\\

 In this, the CH_3COOH is a weak acid so, it not completely dissociated.

CH_3COONa \ (aq) \ \ and \ NaCl were strong electrolytes they are completely dissociated.

The HCl is a strong acid so, it is completely dissociated So, the net ionic equation is:

CH_3COO^{-} \ (aq) + H^{+}\ (aq) \longrightarrow CH_3COOH \ (aq)

8 0
2 years ago
John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det
navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0
2 years ago
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