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Lemur [1.5K]
4 years ago
8

Consider a homogeneous gallium arsenide semiconductor at T 300 K with Nd 1016 cm3 and Na 0. (a) Calculate the thermal-equilibriu

m values of electron and hole concentrations. (b) For an applied E-fi eld of 10 V/cm, calculate the drift current density. (c) Repeat parts (a) and (b) if Nd 0 and Na 1016 cm3 .

Chemistry
1 answer:
solmaris [256]4 years ago
8 0

Answer:

a) The electron carrier concentration is 10^{16}cm^{-3}

   The hole carrier concentration is 3.24 * 10^{-4}cm^{-3}

b) The drift current density is given as 136 Acm^{-2}

c) Repeated part a the electron carrier concentration is 3.24 * 10^{-4}cm^{-3}

  Repeated part a the holes carrier concentration is 10^{16}cm^{-3}

  Repeated part b the electron density is 6.4Acm^{-2}

Explanation:

The explanation is shown on the first , second ,third and fourth image

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What do we call the study of weather? A.forcasting B.meteorology​
svp [43]

The correct answer is B. meteorology. The study of weather is called meteorology, derived from the Greek word for "of the atmosphere".

5 0
3 years ago
In the reaction H2SO4 + 2 NaOH -> Na2SO4 + 2H2O, an equivalence point occurs when 28.54 mL of 0.2075 M NaOH is added to a 34.
damaskus [11]
            moles NaOH = c · V = 0.2075 mmol/mL · 28.54 mL = 5.92205 mmol
            moles H2SO4 = 5.92205 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.961025 mmol
Hence
            [H2SO4]= n/V = 2.961025 mmol / 34.23 mL = 0.08650 M
The answer to this question is  [H2SO4] = 0.08650 M

3 0
3 years ago
Guanidin HNC(NH2)2 is a fertilizer. What is the percent by mass of nitrogen in the fertilizer
Katyanochek1 [597]

Answer:

71.1%. , is a fertilizer.

7 0
3 years ago
An unknown compound contains only C , H , and O . Combustion of 5.00 g of this compound produced 9.99 g CO 2 and 4.09 g H 2 O .
Novosadov [1.4K]

Answer:

C₂H₄O

Explanation:

What we want to calculate when determining empirical formulas is to obtain the proportion of the coefficients of the atoms in the molecule. This proportion is the same as the ratio in terms of number of moles, and since we have the quantities in grams  after combustion , we can calculate the number of moles and thus the empirical formula.

Molar mass CO₂ = 44 gmol⁻¹

Molar mass H₂O = 18 gmol⁻¹

# mol CO₂ =  9.99 g / 44 gmol⁻¹ =     0.227 mol CO₂  ⇒  mol C = 0.227

# mol H₂O =4.09 g H₂O / 18 gmol⁻¹ = 0.227 mol H₂O ⇒ mol H = 0.454

Now to determine the mol of oxygen  we would have to use the fact that we have 5.00 g in the combustion, and since we know the mol C and H, it is easy to determine the mass of O. We can not do the relations we did for C and H, since we already have oxygen in the formula.

mass C = 0.227 mol x 12 gmol⁻¹ = 2.72 g C

mass H = 0.454 mol x 1 gmol⁻¹ = 0.454 g H

mass O present in compound = 5.00 g - 2.72 g - 0.454 g = 1.82 g O

mol O = 1.82 g/ 16 gmol⁻¹ = 0.114

Now the composition and proportions are:

mol C = 0.227 / 0.114 = 1.99

mol H = 0.454 / 0.114 = 3.98

mol O = 0.114 / 0.114 = 1

Rounding these numbers we have the empirical formula C₂H₄O

5 0
4 years ago
Calculate the volume of a carbon monoxide when 22 grams of CO(g) exerts a pressure of 99.5 Kpa at 50 C.
WITCHER [35]

Answer:

21.32 L

Explanation:

First, convert grams to moles using molar mass. C has a mass of 12 and O has a mass of 16. CO2 will therefore have a mass of 28 g/mol. Divide 22 by 28 to get about 0.79 moles. Convert C to Kelvin by adding 273; getting 323K. Then use the ideal gas law (PV=nRT) to solve for V (V=nRT/P). This gets you 21.32 L, hope this helps!

7 0
3 years ago
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