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gregori [183]
2 years ago
10

Germanium is a group 4A semiconductor. The addition of a dopant atom (group 3A element) that has fewer valence electrons than th

e host atom results in a p-type semiconductor. The addition of a dopant atom (group 5A elements) that has more valence electrons than the host atom results in an n-type semiconductor. Which of the following elements when used for doping germanium will yield p-type semiconductors, and which elements when used for doping germanium will yield n-type semiconductors?
Chemistry
1 answer:
klasskru [66]2 years ago
7 0

Answer:

Doping with galium or indium will yield a p-type semiconductor while doping with arsenic, antimony or phosphorus will yield an n-type semiconductor.

Explanation:

Doping refers to improving the conductivity of a semiconductor by addition of impurities. A trivalent impurity leads to p-type semiconductor while a pentavalent impurity leads to an n-type semiconductor.

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Calculate the heat change in calories for melting 65 g of ice at 0 ∘c.
Genrish500 [490]

When ice melts, the physicals state changes from solid to liquid. The energy or the heat required (q) required to change a unit mass (m) of a substance from solid to liquid is known as the enthalpy or heat of fusion (ΔHf). The variables; q, m and ΔHf are related as:

q = m * ΔHf

the mass of ice m = 65 g

the heat of fusion of water at 0C = ΔHf = 334 J/g

Therefore: q = 65 g * 334 J/g = 21710 J

Now:

4.184 J = 1 cal

which implies that: 21710 J = 1 cal * 21710 J/4.184 J = 5188.8 cal

Hence the heat required is 5188.8 cal or 5.2 Kcal (approx)

5 0
2 years ago
ethylene glycol used in automobile antifreeze and in the production of polyester. The name glycol stems from the sweet taste of
Luden [163]

<u>Answer:</u> The empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

<u>Explanation:</u>

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

C_xH_yO_z+O_2\rightarrow CO_2+H_2O

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of CO_2=9.06g

Mass of H_2O=5.58g

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

  • <u>For calculating the mass of carbon:</u>

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 9.06 g of carbon dioxide, \frac{12}{44}\times 9.06=2.47g of carbon will be contained.

  • <u>For calculating the mass of hydrogen:</u>

In 18g of water, 2 g of hydrogen is contained.

So, in 5.58 g of water, \frac{2}{18}\times 5.58=0.62g of hydrogen will be contained.

  • Mass of oxygen in the compound = (6.38) - (2.47 + 0.62) = 3.29 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Carbon =\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{2.47g}{12g/mole}=0.206moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.62g}{1g/mole}=0.62moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{3.29g}{16g/mole}=0.206moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.206 moles.

For Carbon = \frac{0.206}{0.206}=1

For Hydrogen  = \frac{0.62}{0.206}=3

For Oxygen  = \frac{0.206}{0.206}=1

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1 : 3 : 1

Hence, the empirical formula for the given compound is C_1H_{3}O_1=CH_3O

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is:

n=\frac{\text{molecular mass}}{\text{empirical mass}}

We are given:

Mass of molecular formula = 124 amu = 124 g/mol

Mass of empirical formula = 31 g/mol

Putting values in above equation, we get:

n=\frac{124g/mol}{31g/mol}=4

Multiplying this valency by the subscript of every element of empirical formula, we get:

C_{(1\times 4)}H_{(3\times 4)}O_{(1\times 4)}=C_4H_{12}O_4

Thus, the empirical and molecular formula for the given organic compound is CH_3O and C_4H_{12}O_4

3 0
2 years ago
Your little sister asks you a scientific question: "Does chocolate milk come from brown cows?" In order to answer the question,
aksik [14]

Answer:

Hey there,

Your answer is iii. Brown cows produce white milk. (1 point) <u>and</u> ii. Brown cows never produce chocolate milk. (1 point)

You already know from the basic understanding that chocolate comes from beans, not cows, and is added to milk, and thereby- brown cows do not produce chocolate milk. You also know that they produce white milk.

Wbob1314

if you have any questions I <em><u>INSIST</u></em> you ask them in the comments.

5 0
2 years ago
Two substances in a mixture differ in density and particle size. These properties can be used to
lianna [129]

Answer:

(1) separate the substances

(2) chemically combine the substances

(3) determine the freezing point of the mixture

(4) predict the electrical conductivity of the mixture

Explanation:

3 0
2 years ago
Freon-11, CCl3F has been commonly used in air conditioners. It has a molar mass of 137.35 g/mol and its enthalpy of vaporization
PilotLPTM [1.2K]

Answer:

180.56 Kilo joules of energy is removed in the form of heat when 1.00 kg of freon-11 is evaporated.

Explanation:

Molar mass of freon-11 = 137.35 g/mol

Enthalpy of vaporization of freon-11= \Delta H_{vap}=24.8 kJ/mol

Mass of freon-11 evaporated = 1.00 kg = 1000 g

Moles of freon-11 evaporated = \frac{1000 g}{137.35 g/mol}=7.2807 mol

Energy in the form of heat removed when 1.00 kg of freon-11 gets evaporated:

7.28067 mol\times \Delta H_{vap}=7.2807 mol\times 24.8 kJ/mol

=180.56 kJ

0 0
2 years ago
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