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3 years ago

I’ll give Brainliest please answer also i know this isn’t chemistry but brainly doesn’t have science

Chemistry

2 answers:

Marysya12 [62]3 years ago

6 0

the answer would be iron. divide mass by volume to get density.

morpeh [17]3 years ago

6 0

Answer:

B. Iron

Explanation:

Correct

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3 years ago

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4 years ago

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3 years ago

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(a) Find the concentration of electrons and holes in a sample of germanium that has a concentration of donor atoms equal to 1015

matrenka [14]

Answer:

a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.

b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.

For clear view of the answers: Please refer to calculation 5 in the attachments section.

Explanation:

So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).

Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:

So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.

So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o). (p subscript o)

Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.

Good Luck Everyone! Hope you will understand.

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3 years ago

A sample weighing 3.00 g of a compound that contains only C, H, and O is completely burned in air to form 4.30g of CO2 and 2.35g

konstantin123 [22]

Answer:

CH3O

Explanation:

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 4.30/44 = 0.0977 mole

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0977= 1.173g

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 2.35/18 = 0.131 mole

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.131 = 0.262 mole

The mass of hydrogen is thus 0.262* 1 = 0.262g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 3 - 0.262 - 1.173 = 1.565g

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 1.565/16 = 0.0978moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that of carbon

H = 0.262/0.0977 = 2.7 = 3

O = 0.0978/0.0977 = 1

C = 0.0978/0.0977= 1

We multiply through by 3 to give CH3O

3 0

3 years ago