How many moles of atoms are in 6.00 g of 13C?

Answers

Answer 1
Answer:

There are 0.462 moles of atoms in 6.00 g of carbon-13.

Further Explanation:

Moles, Atomic mass and Molecular mass  

  • 1 mole of a pure substance contains a mass that is equal to the relative atomic mass or molecular mass of the substance.
  • Therefore; molar mass is given as grams per mole of a substance  
  • Hence;

Molar mass = mass of a substance/ Number of moles

     g/mol = g /mole

  • Thus:

M = m/n; where M is the molar mass, m is the mass and n is the number of moles

  • From this relationship we can therefore, calculate mass by multiplying the number of moles by molar mass of a substance.
  • That is; Mass = moles x molar mass
  • To calculate number of moles;

We have; n = m/M

  • Number of moles = Mass of the substance/ Molar mass

In our case;

Mass = 6 g of Carbon-13

Molar mass = 13.0 g/mol

Since; Number of moles = Mass/ molar mass

Thus;

Moles = (6.0 g)/ (13.0 g/mol)

          = 0.462 moles  

Keywords: Moles, Molecular mass, relative atomic mass  

Learn more about:

Level: High school

Subject: Chemistry

Topic: Moles

Sub-topic: Moles, molecular mass and mass of a pure substance

Answer 2
Answer:

Answer: 0.462 moles

Explanation: 13C indicates an isotope of carbon and its mass number is 13. It means the mass of 1 mol of 13C is 13 gram.

The question asks to calculate the number of atoms present in 6.00 grams of 13C.

To calculate the number of moles we divide the given grams by the mass of 1 mol of the element. The set could be shown easily using dimensional analysis as:

6.00gram((1mol)/(13gram))

= 0.462 moles

So, there will be 0.462 moles of atoms in 6.00 grams of 13C.


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Answers

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Explanation:

  • A positron is a small particle which contains a +1 charge and its mass is equal to the mass of an electron, that is, 9.109 * 10^(-31) kg.

A positron is represent by the symbol ^(0)_(+1)\beta.

For example, ^(37)_(19)K \rightarrow ^(37)_(18)Ar + ^(0)_(+1)\beta

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The decay reaction is as follows.

           ^(16)_(8)O \rightarrow ^(16)_(7)N + ^(0)_(+1)\beta

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The reaction will be as follows.

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Therefore, we can conclude that out of the given options positrons are spontaneously emitted from the nuclei of potassium-37.

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Answers

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Answers

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Answers

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Answers

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Answers

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