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jonny [76]
3 years ago
5

What information is needed to calculate the average atomic mass of an element?(1 point)

Chemistry
1 answer:
g100num [7]3 years ago
7 0

Answer:

Taking into account the definition of average atomic mass and isotopes of an element, the information that you need is the masses of its isotopes and their percent abundances.

Each chemical element is characterized by the number of protons in its nucleus, which is called the atomic number Z.

But in the nucleus of each element it is also possible to find neutrons, whose number can vary. The atomic mass (A) is obtained by adding the number of protons and neutrons in a given nucleus.

The same chemical element can be made up of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called isotopes of the element.

The atomic mass of an element is the weighted average mass of its natural isotopes. Therefore, the atomic mass of an element is not a whole number.

The weighted average means that not all isotopes have the same percentage.

In other words, the atomic masses of chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element, taking into account the relative abundance of each of them.

Explanation:

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3 0
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Using the following balanced chemical equation 8 H2 + S8à 8 H2S. Determine the mass of the product (molar mass = 34.08g/mol) if
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Answer: 7.29 g of H_2S will be produced from the given masses of both reactants.

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}    

{\text{Moles of} H_2}=\frac{1.35g}{2.01g/mol}=0.672moles

\text{Moles of} S_8=\frac{6.86g}{256.5g/mol}=0.0267moles

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According to stoichiometry :

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Thus 0.0267 moles of S_8 will require=\frac{8}{1}\times 0.0267=0.214moles  of H_2

Thus S_8 is the limiting reagent as it limits the formation of product and H_2 is the excess reagent.

As 1 mole of S_8 give = 8 moles of H_2S

Thus 0.0267 moles of S_8 give =\frac{8}{1}\times 0.0267=0.214moles  of H_2S

Mass of H_2S=moles\times {\text {Molar mass}}=0.214moles\times 34.08g/mol=7.29g

Thus 7.29 g of H_2S will be produced from the given masses of both reactants.

5 0
4 years ago
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