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boyakko [2]
3 years ago
12

PLEASE HELP Determine the number of neutrons for the given isotopes:

Chemistry
2 answers:
Julli [10]3 years ago
4 0

Answer:

Helium: 1

Carbon: 2

Nitrogen: 8

Strontium: 52

Tellurium: 71

Explanation:

An isotope is a<em>tom that belongs to the same chemical element as another, has the same atomic number, but different atomic mass</em>

So the data they give us is the mass number of the isotopes

A= mass number

Z=  atomic number ( number of protons in the nucleus).

IS the number that appears in the periodic table

For example, the Z of neon is 10 and the Z of the oxygen is 8

m= number of neutrons

A= Z+ m

m=A-Z

Helium:

m=A-Z\\ m=3-2\\m=1

Carbon:

m=A-Z\\ m=14-12\\m=2

Nitrogen:

m=A-Z\\ m=15-7\\m=8

Strontium:

m=A-Z\\ m=90-38\\m=52

Tellurium:

m=A-Z\\ m=123-52\\m=71

zepelin [54]3 years ago
3 0
<span>Helium = 1
Carbon = 8
Nitrogen = 8
Strontium = 52
Tellurium = 71

If you look on a periodic table, on each element there is a number on the top left. This represents the number of protons in an atom. Protons have a mass of 1 (in relative to Carbon-13)

If we take nitrogen-15 for example; The number 15 tells you that the isotope has a mass of 15. Now if you look on the periodic table, Nitrogen has a proton number of 7. Only protons and neutrons have a mass, electrons are considered to be negligable. Therefore the number of neutrons Nitrogen-15 contains is 15 - 7 = 8 </span>
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3 years ago
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4 0
3 years ago
A sample of a compound is determined to have 1.17 g of carbon and 0.287 g of hydrogen. what is the correct representation of the
yarga [219]

CH3 is the empirical formula for the compound.

A sample of a compound is determined to have 1.17g of Carbon and 0.287 g of hydrogen.

The number of atom or moles in the compound is

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This compound contains 0.097411 mol of carbon and 0.28474 mol of Hydrogen.

So we can represent the compound with the formula C0.974H0.284.

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we can divide 0.284 by 0.0974.

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For a detailed study of the empirical formula refer given link brainly.com/question/13058832.

#SPJ1.

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