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vekshin1
4 years ago
8

If an element has 13 protons, how many electrons are in its outer shell?

Chemistry
1 answer:
stealth61 [152]4 years ago
7 0
There will be three electrons in the outer shell.
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Calculate the mass defect for the formation of phosphorus-31. The mass of a phosphorus-31 nucleus is 30.973765 amu. The masses o
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<u>Answer:</u> The mass defect for the formation of phosphorus-31 is 0.27399

<u>Explanation:</u>

Mass defect is defined as the difference in the mass of an isotope and its mass number.

The equation used to calculate mass defect follows:

\Delta m=[(n_p\times m_p)+(n_n\times m_n)]-M

where,

n_p = number of protons

m_p = mass of one proton

n_n = number of neutrons

m_n = mass of one neutron

M = mass number of element

We are given:

An isotope of phosphorus which is _{15}^{31}\textrm{P}

Number of protons = atomic number = 15

Number of neutrons = Mass number - atomic number = 31 - 15 = 16

Mass of proton = 1.00728 amu

Mass of neutron = 1.00866 amu

Mass number of phosphorus = 30.973765 amu

Putting values in above equation, we get:

\Delta m=[(15\times 1.00728)+(16\times 1.00866)]-30.973765\\\\\Delta m=0.27399

Hence, the mass defect for the formation of phosphorus-31 is 0.27399

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