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Norma-Jean [14]
3 years ago
10

Tin has many possible isotopes. three of these isotopes are 115sn, 117sn, and 126sn. what is the mass number of each? how many p

rotons, neutrons, and electrons are present in each?
Chemistry
1 answer:
Sunny_sXe [5.5K]3 years ago
6 0
Isotopes of an atom is the same atom but the difference between them is that the number of neutrons in the nucleus would differ, depending on the mass number. So let's take a look at your problem and breakdown Tin.

A stable atom of Tin (Sn):
Atomic number: 50
Protons: 50
Electrons:50
Neutrons: 69
Mass Number: 119 (rounded-off)

The Mass number is equal to the sum of protons and neutrons.
Now in an isotope, everything is the same, except for the number of neutrons. We can solve for how many neutrons an isotope has base on the mass number. 

A = p + n
Where A= Mass number
            p = proton
            n = neutron

To get the number of neutron just get the difference between the mass number and proton.
n = A - p

For your given, the mass number is already indicated.

*Note that isotopes have the same number of electrons, protons and atomic number as the original atom. 
 
1. 115 Sn or Tin 115 
Atomic number: 50 
Protons:    50
Electrons: 50 
Neutrons: ?
Mass number: 115

Now we get the number of neutrons:

n = A - p
n = 115 - 50
n  = 65

So 115Sn has a mass number of 115, 50 protons, 50 electrons and 65 neutrons.

2. 117 Sn or Tin-117
Atomic number: 50 
Protons:    50
Electrons: 50 
Neutrons: ?
Mass number: 117

Now we get the number of neutrons:

n = A - p
n = 117 - 50
n  = 67

So 117Sn has a mass number of 117, 50 protons, 50 electrons and 67 neutrons.

3. 126 Sn or Tin-126
Atomic number: 50 
Protons:    50
Electrons: 50 
Neutrons: ?
Mass number: 126

Now we get the number of neutrons:

n = A - p
n = 126 - 50
n  = 76

So 126Sn has a mass number of 126, 50 protons, 50 electrons and 76 neutrons.
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The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C
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<u>Answer:</u> The percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of _{17}^{35}\textrm{Cl} isotope be 'x'. So, fractional abundance of _{17}^{37}\textrm{Cl} isotope will be '1 - x'

  • <u>For _{17}^{35}\textrm{Cl} isotope:</u>

Mass of _{17}^{35}\textrm{Cl} isotope = 35 amu

Fractional abundance of _{17}^{35}\textrm{Cl} isotope = x

  • <u>For _{17}^{37}\textrm{Cl} isotope:</u>

Mass of _{17}^{37}\textrm{Cl} isotope = 37 amu

Fractional abundance of _{17}^{37}\textrm{Cl} isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775

Percentage abundance of _{17}^{35}\textrm{Cl} isotope = 0.775\times 100=77.5\%

Percentage abundance of _{17}^{37}\textrm{Cl} isotope = (1-0.775)=0.225\times 100=22.5\%

Hence, the percentage abundance of _{17}^{35}\textrm{Cl} and _{17}^{37}\textrm{Cl} isotopes are 77.5% and 22.5% respectively.

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According to the Bronsted Lowry conjugate acid-base pair, an acid is a substance that donates protons and a base is a substance that accepts protons.

According to Lewis concept, an acid is the substance which accepts electron pairs and abase is a substance which donates electron pairs.

Thus a substance that increases the hydroxide ion concentration of a solution is an Arrhenius base.

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