Question

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an ion with the mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion. 

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Answer 1

Answer :

• Symbol of the ion : $\rm{^{37}_{17}CI^-.}$

Step-by-step explanation :

Given that,

• Mass number of ion = 37.
• No. of neutrons is 11.1% more than no. of electrons.

$\hrulefill$

Suppose the no. of electrons in the ion = x

Then, no. of neutrons = $\rm{x+\dfrac{11.1}{100}x=1.111x}$

No. of electrons in the neutral atom = x - 1

No. of protons in the neutral atom = x - 1

$\rule{350}{3}$

Now, we are given that mass number = 37. But,

Mass number = no. of neutrons + no. of protons

$\implies\rm{37=1.111x+x-1}$

$\implies\rm{2.111x=37+1=38}$

$\implies\rm{x=\dfrac{38}{2.111}=18}$

∴ No. of protons = Atomic no. =

$\implies\rm{x-1=18-1}$

$\implies\bold{17}$

Hence, the symbol of the ion is $\bold{^{37}_{17}CI^-.}$

Answer 2

Answer:

An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find symbol of the ion. Hence, the symbol of the ion will be . 3717Cl-1.