An ion with mass number 37 possesses unit negative charge.ifbthe ion contains 11.1 % more neeutons than electrons.find the symbol of the ion
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the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present
= x + 11.1% of x
= x + 0.111 x
= 1.111 x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
∴ Number of protons in the neutral atom = x – 1
Therefore 37 = 1.111x + x -1
Or 2.111 x = 38
Or x = 18
Therefore no of protons = atomic no = x -1 = 18 – 1 = 17
∴The symbol of the ion is Cl-^37by17
the number of electrons in the ion carrying a negative charge be x.
Then,
Number of neutrons present
= x + 11.1% of x
= x + 0.111 x
= 1.111 x
Number of electrons in the neutral atom = (x – 1)
(When an ion carries a negative charge, it carries an extra electron)
∴ Number of protons in the neutral atom = x – 1
Therefore 37 = 1.111x + x -1
Or 2.111 x = 38
Or x = 18
Therefore no of protons = atomic no = x -1 = 18 – 1 = 17
∴The symbol of the ion is Cl-^37by17
Answered by
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Suppose no. of electrons = x
no. of neutrons =
No. of electrons in the neutral atom =
No. of protons in the neutral atom =
Mass no. = no. of neutrons + no. of protons
ie,
No. of protons = Atomic no. =
Hence the symbol of the ion is:
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