Question

Question #d0a6c

Answer 1

`""_17^37"Cl"`

Explanation:

For starters, take a look in the Periodic Table and find the atomic number of sulfur, `"S"`.

`Z_ "sulfur" = 16`

Since your unknown atom contains `1` more proton than the sulfur-36 isotope, which contains `16` protons inside its nucleus, you can say that you're dealing with the element that has

`Z = 16 + 1 = 17`

This element is chlorine, `"Cl"`, the element that follows sulfur in the Periodic Table.

Now, the number of neutrons present in the sulfur-36 isotope is given by the mass number of the isotope, `A`, and the atomic number of the element, `Z`.

`color(blue)(ul(color(black)("no. of neutrons" = "mass number " - " atomic number")))`

For the sulfur-36 isotope, you have

`{(A = 36), (Z = 16) :}`

This means that this isotope contains

`"no. of neutrons" = A - Z`

`"no. of neutrons" = 36 - 16 = 20`

You can thus say that the chlorine isotope, which contains the same number of neutrons as the sulfur-36 isotope, will have a mass number equal to

`A = Z + "no. of neutrons"`

`A = 17 + 20 = 37`

This means that the unknown isotope is chlorine-37, one of two stable isotopes of chlorine. In isotope notation, which uses the atomic number, the mass number, and the chemical symbol of the element

you can represent chlorine-37 as

`""_17^37"Cl" -> {("mass number" = 37), ("atomic number" = 17) :}`