Write the shorthand electron configurations for the following atoms. Enter full electron configuration in order electrons fill the subshells with number of electrons in the subshells in parentheses. For example, write the electron configuration for Li as [He]2s(1). The answer is case sensitive.

Explanation:

The elements are;

S K, Ti, Sn,

Shorthand electronic configuration basically involves inserting the closest noble gas element before the element in the configuration.

Sulphur (S): This element has 16 atoms. The noble gas before this element is Neon. So the shorthand electronic configuration is written as;

[Ne] 3s² 3p⁴

Potassium (K): This element has 19 electrons, The noble gas before this element is Argon. So the shorthand electronic configuration is written as;

[Ar] 4s¹

Titanium (Ti): This element has 22 electrons. The noble gas before this element is Argon (18) - remaining 4 electrons. The shorthand is given as;

[Ar] 3d² 4s²

Tin (Sn): This element has 50 electrons. The noble gas before this element is Krypton (36) - remaining 14 electrons. The shorthand is given as;

[Kr] 4d¹⁰ 5s² 5p²

answer seismograph;

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally