Answer:
orbitals s p d f
l(orbital number) 0 1 2 3
n l Er= n + l
7f 7 3 7+3 = 10
5d 5 2 5+2 = 7
3s 3 0 3 + 0 = 3
6p 6 1 6 + 1 = 7
4f 4 3 4 + 3 = 7
6d 6 2 6 + 2 = 8
The correct answer is C. Dr Smith, who measures 13.5 g/mL on e2020.
The answer to this question is C: the ability to memorize
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Answer:
(a) ⁴¹₁₉K
(b) ¹⁰⁶₄₆Pd
(c) ¹²⁵₅₂Te
(d) ⁸⁸₃₈Sr
Explanation:
The identity of an element is its atomic number, by convention we write the atomic mass as superscript and the and the atomic number as subscript to the left of the element .
(a) Z = 19 A = 41 symbol: ⁴¹₁₉K
(b) Z = 46 A = 106 symbol: ¹⁰⁶₄₆Pd
(c) Z = 52 A = 125 symbol: ¹²⁵₅₂Te
(d) Z = 38 A = 88 symbol: ⁸⁸₃₈Sr
Answer:
Ans: Across a row, the nuclear charge increases while the shielding effect of the inner core remains the same. According to Coulomb's law, increasing nuclear charge creates a greater force, so the valence electrons are held more tightly. The greater the ionization energy, the more difficult it is to remove an electron.
Explanation: