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3 years ago

Pls, Help!!!!!!!

Chemistry

1 answer:

yKpoI14uk [10]3 years ago

4 0

Answer:

Element A = Oxygen

Element H =

Element B = Aluminum

Element J = Magnesium

Element C = Selenium

Element L = Carbon

Element D = Sodium

Element Q = Francium

Element F = Antimony

Element R = Calcium

Element G = Chlorine

Element S = Tellurium

Explanation:

Element A is Oxygen because: oxygen 6 valence electrons ; is a gas at room temperature ; and is transported in blood to cells.

Element H is Neon because: Neon is a noble gas ; qppears as red light when charged with electricity (Neon light signs) and it has the second highest Ionization energy of the elements

Element B is Aluminum because: Aluminum is a metal and its ion has charge of +3. It is also located on the borders of the Metalloid staircase .

Element J is Magnesium because its ion has charge of 2+ and is isoelectronic with Neon because it loses two electrons to now have 10 electrons.

Element C is Selenium because its ion that has a charge of -2 is formed by gaining two electrons in order to have 36 electrons which is isoelectronic with Kr ypton

Element L is Carbon because carbon has the smallest atomic radius of any member in the Carbon family because it is the first member of the family and atomic radius increases on going down the group.

Element D is Sodium because its ion has charge of +1 and it has 2 inner core levels , the 1 and 2 energy levels.

Element Q is Francium because it has the largest radius and lowest ionization energy of any element

Element F is Antimony. It is a member of Nitrogen family and has the second highest ionization energy level in family .

Element R is calcium because its on has charge of +2 which is isoelectronic with Argon . Calcium also has atomic radius is larger than Ar gon.

Element G is Chlorine. It has the second to the smallest radius of elements in the 3rd period as the second to the last element in the period because atomic radius decreases across a period from left to right.

Element S is Tellurium. It has atomic mass larger than Iodine just to the right of it and is found in the 5th period

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$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

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