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
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
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Explanation:
