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:
Channeling often occurs in a packed tower This phenomenon takes place when the. ... ------------EFFECT IN TOWERS AND COLUMNS Towers or columns are the ... instruments such as HPLC(high performance liquid chromatography-columns )
Explanation:
The solubility of PbBr₂(s) with the presence of 0.500 M of KBr is
2.64 x 10⁻⁵ M.
Answer:
See explanation.
Explanation:
Hello,
In this case, we could have two possible solutions:
A) If you are asking for the molar mass, you should use the atomic mass of each element forming the compound, that is copper, sulfur and four times oxygen, so you can compute it as shown below:

That is the mass of copper (II) sulfate contained in 1 mol of substance.
B) On the other hand, if you need to compute the moles, forming a 1.0-M solution of copper (II) sulfate, you need the volume of the solution in litres as an additional data considering the formula of molarity:

So you can solve for the moles of the solute:

Nonetheless, we do not know the volume of the solution, so the moles of copper (II) sulfate could not be determined. Anyway, for an assumed volume of 1.5 L of solution, we could obtain:

But this is just a supposition.
Regards.