n = 5
l = 0,1,2,3,4
ml = -4,-3,-2,-1,0,1,2,3,4,
ms = +1/2 and -1/2
This immense mountain range began to form between 40 and 50 million years ago, when two large landmasses, India and Eurasia, driven by plate movement, collided. Because both these continental landmasses have about the same rock density, one plate could not be subducted under the other.Mar. 9, 2015
Answer:
0.0159m
Explanation:
9 M
Explanation:
Lead(II) chloride,
PbCl
2
, is an insoluble ionic compound, which means that it does not dissociate completely in lead(II) cations and chloride anions when placed in aqueous solution.
Instead of dissociating completely, an equilibrium rection governed by the solubility product constant,
K
sp
, will be established between the solid lead(II) chloride and the dissolved ions.
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
Now, the molar solubility of the compound,
s
, represents the number of moles of lead(II) chloride that will dissolve in aqueous solution at a particular temperature.
Notice that every mole of lead(II) chloride will produce
1
mole of lead(II) cations and
2
moles of chloride anions. Use an ICE table to find the molar solubility of the solid
PbCl
2(s]
⇌
Pb
2
+
(aq]
+
2
Cl
−
(aq]
I
−
0
0
C
x
−
(+s)
(
+
2
s
)
E
x
−
s
2
s
By definition, the solubility product constant will be equal to
K
sp
=
[
Pb
2
+
]
⋅
[
Cl
−
]
2
K
sp
=
s
⋅
(
2
s
)
2
=
s
3
This means that the molar solubility of lead(II) chloride will be
4
s
3
=
1.6
⋅
10
−
5
⇒
s
= √
1.6
4
⋅
10
−
5 =
0.0159 M
The missing data is the value of I.E and E.A for Cl and F, which are, Cl: I.E =1255 kJ/mol, E.A = -348.7 kJ/mol, F: I.E = 1678 kJ/mol, E.A = -327.8 kJ/mol.
Answer:
3.2
Explanation:
The electronegativity is the capacity of an atom to attract an electron in a bond. As higher is it, as higher the electron will be attracted. The ionization energy is the minimum energy necessary to remove one electron of the atom, transforming it on a cation. On the other hand, the electron affinity is the energy that is lost when the atom gains an electron and is transformed on an anion.
Calling the constant as K, and knowing that the electronegative of fluorine (F) is 4.0:
4 = K*(1678 - (-327.8))
2005.8K = 4
K = 2.00x10⁻³ mol/kJ
Thus, for chlorine (Cl):
Electronegativity = 2.00x10⁻³*(1255 - (-348.7))
Electronegativity = 3.2
