Answer:
I don't know what you're saying cuz give me the brainless answer please
Answer:
Both A and B will be unreactive!
Explanation:
10=2, 8
18=2, 8, 8
A) in pure water :
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0
change +X +X
Equ X X
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X* X
∴X = √2.1 x 10*-10
∴X = 1.4 x 10^-5
∴ the solubility = X = 1.4 X 10^-5
B) In 1.6 x 10^-3 m Na2CrO4
by using ICE table:
According to the reaction equation:
BaCrO4(s) → Ba^2+(aq) + CrO4^2-(aq)
initial 0 0.0016
Change +X +X
Equ X X+0.0016
when Ksp = [Ba^2+][CrO4^2-]
by substitution:
2.1 x 10^-10 = X*(X+0.0016) by solving for X
∴ X = 1.3 x 10^-7
∴ solubility =X = 1.3 x 10^-7
Answer: im thinking its gonna be d.C2H6 and also
the explanation is on the research i had did before i had answered this question so i really hope this help :)
Explanation:
Ar = van de waals forces or london forces
C
H
4
= van de waals forces or london forces
HCl=permanent dipole-dipole interactions
CO = permanent dipole-dipole interactions
HF = hydrogen bonding
N
a
N
O
3
= permanent dipole-dipole interactions
C
a
C
l
2
= van de waals forces or london forces
Vanadium
V 1s²2s²2p⁶3s²3p⁶4s²3d³
