Synthesis and Applications of Nano-structured Iron Oxides Hydroxides
Ca(NO 3 ) 2 YES All nitrates are soluble! Zn(OH) 2 NO Hydroxides are insoluble except for sodium and...
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Transcript of Ca(NO 3 ) 2 YES All nitrates are soluble! Zn(OH) 2 NO Hydroxides are insoluble except for sodium and...
![Page 1: Ca(NO 3 ) 2 YES All nitrates are soluble! Zn(OH) 2 NO Hydroxides are insoluble except for sodium and potassium ions! K 3 PO 4 YES Phosphates are insoluble.](https://reader036.fdocuments.us/reader036/viewer/2022062721/56649f265503460f94c3c9eb/html5/thumbnails/1.jpg)
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Ca(NO3)2
YESAll nitrates are soluble!
Zn(OH)2
NOHydroxides are insoluble except for sodium and potassium ions!
K3PO4
YES Phosphates are insoluble except for sodium, potassium, and ammonium ions!
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Ene
rgy
to b
oil (
kJ/m
ol)
1100
Ion-ion
700
500
Covalent bonds
100
60
Intermolecular forces
0.1
H-bonds
Polar bonds
Non-polar bonds
(intramolecular forces)
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Why so strong?
-Small size of hydrogen
-High electronegativity of O, N, and F
What are they?
-A network between H and either N, O, or F molecules
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NameName Simple CubicSimple Cubic Body-Body-Centered Centered CubicCubic
Face-Face-Centered Centered CubicCubic
# atoms/unit # atoms/unit cellcell
11 22 44
VolumeVolume
e = edge e = edge lengthlength
r = cell r = cell
radiusradius
ee33 = (2r) = (2r)33 = 8r = 8r3 3 ee33 = =
(4r/√3)(4r/√3)3 3
ee33 = =
(4r/√2)(4r/√2)33
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For:
3A(aq) + B(s) ↔ 2C(g) + D(l) + E(aq)
** Only gaseous and dissolved particles are expressed in the equilibrium expression because their concentrations can vary (whereas solids and liquids cannot)
2
3
[ ] [ ]
[ ]c
C EK
A
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If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
3A(aq) + B(s) ↔ 2C(g) + D(l) + E(aq)
If the concentration of A is raised, as the mixture returns to equilibrium a portion of all of the reactants are consumed, and as a result, the concentrations of C, D, and E will increase
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N2 + O2 ↔ 2NOI
C
E
0.25 0.25 0.0042
-x -x +2x
0.25 0.25 0.0042 + 2x
Kc = 1.7 x 10-3 = (0.0042 + 2x)2 / (0.25)(0.25)
Eventually… x = 0.0030 M change
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Would a precipitate form?
AgCl(s) ↔ Ag+(aq)+ Cl-(aq)
Ksp = 1.8 x 10-10
Q = [Ag+][Cl-]
25 mL of 0.1M AgNO3 solution is added to 100 mL of 0.0050M NaCl solution
If Q<Ksp, no precipitation
If Q>Ksp, precipiation
[Ag+] = 0.10M(25mL/125mL)
[Cl-] = 0.005M(100mL/125mL)
Therefore Q>Ksp, and this is expected to precipitate
Q = (0.100)(0.005) = (5.0 x 10-4)
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Molar Concentration “molarity”
M = # moles/vol. of solution
Molal Concentration “molality”
m = # moles solute / mass of solvent (in kg)
Mass Fraction
= mass solute / mass total
Ppm (parts per million)
= mass solute (in mg) / mass total (in kg)
Mole Fractions
XA = molesA/molesA + molesB +…
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Cd(OH)2 Ksp = 1.2 x 10-14
Ca(OH)2 Ksp = 7.9 x 10-6
Ca2+
Cd2+
0.10 M Cd2+
0.10 M Ca2+
At what concentration of OH- will one of the ions precipitate?
(OH-) is more attracted to Cd2+, because, as the equations are similar in structure (1 cation, 2 anions) they are comparable and Cd(OH)2 would precipitate first
Cd(OH)2 Cd2+(aq) + 2OH-
(aq)
0.10 ~0
+x +x
Ksp = 1.2*10-14 = (0.10)(x)2
X2 = 12*10-12
x = 3.5*10-6
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Freezing Point Depression
ΔTf = iKfm
Boiling Point Elevation
ΔTb = iKbmi = # dissociated particles in empirical formula (Van’t Hoff factor)
Kf or Kb = boiling point elevation / freezing point depression constant
m = (n mol solute / mass solvent (kg))
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Volatile SolutionPvaptotal = XAPvapA + XBPvapB +…
Non-volatile Solution
PvapA = XAPvapA + 0
volatile= changes to gas
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“the solubility of a gas is proportional to the pressure of the gas”
Solg = KHenryPg
Osmotic Pressure
Π = (n/v)RT
R = 0.082057 L · atm / K · mol
C = (n / v) = (mol/L)
T = temperature in Kelvin
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