O8-#51-52
BASE
ACID SALT
WATER
KOH + HCl  KCl + H2O
51)
COMBINE THE METAL K FROM THE BASE WITH THE NON METAL Cl FROM
THE ACID TO FORM YOUR SALT. THE OH FROM THE BASE AND THE H FROM
THE ACID TO FORM WATER. ALWAYS CHECK YOUR SALT WITH CRISS CROSS
TO HAVE THE CORRECT FORMULA.
52)THE PHRASE EXACTLY NEUTRALIZED INDICATES TITRATION.
MAVA
= MBVB
(1.22)(10.00)=(X)(15.650)
O8-#53-55
53) FREEZING:THE LIQUID ACID IS
COOLED FROM 75,
HAS A PLATEAU AT 69.3 AND THEN
DROPS TO 65. 54)YOUR GRAPH WILL BE:
O8-#55
55) STEARIC ACID C18H36O2
C
18
x 12. = 216
H
36
x
1. =
36
O
2
x 16. =
32
284 g/mol
O8-#55-58
56)
mol = given mass/ GFM
Mol = 20.0/85.0 = 0.235 mol.
O8-#55-57
57)
(20o,88g) in 100 of water
(20o,20g) in 100 of water
1) AT 20 DEGREES, 20 GRAMS IS DISSOLVED ACCORDING TO THE
QUESTION CAPTION.
2) AT 20 DEGREES SATURATION (ON THE CURVE ) IS ABOUT 88 g.
3) ABOUT 68 GRAMS MORE CAN DISSOLVE IN THIS SOLUTION TO
SATURATE IT.
O8-#55-57
∂+
59)
∂2-methylpropane
alkane
2-iodo-2 methylpropane
halocarbon or halide
60) ALL OF THE BONDS IN 2 METHYL PROPANE ARE SINGLE,AS IN ALL
ALKANES, CNH(2N+2)
C4H (2(4)+2)
C4H10
61) 2 METHYL PROPANE IS NONPOLAR (WEAK VAN DER WAALS)
WITH A LOWER BOILING POINT THAN THE POLAR
(STRONG DIPOLE ATTRACTION) 2-IODO 2 METHYL PROPANE.
NOTE THE ELECTRONEGATIVE IODINE IS THE NEGATIVE POLE.
O8-#55-57
62) AT EQUILIBRIUM, THE CONCENTRATIONS OF ALL PRODUCTS AND
REACTANTS ARE CONSTANT (PLATEAU), AS IN THE BEGINNING OF THIS
GRAPH.
63) AS H2 IS ADDED:
1) THERE ARE MORE COLLISIONS AND THEREFORE MORE EFFECTIVE
COLLISIONS WITH N2 WHICH :
A) INCREASES EFFECTIVE COLLISIONS.
B) INCREASES THE FROWARD RATE.
C) PRODUCES MORE PRODUCT.
D) SHIFTS THE REACTION TO THE RIGHT.
2) NOTE, THE CONCENTRATION IS INCREASED ON A SIDE OF THE
REACTION, EVERYTHING ON THAT SIDE DECREASES, AND ALL ON THE
OTHER SIDE INCREASES.
3) THE HYDROGEN INCREASES (DUE TO YOU ADDING SOME) AND
THEN DECREASES, THE NITROGEN DECREASES AND THE AMMONIA
INCREASES.
N2
+ H2

DECREASED INCREASED
THEN DECREASED
NH3
INCREASED
O8-#55-57
64)
1)WHENEVER YOU ADD PARTICLES OF A REACATANT TO AN EXISTING
EQUILIBRIUM, YOU CAUSE MORE COLLISIONS BETWEEN REACTANTS
WHICH INCREASE THE FORWARD RATE. THE REACTION RUNS RIGHT.
2) WHENEVER YOU ADD PARTICLES OF PRODUCTS TO AN EXISTING EQ, YOU
INCREASE COLLISIONS BETWEEN PRODUCTS AND INCREASE THE
REVERSE RATE…REACTION RUNS LEFT.
65) FROM TABLE “E” CARBONATE IS
IS H. THERFORE NaHCO3
66 )ETHANOIC ACID, 2 CARBON ORGO
ACID.
CO3 2- AND SODIUM IS Na, HYDROGEN
O8-#55-57
67)USE THE FORMULA DENSITY = MASS/VOLUME
GIVEN :V = 0.20 L, D = 1.8g/L
1.8 g/L = X/0.20L
X= 0.36 g
OR USE m =VD
68)
THE TWO INDICATORS
OVERLAP (BETWEEN
GREEN LINES) AT ABOUT
7.5 AND 8.3
69) THE CONCENTRATION IS 1 X 10 –pH THEFORE A pH of 7 is 1 x 10-7
pH = - log of [H+]
70) FOR ANY GAS, AS TEMP AND SOLUBILITY ARE INVERSE, OXYGEN IS
LESS SOLUBLE AT HIGH TEMPS. CURVE IS NEGATIVE SLOPE.
71) ( 3 POINTS)
PPM = MASS SOLUTE
X 106
MASS SOLUTION
PPM = 2.7 x 10-2 X 106 = 7.1 PPM…WHICH IS HEALTHY.
3800 g
72)
410 434
486
656
73) AS EXCITED ELECTRONS FALLBACK THEY EMIT ENERGY AS A
SPECIFIC ( FOR THE ELEMENT) SPECTRUM.
74)
Cd + NiO2 + 2 H2O  Cd(OH)2 + Ni(OH)2
EACH 0
75)
+4 -2
ANODE(0  +2)
+1 -2
+2 -2 +1
+2 -2 +1
Cd0  Cd+2 + 2 e-
2e- + Ni+4  Ni +2
76)IN THIS REACTION Cd IS OXIDIZED WHICH INDICATES IT IS THE MORE
ACTIVE METAL , METALS OBSERVED TO BE MORE ACTIVE WILL APPEAR
ON TOP OF TABLE J.
77)
241
=
Am 
95
=
4
+
He +
2
+
237
X
93
78) Fr-220 HAS A HALF LIFE OF 28 SECONDS AND WOULD NOT LAST
VERY LONG IN A DETECTOR.
79) THE SMOKE INTERRUPS THE IONS FORMED BY
THE ALPHA PARTICLES.
FOR THE ABOVE COMBUSTION: IN kcal
1) HOW MUCH HEAT IS EVOLVED PER MOLE OF WATER?
2) HOW MUCH ENERGY WILL BE OBTAINED BY THE COMBUSTION OF 5 MOLES
OF PROPANE?
3) WHAT IS THE MOLE RATIO OF WATER TO PROPANE?
4) WHAT IS THE MOLE RATIO OF HYDROGEN TO CARBON IN PROPANE?
5) HOW MANY kJ is in a kcal?