O8-#51-52 BASE ACID SALT WATER51) KOH + HCl KCl + H2O

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

284 g/mol

O8-#55-58

56)mol = given mass/ GFM

Mol = 20.0/85.0 = 0.235 mol.

C 18 x 12. = 216

H 36 x 1. = 36

O 2 x 16. = 32

O8-#55-57 57)

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.

(20o,88g) in 100 of water

(20o,20g) in 100 of water

O8-#55-57

59) ∂+

∂- 2-methylpropane 2-iodo-2 methylpropane alkane 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 NH3

DECREASED INCREASED INCREASED

THEN DECREASED

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 CO3 2- AND SODIUM IS Na, HYDROGEN IS H. THERFORE NaHCO3

66 )ETHANOIC ACID, 2 CARBON ORGOACID.

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 +4 -2 +1 -2 +2 -2 +1 +2 -2 +1

75) ANODE(0 +2) 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)

Am He + X

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.

241 = 4 + 237

95 = 2 + 93

FOR THE ABOVE COMBUSTION: IN kcal1) 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?