JEOPARDY
Stoichiometry,
Atomic/Molecular
Structure
Energetics,
Hess’s Law
Acids and
Bases
Gas Law
Miscellaneous
100
100
100
100
100
200
200
200
200
200
300
300
300
300
300
400
400
400
400
400
500
500
500
500
500
100: STOICHIOMETRY,
ATOMIC/MOLECULAR STRUCTURE
How many atoms are in 23 g of
potassium?
Answer
200: STOICHIOMETRY,
ATOMIC/MOLECULAR STRUCTURE
A sheet of paper has a density of 0.2 g/cm3,
a surface area of 93.5 in2, and a weight of
2.0 mg. What is the thickness of the sheet?
Answer
300: STOICHIOMETRY,
ATOMIC/MOLECULAR STRUCTURE
Fill in the following table:
Symbol
Protons
27Al
43
Electrons
Neutrons
Mass
Number
Answer
37
55
48
400: STOICHIOMETRY,
ATOMIC/MOLECULAR STRUCTURE
Which reagent is the limiting reactant when
0.450 mol Al(OH)3 and 0.450 mol H2SO4 are
allowed to react?
Answer
500: STOICHIOMETRY,
ATOMIC/MOLECULAR STRUCTURE
Write the following names of the
compounds:
Potassium Phosphide
Calcium Hydride
Lithium Phosphate
Answer
100 STOICH. ANSWER
3.5 x 1023 atoms of potassium
200 STOICH. ANSWER
300 STOICH. ANSWER
Symbol
27Al
98Tc
85Rb
Protons
13
43
37
Electrons
13
43
37
Neutrons
14
55
48
Mass
Number
27
98
85
400 STOICH. ANSWER
500 STOICH. ANSWER
Potassium Phosphide= K3P
Calcium Hydride= CaH2
Lithium Phosphate= Li3PO4
100: ENERGETICS
2 O3(g)  3O2 (g) ΔH= -284.6 kJ
What is the enthalpy change for
this reaction per mole of O3?
Answer
200: ENERGETICS
2 CH3OH 2 CH4 + O2 ΔH= 252.8 kJ
How much heat is transferred when 24.0 g of
CH3OH is decomposed by this reaction at constant
pressure?
Answer
300: ENERGETICS
The specific heat of octane is 2.22 J/g-K. How
many J of heat are needed to raise the temperature
of 80.0 g of octane from 13-60 °C?
Answer
400: ENERGETICS
Using Hess’s Law: Find the ΔH for the reaction
PCl5(g) → PCl3(g) + Cl2(g), given the following reactions:
P4(s) + 6Cl2(g) → 4PCl3(g) ΔH = -2439 kJ
4PCl5(g) → P4(s) + 10Cl2(g) ΔH = 3438 kJ
Answer
500: ENERGETICS
Find the ΔH for the reaction 2CO2(g) + H2O(g) → C2H2(g) +
5/2O2(g), given the following reactions:
C2H2(g) + 2H2(g) → C2H6(g)
ΔH = -94.5 kJ
H2O(g) → H2(g) + 1/2O2 (g)
ΔH = 71.2 kJ
C2H6(g) + 7/2O2(g) → 2CO2(g) + 3H2O(g) ΔH = -283 kJ
Answer
100 ENERGETICS ANSWER
ΔH= -142.3 kJ/mol O3
200 ENERGETICS ANSWER
300 ENERGETICS ANSWER
q= mcΔT
q= (80.0 g)(2.22 J/g-°C)(60-13°C)
q= 8350 J
400 ENERGETICS ANSWER
500 ENERGETICS ANSWER
100: ACIDS AND BASES
What is the pH of a 0.04 M HCl solution?
Answer
200: ACIDS AND BASES
In the following equation, lable the acid, base,
conjugate acid and conjugate base.
HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)
Answer
300: ACIDS AND BASES
What is concentration of hydroxide ions if
the pH is 8.9?
Answer
400: ACIDS AND BASES
Find the pH of a solution containing 0.33 M Acetic
Acid (HC2H3O2). The Ka value for acetic acid is
1.8 x 10-5.
HC2H3O2 ⇔ C2H3O2- + H+
Answer
500: ACIDS AND BASES
Caffeine is a base with a formula C8H10N4O2. If
there is 0.0032 g of caffeine in a solution of 1 L,
what will the pH of the solution be? Caffeine has a
Kb of 4.1 x 10-4.
Answer
100 ACIDS/BASES ANSWER
pH= -log (.04)= 1.40
200 ACIDS/BASES ANSWER
HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)
HC2H3O2= acid
H2O= base
H3O+= conjugate acid
C2H3O2-= conjugate base
300 ACIDS/BASES ANSWER
What is concentration of hydroxide ions if the pH
is 8.9?
14 - 8.9= 5.1= pOH
[OH-]= 10-5.1= 7.9 x 10-6 M
400 ACIDS/BASES ANSWER
1.8 x 10-5= x2/0.33
x= 0.0024= [H+]
-log(.0024)= 2.61
500 ACIDS/BASES ANSWER
0.0032 g * (1 mol/166.18 g)= 1.92 x 10-5 mol/ 1L=
1.92 x 10-5 M
4.1 x 10-4= x2/ 1.92 x 10-5
x= 8.9 x 10-5 = [OH-]
pOH= -log (8.9 x 10-5 )= 4.05
14 - pOH= pH= 9.95
100: GAS LAW
What are the conditions of STP?
Answer
200: GAS LAW
If 0.03 mol of gas is in a 250 mL container at
450 K, what will the pressure be?
Answer
300: GAS LAW
What volume is occupied by 2.1 g of He at a
pressure of 5.6 atm at 298 K?
Answer
400: GAS LAW
If a gas has a pressure of 4.5 atm in a 1 L
container and you transfer it to a 2 L
container, what will the new pressure be?
Answer
500: GAS LAW
If a gas is in a 520 mL container at 298 K and you
raise the temperature to 450 K, what volume will
the gas occupy?
Answer
100 ANSWER GAS LAW
Standard Temperature and Pressure:
1 atmosphere, 273 K
200 ANSWER GAS LAW
If 0.03 mol of gas is in a 250 mL container at 450 K, what will
the pressure be?
PV=nRT
P(0.250 L)= (.03 mol)(0.0821L*atm/mol*K)(450 K)
P=4.4 atm
300 ANSWER GAS LAW
2.1 g He*(1 mol/4 g He)= 0.525 mol He
PV=nRT
(5.6)V= (.525)(0.0821)(298)
V= 2.3 L
400 ANSWER: GAS LAW
P1V1= P2V2
(4.5 atm)(1 L)= P2 (2 L)
P2= 2.25 atm
500 ANSWER GAS LAW
V1/T1=V2/T2
(520 mL)/(298 K) = V2/ (450 K)
V2= 785 K
100: MISC.
The electron configuration [Ar]4s23d104p2
represents what element?
Answer
200: MISC.
Draw the Lewis Structure of BeF2
Answer
300: MISC
What is the wavelength of a radiation that
has a frequency of 7.9 x 10-5 seconds?
Answer
400 MISC.
Write all the possible quantum (n, l, ml, ms)
numbers for n=2.
Answer
500 MISC.
What is the value of the rate constant, k, for the
reaction?
2A + B2 2 AB
Answer
100 ANSWER MISC.
Ge
200 ANSWER MISC.
300 ANSWER MISC.
c= λν
3 x 108 m/s= λ (7.9 x 10-5 s)
λ= 3.8 x 1012 m
400 ANSWER MISC.
n= 2
l= 0, 1
For l= 0 ml= 0 and ms= ±1/2
For l= 1, ml= -1, 0, or 1 and ms= ±1/2 for each of the
ml values
500 ANSWER MISC.
• Trial 12: A stays same, B2 increases by x 4 and rate
increases by x 4 so B2 is order 1
• Trial 1 3: A halves, B2 stays same, rate stays same so A
is 0 order
• Rate= k[B2]
• 0.34 M/s= k (0.43M)
k=0.79 s-1