JEOPARDYStoichiometry,

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:

Answer

Symbol 27Al

Protons 43

Electrons 37

Neutrons 55 48

Mass Number

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

3.5 x 1023 atoms of potassium

100 STOICH. ANSWER

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

Potassium Phosphide= K3P

Calcium Hydride= CaH2

Lithium Phosphate= Li3PO4

500 STOICH. ANSWER

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) àH3O+ (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) àH3O+ (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 12: 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