2010-2011 Chemistry
SOL Blitz
Question #1
• Which of the following atoms
contains 30 protons, 40 neutrons and
28 electrons?
• A: 70 Zn+2
• B: 70 Zn-2
• C: 40 Zn+2
• D: 30 Zr-2
Answer to Q#1
• Letter A: 70 Zn+2 is the answer.
• Why? The format for chemical symbols
is A Xcharge
z
• Remember, A = mass, which equals
protons + neutrons
• Z = atomic number = protons
• Charge = protons – electrons
Question #2
• The element boron has only two stable
isotopes. One stable isotope has a mass
number of 10 and the other has a mass number
of 11. Which of the following could be the
atomic weight of the element?
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A:
B:
C:
D:
9.5
10.8
11.7
12.4
Answer to Q#2
• Letter B: 10.8 is the answer
• Why? The mass of an element on
the periodic table is the
weighted average of all the
isotopes of that element
Question #3
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An atom of Argon—40 contains—
A: 18 protons and 18 neutrons
B: 18 protons and 22 neutrons
C: 18 protons and 40 neutrons
D: 20 protons and 20 neutrons
Answer to Q#3
• Letter B: 18 protons and 22 neutrons
• Why? The 40 in Argon-40 is the
mass of the isotope
• Remember mass = protons + neutrons
(electrons have no mass)
• Argon has an atomic number of 18, so
there 18 protons
Isotope/ Ion Quick Review
• Isotopes of an element have the same atomic
number (protons) but different masses (hence
different neutrons)
• Both protons and neutrons have a mass of 1
• Electron’s mass is not taken into account in
the mass of an atom
• Ions are anything with an electrical charge
• Positively charged ions have lost electrons
• Negatively charged ions have gained electrons
Question #4
• A scientist comparing K and Ca would
find that K has a—
• A: lower electronegativity and a
smaller atomic radius
• B: higher electronegativity and a
smaller atomic radius
• C: lower electronegativity and a
larger radius
• D: higher electronegativity and a
larger atomic radius
Answer to Q#4
• Letter C: lower electronegativity and a
larger atomic radius
• Why? Electronegativity gets higher as
you go across a period and atomic radius
gets smaller as you go across a period
• K is farther left so it must have a lower
electronegativity and larger atomic
radius
Question #5
• Which of the following elements is
the most reactive?
• A: Cs
• B: Sr
• C: Mg
• D: Rb
Answer to Q#5
• Letter A: Cs
• Why? All these elements are
metals—the most reactive metals are
the largest
• Atomic radius gets larger as you go
down a group
• Cs is farthest down so it is the
largest and most reactive
Trends Quick Review
• Trends as you go down a group: larger
atomic radius, lower electronegativity,
lower and ionization energy (easier to
lose an electron)
• Trends as you go across a period:
smaller atomic radius, higher
electronegativity, and higher ionization
energy (harder to lose an electron)
• Remember IE is the energy it takes to
lose an electron
• Electronegativity is the ability to gain an
electron
Quick Review Continued
• Metal reactivity: Largest metals =
most reactive metals.
• Remember metals like to lose
electrons
• Non-metal reactivity: smallest nonmetals = most reactive non-metals.
• Remember non-metals like gain
electrons
• Noble gases are non-reactive—they
have a full valence shell
Question #6
• A bond between an element of group
2 and an element of group 17 will be:
• A: ionic
• B: Nonpolar covalent
• C: Polar covalent
• D: Metallic
Answer to Q#6
• Letter A: Ionic
• Why? Group 2 elements are metals
(also known as alkaline earth metals)
and group 17 (or 7) elements are nonmetals (also known as halogens)
• Metals + non-metals = ionic bonding
Bonding Quick Review
• Covalent bonding = sharing of
electrons between non-metals
• Ionic bonding = transferring of
electrons between metal + non-metal
• Nonpolar covalent bond = equal
sharing of electrons between nonmetals
• Polar covalent bond = unequal sharing
of electrons between non-metals
Question #7
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What is the name for Cu2S?
A: Copper sulfide
B: Copper (I) sulfide
C: Dicopper sulfide
D: Dicopper monosulfide
Answer to Q#7
• Letter B: Copper (I) sulfide
• Why? Copper is a transition (d block)
metal, so you must use roman
numerals to indicate its charge
• Look behind S for the charge of Cu.
• Charge is +1 so copper (I) sulfide
Question #8
• Which of the following is the formula
for carbonic acid?
• A: HC
• B: HCO
• C: HO
• D: H2CO3
Answer to Q#8
• Letter D: H2CO3
• Why? It’s an acid so H must come
first. Carbonic = carbonate = CO3-2
• H+1 + CO3-2 = H2CO3
Question #9
• What is the molecular formula of
tetraphosphorus decoxide?
• A: PO
• B: P4O
• C: P4O10
• D: PO10
Answer to Q#9
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Letter C: P4O10
Why? Tetra = 4 and deca = 10
So P4O10
Remember, when using prefixes don’t
switch—simply write what’s given
Naming Quick Review
• Must use roman numerals to indicate
charge of a transition metal
• Only use prefixes when the compound is
all non-metals
• If an acid has hydro prefix then it’s
H+element
• If no hydro prefix for an acid then it’s
H+polyatomic
Question #10
___C3H4 + ___O2  ___CO2 + ___H2O
• When the equation above is balanced,
what is the sum of the coefficients?
• A: 4
• B: 5
• C: 8
• D: 10
Answer to Q10
• Letter D: 10
• Why? C3H4 + 4O2  3CO2 + 2H2O; so
1+4+3+2 = 10
• Don’t forget the law of conservation
of mass says the reactants side =
products side
Question #11
• Which of the following is an example
of a decomposition reaction?
• A: 2AgCl  2Ag + Cl2
• B: CuO + H2O  Cu(OH)2
• C: AgCl + Mg  MgCl2 + Ag
• D: HCl + Na(OH)  H2O + NaCl
Answer to Q #11
• Letter A: 2AgCl  2Ag + Cl2
• Why? Decomposition is when you
start with one reactant and break
into 2 products
• Letter B = synthesis; letter C = single
replacement and letter D =
neutralization reaction
Question #12
• 12 x 403 =
• What is the answer to the above
problem expressed with proper sig figs?
• A: 5,000
• B: 4800
• C: 4830
• D: 4836
Answer to Q#12
• Letter B: 4800
• Why? When multiplying or dividing,
you use lowest # of sig figs for your
answer
• 12 only has two sig figs, so your
answer can only have two
Question #13
• How many atoms of Na are in 0.300
moles of Na?
• A: 0.0131 atoms
• B: 7.86x1021 atoms
• C: 1.81x1023 atoms
• D: 6.02x1023 atoms
Answer to Q#13
• Letter C: 1.81x1023 atoms
• Why? 0.300 moles Na x 6.02x1023atom
1 mole
= 1.81x1023 atoms
Question #14
• How many moles are in a 342grams of
CaO?
• A: 6.10 moles
• B: 56.1 moles
• C: 1.92x104 moles
• D: 3.67x1024 moles
Answer to Q#14
• Letter A: 6.10 grams
• Why? 342g x 1mole CaO = 6.10 moles
56.08 grams
• 56.08 grams is the molar mass of
CaO from the Periodic Table
Moles Quick Review
• When multiplying/dividing use lowest # of
sig figs for answer
• When adding/ subtracting use lowest # of
decimal places for answer
• MassMoles use mass of Periodic Table =
1 mole
• Moles Particles use 1 mole = 6.02x1023
particles
• Moles Liters at STP use 1 mole = 22.4 L
Question #15
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Be + 2HCl  BeCl2 + H2
Using the above reaction, what mass of
beryllium was consumed in the reaction
if 4.0 moles of HCl were used?
A: 2.0grams
B: 9.0 grams
C: 18.0 grams
D: 36.0 grams
Answer to Q#15
• Letter C: 18.0 grams
• Why?
• 4.0molsHCl x 1mols Be x 9.01gBe
2 mols HCl 1 mole Be
= 18.0 grams Be
Letter #16
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2C2H6 + 7O2  4CO2 + 6H2O
In an experiment, 0.500 mols of C2H6
were reacted with 1.50 moles of oxygen
gas. Which of the following is the
limiting reactant in this experiment?
A: C2H6
B: O2
C: CO2
D: H2O
Answer to Q #16
• Letter B: O2
• Why? 0.500molsC2H6 x 4mols CO2
2 mols C2H6
= 1.00 mols CO2
• 1.50molsO2x4mols CO2 = 0.857molsCO2
7 mols O2
• Since O2 produces less CO2, it must be
the limiting reactant
• C and D shouldn’t even be choices b/c
they are products!
Stoich Quick Review
• Mass A  Mols A Mols B  Mass B
• Go from mols A  mols B using mol to
mol ratio from coefficients of
balanced equation
• Limiting reactant is the one used up
first. It also produces the smaller
amount of product
• % yield = (actual/theoretical) x100
• Can also use stoich to go from
mass/mols of A to liters of B; use
22.4 L = 1mol at STP
Question #17
• Which of the following is NOT a part of
the kinetic molecular theory?
• A: Gases move in a straight,
continual motion
• B: Gases have no volume themselves
• C: Gases participate in inelastic
collisions
• D: Kinetic energy is directly
related to temperature
• E: Gases feel no attractive forces
Answer to Q#17
• Letter C: Gases participate in
inelastic collisions
• Why? Gases actually participate in
elastic collisions.
• The other 4 parts are all TRUE
Gas Laws Quiz Review
• Boyle’s Law  P1V1 = P2V2, where P
and V are inversely related.
• Charles’ Law  V1/T1 = V2/T2
• Gay-Lusac  P1/T1 = P2/T2
• Combined Gas Law  (P1V1)/T1 =
(P2V2)/T2
• Ideal Gas Law  PV = nRT, where R
= 0.0821 (L*atm)/(mol*K)
More Gas Laws Review
• Must always convert temperature
from celsius to KELVIN!
• Absolute zero is -273C or 0K!
• Absolute zero is the point where all
motion slows down and stops
• Dalton’s Law  P1+P2+P3… = Ptotal
Question #18
• Which of the
following phase
changes occurs at
the arrow?
• A: Sublimation
• B: Melting
• C: Evaporation
• D: Condensation
Answer to Q#18
• Letter B: melting
• Why? Graph always
goes S, L, G—look at
pic on right
• The arrow points to
line b/w solid and
liquid
• Only phase changes
there are melting
(SL) or freezing
(LS)
Phases Quick Review
• 6 phases changes
– Melting (SL)
--Freezing (LS)
– Evaporation (LG) --Condensation (GL)
– Sublimation (SG) --Deposition (GS)
• Molar heat of fusion = amount of energy needed
to melt 1 mole of a substance
• Molar heat of vaporization = amount of energy
needed to evaporate 1 mole of a substance
• Specific heat capacity = amount of energy needed
to raise the temperature of 1 gram of a
substance by 1˚C use q = mxCpxΔT
IMF Quick Review
• Four different intermolecular forces
– London dispersion (weakest) = nonpolar
covalent molecules
– Dipole-dipole = polar molecules
– Hydrogen bonding = H connected to N, O or
F bonding to another O, N or F
– Ionic (strongest) = metal + non-metal
• Don’t forget to do the arrow test to
determine polarity of a molecule
Question #19
• How many moles of MgCl2 are present
in 3.00 liters of a 0.150M MgCl2
solution?
• A: 0.0500 moles
• B: 0.450 moles
• C: 1.35 moles
• D: 20.0 moles
Answer to Q#19
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Letter B: 0.450 moles
Why? Molarity (M) = mols/liters
So 0.150 M = x/3.00 L
(0.150M)(3.00L) = x
X = 0.450 moles
Solutions Quick Review
• Molarity(M)=mols of solute/L of solution
• Molality(m)=mols of solute/kg of solvent
• Freezing point depression:
ΔTf=(-Kf) (m)(n)
• Boiling point elevation ΔTB=(Kb)(m)(n)
• Remember if the solute is covalent, n = 1
Equilibrium Quick Review
• Write K expressions as products over
reactants. Remember the coefficients
in the balanced equation are used as
exponents in the K expressions
• Solids and liquids are not included!
• Catalysts lower the activation energy of
a reaction
• Enthalpy (ΔH) and entropy (ΔS)
determine the spontaneity of a reaction
• - ΔH (exothermic) and +ΔS (more
disorder) means a spontaneous reaction
Question #20
• PCl5(g) + heat  PCl3(g) + Cl2(g)
• Which of the following will cause an
increase in the equilibrium
concentration of Cl2 gas?
• A: The addition of PCl3 gas
• B: The removal of PCl5 gas
• C: A decrease in temperature
• D: An increase in temperature
Answer to Q#20
• Letter D: An increase in temperature
• Why? Heat is a reactant, so increase
reactant side, equilibrium shifts to
product side
• Letter A causes shift to L; letter B
causes shift to L; and letter C causes
shift to L
Acid/Base Quick Review
-log[H+]
• pH =
• pOH = -log
[OH-]
• pH + pOH = 14
• [H+][OH-] =
1x10-14
• [H+] = 10^(-pH)
• [OH-] = 10^(pOH)
• pH < 7 = acid
• pH > 7 = base