ACS Study Guide Solutions for Practice Questions
Atomic Structure
1. D – definition of a neutral atom
2. D – can identify based on atomic number (number of protons) which is the bottom left
number. Top left number is the mass number (protons + neutrons)
3. D – protons and neutrons occupy next to no volume, they are concentrated in the nucleus.
Only electrons have volume/take up space in the atom.
4. B – mass number minus the atomic number = number of neutrons
5. C – isoelectronic means to have the same number of electrons
6. A – same as 6
7. C – remember that positively charged ions lost electrons from neutral atom and
negatively charged ions gained electrons
8. B – definition
9. A – inert means unreactive  noble gas configuration. Ions always have noble gas
configurations.
10. A – sodium ion is positively charged
11. C – deuterium is an isotope of hydrogen that has 1 proton and 2 neutrons. Also, you can
eliminate the other choices based on the fact that you have never heard those terms
before.
12. C – use weighted average isotope formula
13. B – isoelectronic means same number of electrons, isotopic means same element but
different mass number
14. A – an alpha particle is a helium nucleus (2 protons, 2 neutrons), so the mass number
should decrease by 4 and the number of protons (atomic number) should decrease by 2
15. B – like charges repel, so nucleus will only attract the alpha particles at high speeds; A
doesn’t make sense since that would naturally happen without extra speed, C is wrong
because speed doesn’t affect accuracy
16. C – experiments are found in first chapter of your textbook. A few to know: Thomson
discovered the electron, Millikan discovered the charge on an electron, and Rutherford
discovered the proton through a gold foil experiment
17. A – valence electrons are those in the highest energy level (highest n value)
18. B – 2d does not exist
19. D – use your periodic table and s, p, d, and f blocks to determine ground-state electron
configurations
20. D – missing the 4s shell tells you this is an ion with a +2 charge
21. B – maximum 2 electrons in ANY orbital
22. D – draw orbital box diagram, remember that every orbital must have one electron before
they can be paired
23. A – paramagnetic means that the species has unpaired electrons
24. D – fill all positive spins before pairing and filling negative spins
25. D – definition
26. A – highest frequency means highest energy jump, remember that lower energy level
transitions have larger energy values on the Bohr diagram
27. A – last quantum number must be +/- ½, ml value ranges from – to + l
28. D – electropositive is a fancy term for a positively charged ion, these lose electrons and
become smaller
29. C – definition
30. A – the helium ion has 1 proton and 1 electron, exactly the same composition as the
hydrogen atom
Molecular Structure and Bonding
1. A – draw the Lewis structure for the final compound and use your knowledge of
molecular geometries to determine this. FOR THE ACS, geometry or shape ALWAYS
refers to the molecular geometry. If they want the electronic geometry, they will specify.
In other words, always take lone pairs into account.
2. D – draw structure
3. C – draw Lewis structure
4. D – draw the Lewis structures and see two have the same geometry
5. C – planar means flat, that all atoms are in the same plane  anything with tetrahedral
geometry has one atom sticking out of the plane and cannot be planar
6. D – draw Lewis structure
7. A – use molecular geometries
8. A – draw Lewis structure
9. A – draw Lewis structures and remember that lone pairs on the central atom compress
bond angles  the more lone pairs or more lone electrons something has, the more the
bond angles are compressed
10. C – draw the Lewis structures
11. D – this is the definition of resonance, which is the concept being tested. Look always for
the MOST CORRECT answer, and not just the one that is correct on its own.
12. B – theory that explains covalent bonding is hybridization
13. A – hybridization + Lewis structure
14. B – lone pairs = polar, larger dipole moment
15. C – tricky because this is a coordinate covalent compound. All of the individual SiO2
molecules are bonded to other ones, making the geometry tetrahedral and the angles
about 109.
16. C
17. B – only answer that is true and pertains to the idea of polarity
18. C – add all valence electrons
19. B – draw Lewis structures
20. C – only difference between all of the molecules
21. B – molecule needs to have lone pairs to shift or isomerize between two forms
22. D – look for the polyatomic ions (those are connected within by covalent bonds, but to
other ions by ionic bonds)
23. D – single bonds are longer than double bonds  ethane has single bonds, benzene has a
mix of double and single bonds, ethene has all C-C double bonds
24. C – we want formal charge to be as close to zero as possible (formal charge = valence
electrons – lone pair electrons - # of bonds)
25. C – know through practice
26. A – isomers have same molecular formula (# of atoms) but different connectivity
27. A – single bonds have one sigma bond, double bonds have one sigma and one pi, triple
bonds have one sigma and two pi
28. B – cubes have six faces, so an atom in the middle of the cube would be in contact with 6
faces
29. D – definition
30. B – complete the Lewis structure they have started for you
Stoichiometry
1. A – follow the steps to find an empirical formula (assume 100 grams, turn percentages
into grams, convert grams to moles, divide by the smallest number of moles to get
subscripts)
2. C – follow steps to find empirical formula, remember if you end up with fractions in your
subscripts you must multiply through by a factor to get rid of them (multiply anything
with .25 and .75 by 4, .5 by 2, and .33 and .66 by 3)
3. D – empirical formula, since you are missing percentages you must find those first before
you can follow the steps in #1
4. C – percent by mass formula = (mass of element * how many of that element is present) /
total mass of compound
5. C – same as 4
6. B – convert grams to moles using the molar mass, multiply by the percentage of carbon
in the sample
7. C – convert grams to moles using molar mass, then moles to atoms using Avagadro’s
number
8. C – this is a tricky one. Convert molecules given to moles using Avagadro’s number.
Anytime you are have both grams and moles values for the same compound, you can
divide grams by moles to get the molar mass. Subtract the mass of xenon from this molar
mass, then divide by the mass of fluorine.
9. C – convert molecules to moles. For every 1 mole of compound, there are 12 moles of
carbon, so you can find the moles of carbon in the same. Convert those moles to grams of
carbon.
10. D – write a balanced equation for this reaction (4 M + 3 O2  2 M2O3). Convert from
moles of metal oxide to moles of metal using mole to mole ratios. Calculate the molar
mass by dividing grams given by moles calculated and use calculated molar mass to
determine identity of metal.
11. A – convert molecules to moles, then divide grams by moles to get molar mass to identify
compound.
12. D – definition
13. A – convert grams to moles and moles to atoms for each value given and compare.
14. D – convert molecules to moles, then divide grams by moles
15. C – write balanced equation (2 KClO3  2 KCl + 3 O2). Convert grams to moles, use
molar ratios, then convert back from moles to grams.
16. C – use molar ratios to covert to aluminum chloride then covert from moles to grams.
17. A – convert from grams to moles, use molar ratios to convert to oxygen gas, then convert
from moles to molecules.
18. A – use value of S moles and molar ratios to calculate how much fluorine was used
during the reaction, then subtract this value from the amount that you started with.
19. B – this is a tricky one. Write balanced equation (M2O + H2  2 M + H2O). Substitute x
for the molar mass in all of your conversions, but use the same conversions as any other
problem above. Convert from grams of metallic oxide to moles, then use molar ratios to
convert to metal, then convert from moles to grams – all conversions should equal 1.054
grams, then solve for x.
20. C – use molar ratios to convert oxygen gas to iron
21. D – definition of limiting reagent
22. C –limiting reagent problem, so follow steps. First convert both reactants all the way to
grams of product – whichever one produced less is the limiting reagent and the maximum
yield from the reaction.
23. C – reaction has a 1:1 molar ratio, so you can use M1V1 = M2V2 here
24. A – solution stoichiometry. First multiply molarity by volume to get moles. Convert from
moles to moles using molar ratio, then use molarity given and moles calculated to get the
volume. Remember that molarity is equal to moles of solute / L of solution.
25. C – gas stoichiometry. Trick here is that at STP, 1 mole of gas is equal to 22.4 L. Use this
to convert L to moles, then use mole to mole ratios, then convert from moles to grams.
26. A – determine which is the limiting reagent using steps in 22, then use the technique to
find out how much excess is remaining from 18
27. A – asking for the actual yield produced. Determine the theoretical yield by converting
moles of ammonia to moles of product, then multiple by the percent yield.
28. D – definition
29. B – calculate theoretical yield by converting antimony to grams to product. Remember
that percent yield = actual/theoretical *100.
30. A – same as 29
States of Matter/Solutions
1. B – in phase diagrams like this, the section to the far left is solid, the section at the
bottom is gas, and the section in the middle is liquid ALWAYS
2. D – same as 1
3. B – molecular solids don’t conduct electricity and are soft, ionic solids conduct
electricity, metals are hard
4. B
5. C – melting point is the same thing as freezing point. The colligative property that deals
with this is freezing point DEPRESSION, so the temperature at this point decreases when
the compound has more impurities or solute particles.
6. B – partial pressures law. Multiply each pressure by the mole fraction, then add them
together to get the total pressure
7. B – a body-centered cube has a total of 2 atoms in it. Use dimensional analysis to convert
these two atoms to grams of lithium.
8. A – in a face centered cubic solid, the formula to calculate the radius is r = (2)1/2(edge
length)/ 4. This is not a formula we talked about in lecture, but plug in the edge length
and it should give you the radius of the atom in question.
9. A – definition
10. D – point between a gas and a liquid is called the critical temperature, liquification occurs
at bottom, so at lower temperatures
11. A – look at graph
12. C – use PV = nRT to solve for moles, then divide grams by moles to get molar mass.
Remember to make sure everything is in the right units first!
13. C – gas deviates from ideal behavior at low temperatures and high pressures ALWAYS
14. B – using combined gas law to solve for the final volume. Only trick is that you must
subtract the vapor pressure of water from the original pressure since the gas was collected
over water.
15. C – most solute particles
16. B – molarity is moles of solute / L of solution. Convert grams to moles, then divide by
volume
17. D – multiply molarity by volume to get moles, then convert to grams. Remember you
always need to dissolve your solute and then dilute it.
18. C – use M1V1 = M2V2
19. D – use solubility guidelines
20. B – definition of supersaturated is when a solution prepared, then cooled and the solid
stays in solution  fitting more in the solution than should be possible based on the
solubility.
21. D – only answer choice that, according to the graph, will not dissolve in 100 g of water
22. A – remember that molality = mol of solute/kg of solvent. We are solving for kg of
solvent. First convert grams to moles of solute, then divide moles of solute by the
molality and you will be left will kg.
23. B – the formula for mole fraction is X = mol of solute / mol of total solution. In this case,
X = moles of water / moles of water + ethanol. There are 200 grams of total solution and
95% of that is ethanol, so the solution must be 190 g ethanol and 10 g water. Convert
those to moles and plug into mole fraction expression.
24. B – formula for freezing point depression is T = ikm. Calculate m by converting grams of
ethanol to moles and dividing by kg of solvent (water) in the solution. Plug and chug to
get change in temperature.
25. C – use dimensional analysis here. Cube the edge value given to convert the units to cm3.
Then use the density given to convert this value to grams. Since you know that the unit
cell contains 2 atoms, then you can use Avagadro’s number to convert this 2 atoms into
moles. To find the molar mass, divide the grams value by moles.
26. B - definition
27. C
28. D
29. C – using PV = nRT, see that if both pressure and temperature are doubled, it will not
change anything else since they are directly proportional. Final system should look
exactly the same as initial system.
30. D – 29.2 cm = 292 mmHg, then add the pressure in the manometer to the air pressure to
get total pressure.
Energetics
1. B – using thermal equilibrium (heat that the metal lost is equal to heat that water gained,
final temperature is the same for both). Expand heats to mcT and solve for missing
variable.
2. A – use q = mcT to solve for mass, then covert mass to volume using density given
3. D – not changing the specific heats or the reaction that is taking place will not change the
temperature change
4. C – going up our phase diagram (from solid to gas) requires energy, while going down it
releases energy
5. A – dissolving something breaks bonds, which requires energy and enthalpy should
increase at a semi-constant rate
6. B – when using bond energies, add up all the bonds in the reactants and subtract all the
bonds of the products. You’ll need to draw Lewis structures to see the number and types
of bonds. REACTANTS – PRODUCTS
7. D – steam to water requires a change in state of matter, which involves using the heat of
condensation, vaporization, or fusion
8. C –same as 6
9. C – same as 6
10. A – to calculate the heat released by a bomb calorimeter, use the formula: Q = CT
(multiply the calorimeter constant and the change in temperature). Since we are asked for
molar heat, you have to divide the heat released by the number of moles of ammonium
nitrate in the system.
11. B - definition
12. B – definition
13. B – when using heats of formation to calculate enthalpy, add up all the heats for the
products multiplied by molar coefficients and subtract the reactants from that.
PRODUCTS – REACTANTS
14. A – exothermic released heat (negative), endothermic absorbs heats (positive)
15. D – same as 13
16. C – use Hess’ Law. Remember whatever changes you make to the reactions, you must
also make to the heats before you add them together.
17. C – same as 13
18. D – Hess’ Law
19. A – same as 13
20. C – same as 13
21. C – same as 13
22. C – same as 13
23. C – look for process that becomes more ordered (a decrease in disorder or entropy)
24. B – solid to a gas skips 2 states of matter, so will have more of an entropy change than
liquid to gas to solid to liquid.
25. D – changes from liquid to solid  increasing order
26. A – going from a liquid to gas, so entropy will be positive for sure. Heat needs to be
added to get a liquid to condense, so enthalpy is positive also.
27. D – use G = H – TS formula chart
28. C – heat was removed so enthalpy is positive – there was more heat in the solution to
begin with. Dissolution process always gives a positive entropy change.
29. B – entropy should be positive since we are gaining moles of gas, use Gibbs free energy
formula to predict the spontaneity
30. C – for the reaction to drive in the same direction, enthalpy and entropy changes must
have opposite signs (think of your Gibbs free energy formula)  the only reaction where
this is true is C, entropy is negative and enthalpy is positive.
Dynamics
1. D – since this is a bimolecular reaction, you can read the rate law straight off of the
balanced equation. Both reactants are first order, so if you double each, the rate should
increase by 4.
2. C – must be first order in chlorine, since changing its concentration has the same effect
on the rate. Doubling both increases rate by 8 – chlorine doubled the rate, so the other
reactant must have quadrupled the rate, making it second order.
3. A – write rates of appearance and disappearance expressions and set them equal to each
other to solve for the disappearance of the O3
4. B – activation energy is the difference between the reactants and the top of the hill. Heat
of reaction is difference between reactants and products.
5. C – add individual orders to get overall order
6. C – can use common sense and definition of half-life. Or use first order integrated rate
law to solve for k. Then use t = .693/k formula to solve for the half-life.
7. B – use common sense and definition of half-life. Or use t = .693/k to solve for k and
then plug into integrated rate law (first order) to solve for time.
8. A – definition/graphs of orders
9. B – initial rates method  hold concentration of one reactant constant and observe how
the other reactant has changed the rate. When [water] doubled, rate quadruples, so must
be second order. When [CH3Cl] is doubled, rate doubles, so must be first order.
10. B – same as 9
11. D – A, B, and C have to do with kinetics (making a reaction faster or slower). D will not
change the rate of the reaction but will only shift the direction of the reaction
(equilibrium not kinetics)
12. B – graph that plots a straight line corresponds to the order of the reaction.
13. A – to calculate initial rate = change in concentration / change in time. Pick any two
points on graph and plug into formula.
14. D – definition
15. A – definition, reactions that have smaller activation energies will happen faster than
those with higher energies
16. A – increasing temperature increases rate, increases temperature also increases K for
endothermic reactions (Le Chatelier’s principle)
17. D – definition, look at Boltzmann plots in book or lecture notes
18. B – increases, but doesn’t depend on whether reaction is exothermic or endothermic
19. A – rule about activation energies from #15 flip flops when two reactions are not at the
same, constant temperature
20. A – definition
21. A – remember that the slow step determines rate  if it is the first step, the rate law for
the overall mechanism can be read straight off of the slow step
22. D – draw the potential energy diagram
23. B – definition
24. C – definition
25. B – same as 9
26. B – analyze the graph to see how much time it took for the partial pressure to decrease by
half
27. C – same as 3
28. D – definition, look for activation energy from reverse reaction
29. A – definition, made in one step and consumed in the next
30. C – definition
Equilibrium
D – definition of equilibrium, rates of forward and reverse reactions are the same
C – K = products over reactants raised to their molar coefficients, omit solids and liquids
A – same as 2
D – same as 2
D – write K expression like 2, then plug and chug the values given for concentrations
A – same as 5
D – cannot be used to predict how fast a reaction happens  that is kinetics, not
equilibrium
8. D – definition
9. C – temperature does change the equilibrium constant, and if the constant changes the
concentrations must change as well
10. D – only temperature will change the equilibrium constant
11. B – Le Chatelier’s principle  decreasing pressure shifts to side with more moles of gas
12. B – Le Chatelier’s principle  adding products shifts to reactants
13. A – same as 12
14. B – same as 12
15. A – Le Chatelier’s principle  increasing temperature for an endothermic reaction will
shift reaction to make more products
16. B – same as 15
17. D – same as 10
18. B – same as 15
19. D – set up ICE chart and plug into K expression
20. C – use Kw = [H3O+][OH-] to solve for the amount of hydronium in solution and then take
the –log to solve for pH.
21. A – definition
22. A – set up K expression for this salt and plug in its solubility for the concentrations of the
ions
23. B – same as 22 (“saturated” value is just giving you the value for its molar solubility)
24. B – solve for Q using the concentrations given and compare it to K. The ions left over in
solution are simply the ones that you have more of to start off with
25. C – common ion effect
26. C – set up ICE chart and plug in values for H+ and HA  [A-] = [H+]. Plug these
concentrations into K expression.
27. B – same as 15
28. D – remember that % ionization = [HA] change / [HA] initial. Use this formula to find
change value. Set up ICE chart to find equilibrium values to plug into K expression.
29. C – G = -RTlnK  plug and chug
30. A – Use equation from #29 to find K for this reaction at 298 K. Then use the derived
Arrhenius equation to solve for the K at 1000 K. Then use equation from #29 again to
solve for G at 1000 K.
1.
2.
3.
4.
5.
6.
7.
Electrochemistry and Redox
1. A – assign oxidation numbers for each of the elements referenced. Remember rules for
assigning them (oxygen is always -2 except in peroxide, hydrogen is always +1, anything
in elemental state is 0, anything with a charge is just that number, all oxidation numbers
must add up to charge on compound).
2. D – same as 1
3. B – same as 1
4. C – transition metals fit this definition
5. C – copper goes from +2 to 0 oxidation state  gained electrons  reduction is gain of
electrons
6. A – an oxidation process will require an oxidizing agents, so look for one that is
oxidation (loss of electrons)
7. A – zinc goes from 0 to +2 oxidation state, so it is oxidized which makes it the reducing
agent
8. C – lead is PbO2 goes from +4 to +2, which is reduction  must be reducing agent
9. D – definition
10. D – trick here is balancing electrons first. Mn half-reaction uses 5 electrons, while Fe
half-reaction only uses 1  would need to multiply Fe by 5 to get those electrons
balanced.
11. B – same as 10  ALWAYS balance electrons first.
12. D – reducing agents are more likely to be oxidized, so they have to be a product in these
standardized reactions. Also, remember the higher the E value, the more likely the
reaction proceeds as a reduction. Since we want something more likely oxidized, pick the
lower E value.
13. D – same reasoning as 12
14. D – same as 12
15. B – remember to find the total voltage, we add the voltage from each half-reaction
together, but we have to flip the sign of one E value to make it an oxidation reaction. Flip
the most negative one and add it to the most positive one to get the highest voltage.
16. B – voltaic cell = positive E value  flip the reaction that is most negative and add
values together
17. C – same as 16 except given balanced equation tells you which reaction to flip
18. A – same as 16, except you are solving for one of the half-reaction potentials rather than
the total voltage
19. A – same as 18
20. C – cathode is where reduction takes place, so this is where metal will be plated and mass
will be gained
21. D – half-reaction for oxygen can be found lower down the table (more easily oxidized,
less easily reduced) than fluorine  remember fluorine is the most easily reduced and
will never be oxidized
22. D – look at half-reactions to see products of this reaction
23. A – use dimensional analysis. Some conversions to remember: current (A) = C/s;
Faraday’s constant = 96,500 C/ mole of electron.
24. B – same as 23
25. C – same as 23
26. D – higher molar mass = less moles of electrons transferred per gram  less electricity
needed
27. B – same as 23
28. B – plug and chug into Nernst equation
29. A – B, C, and D are all wrong so use process of elimination
30. C – Le Chatelier’s principle  we want to increase value of E, so make the reaction go
forward. Increasing the concentration of reactants will make the reaction go in the
direction written.
Descriptive Chemistry/Periodicity
1. A – uses the activity series
2. B – always want a solvent that will be as similar as possible to the metal but not exactly
the same
3. C – all other choices will create another molecule or ion when they are dissolved in water
(ammonia will produce the ammonium ion, carbon dioxide will produce bicarbonate,
hydrogen chloride will make the chloride ion)  need one that will be collected in
isolation
4. A – like dissolves like, so looking for something polar like water
5. C – definition
6. D – want to use something inert (unreactive) so balloon wont explode
7. D – all others are traits that transition metals have that you should know  use process of
elimination
8. C – lab experience
9. D – lab experience
10. A - definition
11. B – combustion reaction definition
12. D – definition, what makes up hard water
13. C
14. D – remember elements will share characteristics with other elements in their period 
other elements in gallium’s period have a charge of +3, other elements in selenium’s
period have a charge of -2
15. B – needs a charge of +2 to form the oxide
16. A – Y has a charge of +2, X has a charge of +4, so Z has a charge of -2  YZ
17. A – atomic radii trend
18. A – atomic radii trend
19. B – Z is effective nuclear charge, but that doesn’t really matter. Still use your atomic radii
trend
20. B – effective nuclear charge is the amount of pull the nucleus has on electrons 
increases with increasing positive charge
21. A – anions are largest, then neutral atoms, then cations
22. A – cations are smallest with highest charge
23. C – anions larger than neutral atoms
24. A – atomic radii trend + anions are larger
25. A – electronegativity trend
26. B – electronegativity trend
27. D – electronegativity trend
28. B – ionization energy trend
29. B – ionization energy trend
30. B – huge jump in energy from 2nd to 3rd energy  harder to ionize a third time
Laboratory Chemistry – omit 1, 10, 11, 13
1. A – II would give you a higher percentage of water, I would cause you to lose more water
and give a percentage that is too low
2. D – we are looking specifically for a problem with the use of a buret, the other answer
choices have to do with other parts of the titration
3. B – use M1V1 = M2V2 to solve for volume needed  remember to add extra to dilute,
don’t dump it all in at once
4. B – subtract the two and watch significant figures  use the number in the value with the
least amount, so two decimal places needed
5. C – when measuring, we can extrapolate out one decimal place further than what can be
seen on the instrument. This instrument reads to the tenths place, so we can report a value
to the hundredths place.
6. B – the more demarcations and smaller an instrument is, the more precise it is
7. A – average the first three values, leave out the third one because it is an outlier
8. A – accuracy is how close the reported numbers are to the actual value (given as 1.0000)
and precision is how close the reported numbers are to each other
9. C – A, C, and D will all work but C provides the most detailed procedure
10. B – ammonia is basic and will turn litmus paper blue
11. D – reaction took carbon dioxide out of solution. Carbon dioxide has acidic properties, so
must be a base in the solution for it to react with.
12. C – graph shows us we need 2 moles of base for every one mole of acid
13. B – chlorine needs to oxidize something else. Br-  Br2 is an oxidation reaction. A would
mean that chlorine was oxidized, which is wrong. C is not an experiment. D has nothing
to do with redox reactions.
14. C- purple indicates a positive displacement reaction has occurred
15. D
16. D – must have strontium since it forms a white precipitate with acid. Must have
magnesium or calcium as well since there was no reaction with dichromate. Barium
reacted with dichromate, so it can’t be present.
17. C – safety rules
18. B – use wider tongs
19. D – ALWAYS add acid to water, not the other way around otherwise reaction will
combust
20. D – danger sign, look in safety rules