WA AP Chemistry
Exam 4 MC review Set A 14-15
Comments in red are teacher comments and not part of original work
Chart 1:
Number
1
NaBr
2
K2SO4
3
MgOH
4
Octane
5
CaSO4
1.Rank the compounds in Chart 1 based on their solubility in water (H2O) (first- least
soluble, last- most soluble)
A. 4,3,5,1,2
B. 4,3,1,2,5
C. 1,2,4,3,5
D. 4,3,5,2,1
2. Which of the following would have a higher solubility in water (H2O) than NaBr
3. Which of the following has a solubility in water (H2O) most similar to MgOH
A. NaOH
B. LiOH
C. Ethanol
D. Butane
4. Which of the following is the most soluble in water?
A. decanol
C. dimethyl ether
B. 2-methyl-3-pentene
D. propanol
5. Which of the following orders these compounds by highest boiling point to
lowest boiling point?
A. HCl, MgCl2, AgBr, NO2
B. AgBr, MgCl2, NO2, HCl
C. AgBr, MgCl2, HCl, NO2
D. NO2, HCl, AgBr, MgCl2
6. When the pressure in a sealed container of a gas stays constant, and the
volume increases, what will happen to the average kinetic energy?
A. increase B. decrease C. stay constant D. not enough info provided
Use the following chemicals to help answer questions 7 and 8:
methane; 1-pentanol; 1,2,3-cyclononantriol, silicon dioxide
7. Order the chemicals from lowest boiling point to the highest boiling point
(lowest --> highest):
a. silicon dioxide; methane; 1-pentol; 1,2,3-cyclononanol
b. methane; 1-pentanol; 1,2,3-cyclononanol; silicon dioxide
c. silicon dioxide; 1-pentanol; 1,2,3-cyclononanol; methane
d. methane; 1,2,3-cyclononanol; 1-pentanol; silicon dioxide
8. How many isomers does methane have?
a. <1
b. <4 or >7
c. >10
d. >3 and <10
9. Is the diagram an accurate representation of a solution of NaCl in water?
A. No, because the NaCl atoms do not split apart when it is dissolved.
B. No, because the atoms are not facing the correct directions.
C. Yes, because NaCl is a small molecule.
D. None of the above
10. Identify this molecule. A. 3-­‐hexene B. 3-­‐methylhexane C. 4-­‐methylhexane D. 3-­‐hexanol Use the following table to answer questions 11 and 12
Substance Chemical Formula
1-­‐Propanol C3H8O
Chlorine gas
Cl2
Ethyl methyl ether C3H8O
11. Which substance has the highest boiling point? A. 1-­‐Propanol B. Chlorine C. Ethyl methyl ether D. Both A and C have the same boiling point. 12. What type(s) of intermolecular forces does each substance have? A. 1-­‐Propanol-­‐ London Dispersion forces and dipole-­‐dipole forces. Chlorine gas-­‐ London Dispersion forces.
Ethyl methyl ether-­‐ London Dispersion forces and dipole-­‐dipole forces.
B. 1-­‐Propanol-­‐London Dispersion forces and dipole-­‐dipole forces and hydrogen bonds. Chlorine gas-­‐ none.
Ethyl methyl ether-­‐ London Dispersion forces and dipole-­‐dipole forces.
C. 1-­‐Propanol-­‐London Dispersion forces and dipole-­‐dipole forces. Chlorine gas-­‐ Dipole-­‐dipole forces.
Ethyl methyl ether-­‐ London Dispersion forces and dipole-­‐dipole forces.
D. 1-­‐Propanol-­‐London Dispersion forces and dipole-­‐dipole forces and hydrogenbonds. Chlorine gas-­‐ London Dispersion forces.
Ethyl methyl ether-­‐ London Dispersion forces and dipole-­‐dipole forces.
Use this data to answer questions 13-15
molecule amount in mixture
H2O
10.0g
CH4
20.0g
C8H18
5.0g
H2S
7.0g
13. A mixture of these substances is run through a chromatography column with
a nonpolar substance lining both sides. Rank the substances by the speed they
go through the column from fastest to slowest.
A. C8H18, H2S, H2O, CH4
B. C8H18, H2S, CH4, H2O
C. H2S, H2O, CH4, C8H18
D. H2O, H2S, C8H18, CH4
14. The mixture of the gaseous forms of these substances has a total pressure of
3.00 atm. What is the partial pressure of CH4?
A. 1.83 atm
B. .417 atm
C. .060 atm
D. .608 atm
15. At -100 degrees Celsius only one of the substances is boiling. Which one?
Hint: red circle is bigger than yellow circle (color did not come through and all
look the same – see answers)
A.
B.
C.
D.
16. A constant volume of an ideal gas is heated from 43°C to 75°C. Which of the
following factors will increase?
I. The amount of kinetic energy
II. The distance between molecules
III. The pressure of the gas
A) I only B) II and III only C) I and III only D) I, II, and III
Column chromatography was performed with a certain solvent. The column is made of a
nonpolar material. The chromatogram below shows the results of the chromatography
and pertains to questions 17 and 18.
17. Based on the chromatogram, which are two possible compounds represented by peaks
1 and 2 on the graph?
A) 1 is hexane and 2 is CHCl3
B) 1 is CHCl3 and 2 is hexane
C) Both 1 and 2 are hexane
D) 1 is oil and 2 is water
18. If the two substances have a nearly identical molar mass, which of the two substances
would have a higher boiling point?
A) Substance 1
B) Substance 2
C) They would have the same boiling point
D) There is not enough information given to determine the answer
19. Aaron calculated the boiling points for 4 molecules, HF, NaF, CaF2, and CaO at a
constant pressure. He determined that their boiling points were 19.5°C, 1695°C 1833°C
and 2850°C respectively. He later realized that one of these boiling points was
incorrect. Which one was incorrect?
A. 19.5°C
B. 1695°C
C. 1833°C
D. 2850°C
20. What molecule could this particle diagram represent?
A. H2O
B. NO2
C. CO2
D. Both A and B
21. Tyler combusted 5.00g of Hexanol with 5.00g of oxygen at 160°C in a 1.00L
container. What was the partial pressure of the Co2 produced in the reaction?
A. 9.139 atm
B. 3.702 atm
C. 0.952 atm
D. 2.350 atm
22. Order the following compounds by their relative boiling points. Take into
account their lewis structure and IMF’s. Compounds: MgO, CO, NH3 ,
H2O.
A. MgO , CO , NH3 , H2S
B.
MgO , NH3 , H2S , CO
C.
NH3 , H2S , MgO , CO
D. CO , H2S , NH3 , MgO
23) Based on the phase diagram above, the triple point can be described as:
a. The point in which a solid sublimes at room temperature.
b. The point where an element occupies all three states at STP.
c. The point where an element can coexist in three different phase states.
d. The point where a substance is triple as likely to melt at room temperature.
24) In order to bring Br2 (l) to Br2 (g) it takes 30.91 kJ/mol. To bring I2(l) to I2(g) it
takes 62.44 kJ/mol. Account for why the difference in energy takes place.
a. I2 has double bonds between its diatomic molecules.
b. The LDF forces of Br2 are stronger, making it take less energy.
c. The LDF forces of I2 are more abundant due to its heavier molecule, making
the bonds (IMF) take more energy.
d. Br2 does not react very well with heat.
25.
If 2.0 moles of gas in a sealed glass flask is heated from 25oC to 50oC. Select the
conditions that are true.
Kinetic E
Pressure
n
a
increases
increases
Stays the sames
b
Stays the same
increases
Stays the same
c
decreases
increases
increases
d
Stays the same
increases
increases
26.
Which one of the following molecules has a dipole moment?
a. CI4
b.PF5.
c. SO3
d.NCI3
27. A gaseous compound is 30.4% nitrogen and 69.6% oxygen by mass. A 5.25-gram
sample of the gas occupies a volume of 1.00 liter and exerts a pressure of 1.26
atmospheres at - 4.0 C. What is its molecular formula?
a. NO b. NO2 c.N3O6 d.N2O4
28. Humphrey was given this phase diagram below. He’s a smart kid, so he
already memorized that NaCl boils at 1686 K at 1 atm. With this in mind,
help him analyze the following compounds. Which compound is best
represented by the phase diagram below?
A.
B.
C.
D.
LiCl
Rn
H2O
AgCl
29. Mrs. Fish is swimming in the pond and notices the surface tension on the
water. She wonders what the molecules of surface tensions would look
like. Analyze below - which of the following diagrams best represents
surface tension on water? KEY: Large red circles = O; small silver circles
= H; green circles= other particles.
A.
B.
C.
D.
30. Chemists believe that dodecane, or C12H26 is a great surrogate to various
jet fuels. The researching chemists decided to present this new idea to the
board of directors of American Airlines. However, they realize that they left
some of their data for dodecane in their offices. However, they have
enough knowledge about 1-butanol, 1-decanol and octane. Use the chart
to answer the questions below.
Chemical
Molar Mass (g/mol) Dipole Moment (D)
Boiling Point (K)
C4H10O 1-butanol
74.12 g/mol
1.66
390.4 K
C10H22O 1-decanol
158.28 g/mol
1.68
610 K
C8H18 octane
114.23 g/mol
y
398 K
C12H26 dodecane
x
y
z
I.
Estimate the molar mass of dodecane (no calculator).
A. 144 g/mol (1-decanol minus 16)
B. 184 g/mol (1-decanol + 2 carbons)
C. 164 g/mol (1-decanol + 4 hydrogens)
D. 172 g/mol
II. Given the chart, what should be the dipole moment (y) of dodecane (and
octane)?
A. 1.70
B. 0
C. 1.67
D. Not enough information given
III. What is the best estimate for the boiling point of dodecane?
A. 490 K
B. 398 K
C. 900 K
D. 390.4 K
31.
Image 1 Image 2 Which two compounds are represented by these images? a) 1 is CaCl2 and 2 CO2 b) 1 is SO2 and 2 is CO2 c) 1 is SO2 and 2 is SiO2 d) 1 is CaCl2 and 2 is SiO2 32. Order these in order of increasing boiling point: F2, O2, Ne, and NH3 a) Ne, NH3, CO2, CH3OH b) CO2, Ne, NH3, CH3OH c) Ne, CO2, NH3, CH3OH d) CO2, Ne, CH3OH, NH3 Givens do not match with choices – use answer choices and ignore what is stated in original question 33. This is the amount of energy to increase 20 grams of 30 degrees Celsius solid phosphorus to each indicated temperature. Its melting point is 44 Celsius: 32 Celsius 30.4 Joules 36 Celsius 91.2 Joules 39 Celsius 136.8 Joules 44 Celsius 635.3 Joules What is the specific heat of solid phosphorus in Joules per gram degree Celsius? a) 1.52 b) 0.76 c) 15.2 d) 22.8 Use the following equation and table for questions 34-35
2 X3PO4(aq) + 3 Ca(ClO3)2(aq) -----> Ca3(PO4)2(s) + 6 XClO3(aq)
Mass of filter paper
1.0432g
Mass of unknown phosphate
4.637g
Mass of filter paper and precipitate (1st drying)
9.8645g
Mass of filter paper and precipitate (2nd drying)
9.8622g
34. Assuming that the unknown, X, is a group 1 metal, what is most likely the identity
of the unknown phosphate?
A. Rb3PO4 B. Na3PO4 C. Li3PO4
D. K3PO4
35. What would the boiling point rank be from highest to lowest of the two reagents,
X3PO4 and Ca(ClO3)2, as well as H2O and H2S?
A. X3PO4, Ca(ClO3)2, H2O, H2S
B. Ca(ClO3)2, X3PO4, H2O, H2S
C. Ca(ClO3)2, X3PO4, H2S, H2O
D. Ca(ClO3)2, H2S, X3PO4, H2O
36. The following table contains data about three compounds with their chemical
formulae and boiling points:
Compound
Formula
Boiling point
A
C10H20O2
269ºC
B
C4H10O
34.6ºC
C
C3H8O2
217ºC
D
C6 H6
80.1ºC
Which of these would be most soluble in water at room temperature?
A. A
B. B
C. C
D. D
37. In which of the following scenarios would the ideal gas law give the least
accurate information?
A. Liquid nitrogen evaporates in a half-full sealed container
B. CO2 is collected over water at room temperature
C. Water is boiled in a pressure cooker
D. Natural gas combusts in a bunsen burner
38. Which of the following best represents the dissolution of CO2 in water?
39.A liquid solution of an unknown substance and butanol is separated by boiling
off the unknown, leaving behind liquid butanol. Which of the following could be
the identity of the unknown?
A) propane B) ethyl butyl ether C) decanol D) ethanol
40. Gaseous 1,3 - propanediol is heated in a closed rigid metal container. Which of
the following would occur?
A) The distance between the molecules would increase.
B) The size of the molecules would increase as they sped up.
C) The bonds in between the molecules would break.
D) The kinetic energy of the molecules would increase.
Answers:
1. The correct response is D Agreed, should mention the idea of lattice energy and/or
Coulumb’s law.
A- K2SO4 does not have a higher solublity than NaBr, each ion has a
charge of +1,-1 in NaBr vs in K2SO4, SO4 has a charge of 2- and the 2 K
ions have 1+ (1+) charges. There is a stronger attraction in K2SO4 which
makes it less soluble.
B- CaSO4 clearly is not the most soluble
C- NaBr is certainly not the least soluble
2. The correct response is B
A- Al has a 3+ charge and PO4 has a 3- charge, The anion and cation have a
strong attraction to each other, so they are not likely to separate and dissociate
B- KI has a higher mass than NaBr and the same charges, so it would be more
soluble
C- CH4 is non-polar and therefore is minimally soluble in water, so it would not
be more soluble than NaBr
D- NaCl has the same charges as NaBr but is less massive and therefore is less
soluble. wouldn’t a more polar (ionic) compound dissolve better in water? Cl is
more Eneg than Br and therefore NaCl is more ionic than NaBr
3. The correct response is D
A,B-NaOH and LiOH are soluble in water as all group one hydroxides are soluble
C- Ethanol is polar due to its OH group, therefore, is soluble in water
D- Butane is nonpolar and is minimally soluble in water, MgOH is also minimally
soluble in water probably correct, although ethanol does have some nonpolar
character to it as well.
4. The answer is d, propanol, because letter a is like propanol, but less polar
because it has more carbons, letter b is nonpolar, and letter c is less polar than
letter d because its oxygen is inaccessible for hydrogen bonding. Correct!
5. The answer is b because AgBr and MgCl2 definitely have the highest bp
because of their lattice structures, but AgBr has a higher bp because it is heavier,
so more london dispersion forces. NO2 is next because it is heavier than HCl. A
student may pick letter c instead if they mistake HCl for having hydrogen bonds.
Good point on the H bonds – this is probably correct as the last 2 are polar,
however much lighter
6. The answer is letter a because if the volume increases, so will the
temperature. Average kinetic energy depends on the temperature, so it will also
increase. Correct – nice question!
7. a.This lists the molecules in order of number of atoms. People who do not
know chemistry may think that smaller molecules have lower boiling points.
7.b. This is the correct answer because methane has the least Intermolecular
forces, 1-pentanol has some but it has limited hydrogen bonding and
relatively week LDF, 1,2,3-cyclononanol has more opportunity for hydrogen
bonding and it is large so it is more affected by LDF so there are a lot of LDF,
and silicon dioxide is a network solid which has a lot of intermolecular forces
because of the lattice structure and covalent bonds. Correct
7.c. This is the correct answer in the reverse order. So really the opposite of
the correct answer to trick people.
7d This switches the places of 1,2,3-cyclononanol and 1-pentanol incase
someone does not know a lot about intermolecular forces.
8. A. This could trick people who do not know what isomers are because for
all they know isomers are always less than 1. (A is correct – methane has
no isomers)
8b This could trick people who do not know how to find more than a
few possible isomers.
8c This is the correct answer because it has many possible isomers.
8d. This could trick people who can find some isomers but not as many
as there truly is.
9. A -­‐ This could possibly trick people who do not know chemistry because
people may not know that when a molecule is dissolved the atoms
dissociate.
9b. This is wrong because the diagram is correct and the negative sides of
the water molecules are facing the positive Na ions and the positive sides
of the water molecules are facing the negative Cl ions.
9c This answer does not make sense to anyone who knows chemistry but
to one who doesn’t it could make sense.
9dThis is correct because none of the other options are correct. Correct
10. B. 3-­‐methylhexane. In the longest chain, there are 6 carbons. No double or triple bonds are present and there are no oxygen atoms, meaning it is an alkane. There is one group coming off from the long chain, and when numbering a methyl group, it has to be the lowest number possible, so counting begins from the right. It is on the third carbon from the right making it 3-­‐methylhexane. Correct 11. A. 1-­‐Propanol. Chlorine only has London Dispersion forces, which makes it an obvious wrong answer. Both Ethyl methyl ether and 1-­‐Propanol have the same formula, so they have the same London Dispersion forces, where they differ is that 1-­‐Propanol has accessible atoms to form hydrogen bonds, while in Ethyl methyl ether, the Oxygen atom is stuck in the middle of the other atoms and is not accessible. This allows hydrogen bonding to happen in 1-­‐Propanol. The strong hydrogen bonds make the boiling point of 1-­‐Propanol higher. Correct 12. D. All substances have London dispersion forces, making B. incorrect, but chlorine gas does not have dipole-­‐dipole forces since they have the same electronegativity value, making C incorrect. Both 1-­‐Propanol and Ethyl methyl ether have dipole-­‐dipole forces because of the differences in electronegativity that the atoms in the molecule have, but because the Oxygen atom in 1-­‐Propanol is accessible for bonding but not in Ethyl methyl ether, 1-­‐Propanol has hydrogen bonding while Ethyl methyl ether does not. Correct 13. The answer is C. The chromatography works so the more polar
substances travel faster than the nonpolar substances. H2S and H2O
are polar and therefore travel faster. H2S is more polar because it is
heavier and therefore has more London dispersion forces so it travels
faster than H2O. C8H18 is more nonpolar than CH4 because it has
more hydrocarbon character so it travels slower. CORRECT
14. The answer is A.
Moles H2O= 10g/18.015g=.555 moles
Moles of CH4=20g/16.04g=1.25 moles
Moles of C8H18=5.0g/114.23g=.0437 moles
Moles of H2S=7.0g/34.0809g=.205 moles
.555+1.25+.0437+.205=2.0537 moles
Pressure of CH4= 1.25/2.0537 x 3.00= 1.83 atm
Correct
15. The answer is B. B is CH4. Polar molecules (C and D) have dipole
dipole forces and therefore have higher boiling points than nonpolar
molecules. C8H18 is heavier and therefore has stronger London
dispersion forces and a higher boiling point than CH4. Therefore CH4
has the lowest boiling point and is the only one boiling at -100
degrees Celsius. Explanation is correct, but does not match with
pictures
16. According to Gay-Lussac's Law, temperature and pressure are directly proportional,
so when the temperature increases, the pressure will also increase, so III will increase. I is
also a correct answer because when temperature increases, the speed of molecules also
increases, and thus the kinetic energy also increases according to the kinetic molecular
theory. II does not increase, since the number of molecules and the volume of the gas are
constant the entire time, so the correct answer is I and III only increase. Correct
17. B
Since the column material is nonpolar, the nonpolar compound, represented by 2, would
take longer time to be eluted since it has more attraction to the column. The only option
that has a nonpolar compound as 2 and a more polar compound as 1 is A. Hexane is
nonpolar, and CHCl3 is polar. Correct
18. A
Compound 2 is nonpolar because it has more attraction to the nonpolar column, which is
why it takes more time to be eluted, and compound 1 is polar, or at least more polar than
compound 2. Polar compounds have stronger intermolecular forces than nonpolar
compounds, which is why they have higher boiling points. Since compound 1 is more
polar than compound 2, it has a higher boiling point. Also – have the same LDF (same
MW)
19.
C is the correct answer. 1833 is not the correct boiling point for CaF2. This is
determined because since Calcium has a 2+ charge, CaF2 should have a boiling point
much closer to CaO, in which Calcium also has a 2+ charge, than NaF, in which Sodium
has a 1+ charge. The greater charge of the metal ion causes a higher boiling point
because the intermolecular forces are stronger if there is a higher charge. Decent question
– might have chosen A due to hydrogen bonding
20. A is the correct answer. The particle diagram represents water because water is a
polar molecule. It does not represent NO2 because the size of the Oxygen ion compared
to the size of the Nitrogen ion would be much bigger than the sizes shown in the particle
diagram. It also does not represent CO2 because CO2 is non polar, and therefore does
not contain the dipole-dipole forces that are shown in the particle diagram. Correct –
good use of atomic size and bent shapes
21. B is the correct answer. First. Stoichiometry had to be used to determine that the
products of the reaction include 0.1042 mol CO2, 0.1215 mol H2O, and 0.0315 mol
hexanol. The mole ratio for carbon dioxide (mol CO2/total mol), which was
0.1042/0.2573 which is 0.4051. The ideal gas law then had to be used to determine the
total pressure. pv=nrt. P(1)=(0.2572)(0.08206)(433K). Note: temperature had to be
converted to Kelvin. The total pressure is 9.1388atm. Finally, using Dalton's law of
partial pressure, the mole ration has to be multiplied by the total pressure to determine the
partial pressure of CO2. 9.1388 x 0.4051 = 3.702 atm. If answer A was received, the
student did not multiply by the mole ratio. If answer C was received, the student
multiplied by the number of moles of CO2 instead. Nice question – correct answer
22. The correct answer is B to this question. This is because of their IMF's and
polar forces. The reason MgO has the highest relative boiling point is because of
its extremely strong bonds that need to be broken. It is an ionic bond so it has
the strongest bonds. NH3 has the next highest boiling point because of its
hydrogen bonds. They create LDF's and make the bonds harder to break
therefore increasing the boiling point. For H2S and CO they both are dipoledipole bonds and could be argued either way for their low boiling points, but I
made it clear with the question answers that if you knew MgO had the highest
boiling point, the last two were given for B. Correct
23: The correct answer to this question is C. The reason question B is not
correct is that they three phases are not all present at STP, but rather at .006
atm and 273.2K. STP conditions imply 1 atm and 273K. The reason answer C is
correct is because it only states what the triple point means, rather than adding
an incorrect detail with B. Correct –
24: The correct answer is C to this question. Due to the heavier molecules that
are present in I2, there are more LDF forces which create stronger bonds. The
stronger bonds are then harder to heat up therefore needing more energy for the
reactions. Correct – however bonds do not break in phase changes (should be
IMF)
25. A - When temperature increases, kinetic energy increases, which in turn causes
pressure to increase since the flask has constant volume. The number of moles is constant
because the flask is sealed. A is correct. Correct
26. D - A dipole exists when a molecule has areas of asymmetrical positive and negative
charge. D is the correct answer because of its pyramidal shape with a lone pair on top.
Correct
27. D - Write the mass percent as a mass Mass percent is just mass per 100 g sample. So
30.4 g N and 69.6g O. Then Convert these masses to moles. So 30.4/14=moles of N and
69.6/16=moles of O. Then Find the simplest mole ratios. Divide moles of each element
by the smallest molar amount. Moles of O/ moles of N=2. Then Calculate the number of
moles of gas from the given pressure, volume, and temperature using n = PV/RT. n =
(1.26 atm)×(1 L)/(1 L atm mol-1 K-1 × 277K)=0.045. The molecular weight of the gas is
grams of gas (1.56 g) divided by moles of gas: 5.25g/0.045=116.667. Find the empirical
formula weight by adding up the weights of the atoms in the empirical formula.
NO2=14+16*3=62. 116.67/62=2. So it’s N2O4. Correct
28. The first question presents a phase diagram, and it tests the tester’s ability
to interpret, and pick out the important details. An important detail is the
1820 K on the x-axis. This should stand out as an extremely high
temperature. Thus, the tester should interpret that it takes significant
amounts of heat to boil this substance and turn it into a gas. The tester
could then rule out options B and C, as the highest boiling temperatures
most likely be due to ion-dipole IMFs, the forces found in options A and D.
However, a tester unaware of the ion-dipole IMFs would be tempted to
choose B because of Radon’s large mass. Since large mass is often attributed to
a higher boiling point, the tester would be lured into that answer. However,
Radon is typically a gas, so it’s boiling point is actually very low. Also, in part C, a
tester only aware of dipole-dipole hydrogen bonding would choose C, assuming
that it has the highest possible boiling point.
Between LiCl and AgCl, it is rather difficult to choose the one with a boiling
point of 1820 K. At this point, the tester should use the given information, that
NaCl boils at 1686 K. Since LiCl has a smaller molar mass than NaCl and both
have ion-dipole IMFs, (they do not – both are ion/ion) it should have a lower
boiling point. Also, since AgCl has a larger molar mass than NaCl and both have
IMFs, it should have a higher boiling point. Thus, since 1820 K > 1686 K, it
makes the most logical sense that D. AgCl, is the correct answer. Not sure
about this one – clearly not B or C
29. The second question asks about surface tension. It was not a huge topic,
but the question is ultimately testing the idea of intermolecular forces in a
seemingly convoluted way. Testers are asked to choose the best
representation of water’s surface tension. Answer C is obviously incorrect,
as the molecules are spaced out like a gas. This leaves A, B, and D. A
shows a rigid, solid-like structure to the water molecules. Some individuals
will be tricked by the rigid structure and mistake it as a representation of
‘tension.’ Answer choice D is also incorrect, as there are charges that the
water molecules seem to be attaching to. This makes little to no sense, but
some may be fooled by that diagram. They may be fooled into believing
that surface tension results from small, oppositely charged particles in the
water. This leaves answer B. Although the diagram’s arrows are not
perfectly drawn, it is definitely the best representation of water’s surface
tension.
Corret
30. This last problem introduces a real world problem by talking about
dodecane. Part I starts off relatively easy, by asking for an estimation of
the molar mass of dodecane. The most efficient way to do this would be to
use a given molar mass and add the correct mass of carbons and
oxygens. The correct answer was D. Correct
Part II is a great test for understanding. The alcohols in the first two rows
all have dipole moments, as they have -OH groups that make them have
polar properties. However, alkanes do not have those properties, as they
are non-polar and mostly have a dipole moments of 0. Some people would
choose 1.70, assuming there is a +0.20 pattern through the column.
Others may be deceived by the 1.67, which is an average of the first two
dipole moments in the column. Finally, some would definitely choose
answer D out of confusion. However, the correct answer is B, since the
dipole moment is 0. Correct
Part III is quite difficult, and it requires a clear differentiation between the
polar and non-polar compounds in the chart. Since dodecane has a larger
molar mass than octane, it should have a higher boiling point greater than
398 K. Since B and D are less than or equal to 398 K, the choices narrow
down to A and C. However, how much larger than 398 K should the boiling
point be? Non-polar attraction is typically not as strong as polar attraction,
and dodecane is non-polar, whereas 1-decanol is polar. Thus, it is highly
unlikely that dodecane has a boiling point of 900 K, larger than 1-decanol’s
boiling point of 610 K. Thus, the correct answer is A. Correct
31. Answer d is right. The first one is definitely ionic, CaCl2 is ionic. Image 2 is covalent, and although CO2 and SiO2 are both covalent, SiO2 is a network solid like image 2, CO2 is gas. Correct 32. C is right. Ne has no intermolecular forces because atoms are single and 10 electrons make London dispersion forces low. Although CO2 has a higher weight, NH3 is polar and has hydrogen bonds, which make it above CO2. CH3OH is last because the protruding oxygen is polar, making it easily available for hydrogen bonds. Ne has LDF – otherwise correct 33. B is right. Taking any row, for example, the 36 degree row, 91.2 joules raise 20 grams by 6 degrees. Divide by six will be how many to raise 20 grams one degree. Divide by twenty to raise one gram one degree. The child will know NOT to take the 44 Celsius row because that takes into account the heat of fusion, which the child does not know. Correct answer – D may or may not use the heat of fusion however 34. The answer is d.
9.8622g Ca3(PO4)2 - 1.0432g = 8.819g Ca3(PO4)2
8.819g Ca3(PO4)2 * 1 mol Ca3(PO4)2÷309.97g Ca3(PO4)2= 0.02845mol Ca3(PO4)2
4.637g X3PO4 * 3 mol Ca3(PO4)2 ÷ 2 mol X3PO4* 1 ÷ 0.02845 mol
Ca3(PO4)2 = 244.5 g/mol (molar mass should be 81.2 g/mol from (2)4.637g/0.02845mol
, no answers match this, although the idea behind the rest of the problem is correct)
molar mass of PO4: 94.97 g/mol
3X = 244.5 - 94.97 = 149.5 g/mol
X = 49.83 g/mol
•
This molar mass is the closest to potassium of all the group 1 metals, so d
is the answer.
35. The answer is b. Ca(ClO3)2 and X3PO4 would have the highest boiling points
because they are both ionic compounds, so they have the strongest intermolecular forces
due to their lattice structure. CaClO3 has a higher boiling point than X3PO4 due to a
higher charge, Ca has a 2+ charge and a group 1 metal would have a 1+ charge. Both
H2O and H2S are covalent, but H2O has a higher boiling point because there is hydrogen
bonding present. Correct
36: c. It is one of the lighter compounds in the list, yet it has one of the highest
boiling points. The actual compound, 1,3-propanediol, is miscible in water.
Choice A has a high boiling point mostly due to large LDFs. It is actually very
nonpolar.
Correct 37. A The ideal gas law breaks down at high pressures and low temperatures,
and choice (a) involves both. OK
38. D. It is the only choice that correctly shows the partial formation of carbonic
acid from the dissolving CO2. Interesting question – note that one choice shows
CO2 splitting up which is clearly incorrect
39. D) Ethanol
Ethanol has a lower boiling point than butanol. Propane would not dissolve in
butanol because it is non polar. Ethers, while they do have an oxygen, are only
soluble in non polar solvents. The oxygen is hard to access for dipole dipole
interactions. Decanol has a significantly greater mass than butanol. This increases
the London dispersion forces, making the boiling point higher than that of
butanol's. Ethanol dissolves in butanol and has a lower boiling point due to the
lower mass. Correct
40. D The kinetic energy of the molecules would increase.
The distance would not increase because of the rigid container. The volume is
constant. The size of molecules would never increase. There are not "bonds" in
between molecules, only intermolecular forces that attract and repel. The kinetic
energy would in fact increase because the kinetic energy of molecules is directly
proportional to the temperature. Correct