CHEM FINAL REVIEW

A student runs a reaction in a closed system. In the course of the reaction, 64.7 kJ of heat is released to the surroundings and 14.3 kJ of work is done on the system. What is the change in internal energy (∆U) of the reaction? 1. -50.4 kJ 2. 79.0 kJ 3. 90.4kJ 4. 50.4 kJ 5. -79.0 kJ

1. -50.4 kJ correct The change in internal energy is given by the formula: ∆U = q + w. In this reaction, q is -64.7 kJ and w is 14.3 kJ. The answer is -50.4 kJ.

How many resonance structures are there for the NO−3 polyatomic ion? 1. 3 2. 2 3. 1 4. This molecule does not exhibit resonance. 5. 5 6. 4

1. 3 correct Draw the lewis structure and check how many different ways there are to draw it.

Give the formula for aluminum hydroxide. 1. Al(OH)3 2. AlOH3 3. AlOH 4. AlH3

1. Al(OH)3 correct The aluminum ion is Al3+; the hydroxide ion is OH−. Three OH− are needed to balance the charge of each Al3+, so the formula is Al(OH)3.

A system that can exchange energy but not matter with the surroundings is termed: 1. Closed 2. Open 3. Isolated

1. Closed correct A closed system can exchange energy but not matter with the surroundings.

Which of the following is polar? 1. IF5 2. SF6 3. ICl−4 4. XeF4 5. PCl5

1. IF5 correct XeF4 and ICl−4 have 6 RHED with two RHED being lone pairs situated opposite each other so their effects cancel. Only IF5 has an unbalanced number of lone pairs which places one of the polar I F bonds opposite a lone pair. This unopposed dipole and the lone pair itself make it polar. The others have either 5 or 6 RHED and no lone pairs on the central atom.

A gas is showing a considerable amount of attractive forces. What is the likely value for the compressibility factor? 1. It will be slightly below one. 2. It will be slightly above one. 3. It will be equal to one.

1. It will be slightly below one. correct The compressibility factor (Z) for a gas is equivalent to the value of P V /nRT. This is equal to exactly one if the gas is truly ideal. When attractions between molecules are appreciable, the molecules will attract each other and slightly lower the measured pressure. This will cause the value of Z to drop slightly below one.

The size of an atomic orbital is determined by which quantum number? 1. n 2. ℓ 3. ms 4. mℓ

1. n correct n is the principle quantum number. It is the main energy level or shell and determines the size of the orbitals.

Write the electron configuration for P. (P has atomic number 15)

1s2 2s2 2p6 3s2 3p3

Draw the Lewis structure of xenon difluoride and give the number of lone pairs of electrons around the central atom. 1. 4 2. 3 3. 1 4. 2 5. 5

2. 3 correct

Which of the following is the correct formula for sodium hypochlorite? 1. NaClO4 2. NaClO 3. NaClO3 4. NaCl 5. NaClO2

2. NaClO correct A sodium atom will lose one electron to form a Na+ ion. Hypochlorite is ClO1−. To equalize the charges, Na+ and ClO1− must be present in a 1:1 ratio, so the formula for sodium hypochlorite is NaClO.

Which of the following species is/are paramagnetic? I) Li−2 II) O2 III) H+2 1. II and III 2. I and III 3. I, II and III 4. II only 5. III only 6. I only 7. I and II

3. I, II and III correct Explanation: Li−2 and H+2 both have an odd number of electrons and therefore must be paramagnetic. O2 has 16 total electrons, the last two of which must go into separate degenerate π∗ anti-bonding orbitals.

The best predicted shape and bond angle of SbH3 is 1. tetrahedral; 109.5◦ 2. trigonal planar; 120◦ 3. trigonal pyramidal; 107◦ 4. trigonal pyramidal; 109.5◦

3. trigonal pyramidal; 107◦ correct For the central atom Sb, HED = 4, lone pairs = 1, the electronic geometry is tetrahedral, and the molecular geometry is trigonal pyramidal. Bond angle estimates come from electronic geometry, but get adjusted for lone pairs. The H-Sb-H bond angle is predicted to be less than 109.5◦ .

The viscosity of a liquid depends on which of the following I. strength of intermolecular forces. II. shape of the molecule III. temperature 1. I and II 2. I and III 3. only II 4. only I 5. I, II, and III 6. only III

5. I, II, and III correct The viscosity of a liquid depends on the strength of intermolecular forces and the particulars of the shape of the molecule. In addition the viscosity depends on the temperature. As the temperature increase the molecules have more kinetic energy to overcome the IMF and therefore they flow more easily.

When water condenses, what are the signs for q, w, and ∆Ssys, respectively? 1. +, +, − 2. −, +, + 3. +, −, − 4. +, −, + 5. −, +, − 6. +, +, +

5. −, +, − correct q is negative (-) because water vapor is losing heat to turn into a liquid. If q is negative, w is usually the reverse sign (+). A phase change from gas to liquid means entropy decreases (-)

A low-pressure mercury-vapor lamp has a characteristic emission line at 253 nm. Knowing that this lamp is putting out 11.8 watts of light energy, how many mercury atoms are emitted per second during operation? 1. 5.25 x 1020 atoms 2. 1.08 x 1017 atoms 3. 7.86 x 10−19 atoms 4. 4.73 x 105 atoms 5. 7.11 x 1024 atoms 6. 1.50 x 1019 atoms

6. 1.50 x 1019 atoms correct ν =c/λ =(3 x 10^8)/(253 x 10^−9) = 1.186 x 1015 s^−1) E = hν = (6.626 x 10^−34)(1.186 x 10^15) = 7.857 x 10^−19 J/photon 11.8 W = 11.8 J/photon and for 1 second = 11.8 J (11.8 J)/(7.857 x 10^−19 = 1.50 x 10^19) photons or atoms of Hg

Which of the following is the correct Lewis formula for carbon monoxide (CO)?

C triple bonded to O

What is the lewis structure of BF3?

Check google to verify

Draw the correct Lewis formula for hydrogen cyanide (HCN)?

Google that shit to check homes

Draw the correct Lewis structure of hydroxylamine (NH2OH)

Google that shit to check homes

In an atom, what would be the maximum number of electrons having the quantum numbers n = 6 and ℓ = 2? 1. 10. 2. 5. 3. 6. 4. 8. 5. 72

1. 10. correct Ten electrons can have this set of quantum numbers. Where ℓ = 2, five values of mℓ are possible, namely, 2, 1, 0, −1, and −2. Each one of these mℓ values represents a single orbital, so there are 5 orbitals. Since each orbital can accomodate 2 electrons, 10 electrons can have this set of n and ℓ values.

Consider 2 Al(s) + 6 HCl(ℓ) → 2 AlCl3(s) + 3 H2(g), the reaction of 4.5 mol Al with excess HCl to produce hydrogen gas. What is the pressure of H2(g) if the hydrogen gas collected occupies 14L at 300K? 1. 11.9 atm 2. 0.0763 atm 3. 1.07 atm 4. 0.233 atm 5. 5.28 atm 6. 7.9 atm

1. 11.9 atm correct V = 14 L T = 300 K nH2 = 4.5 mol First calculate the number of moles of H2 produced: ? mol H2 = 4.5 mol Al ×(3 mol H2)/(2 mol Al) = 6.75 mol H2 PV=NRT P=NRT/V P= [(6.75 mol)(0.0821 L·atm/mol·K)(300 K)]/(14 L) = 11.8752 atm

For the reaction 2 HCl + Na2CO3 → 2 NaCl + H2O + CO2 179.2 liters of CO2 is collected at STP. How many moles of NaCl are also formed? 1. 16.0 moles 2. 6.0 moles 3. 12.5 moles 4. 4.0 moles 5. 32.0 moles 6. 8.0 moles

1. 16.0 moles correct VCO2 = 179.2 L At STP we can use the standard molar volume, 22.4 L/mol. (179.2 L / 22.4 L/mol) = 8.00 mol CO2 8.00 mol CO2 × (2 mol NaCl / 1 mol CO2) = 16.0 mol NaCl

Calculate the approximate boiling point of chloroform, CHCl3, given the following data: ∆Hvap = 31.4 kJ mol−1 ∆Svap = 93.6 J mol−1 K−1 1. 335 K 2. 298 K 3. 665 K 4. 59.3 K 5. 0.34 K

1. 335 K correct ∆G = ∆H − T∆S ∆Hvap−T∆Svap = 31.4×1000−T ×93.6 =0 T = 335 K

What is the root mean square speed of carbon dioxide molecules at 98◦C? 1. 459 m · s−1 2. 45.6 m · s−1 3. 236 m · s−1 4. 153 m · s−1 5. 574 m · s−1

1. 459 m · s−1 correct T = 98◦C + 273.15 = 371.15 K MW = 44.0095 g/mol = 0.0440095 kg/mol R = 8.314 N/K·mol The root mean square speed is v=√(3RT/MW) =458.635 m/s.

A sample of nitrous oxide gas (NO) has a density of 12 g L−1. What pressure does the sample exert at 27 ◦C? 1. 9.9 atm 2. 1.0 atm 3. 997.9 atm 4. not enough information 5. 61.6 atm

1. 9.9 atm correct By thoughtful substitutions and rearrangement, the ideal gas law can be used to relate the mass density of a gas to its pressure. PV = nRT Recalling that a number of moles (n) is equal to a mass (m) divided by a molecular weight (M.W.), we can substitute into the ideal gas law. n = m / M.W. PV = (m / M.W.) RT This can be rearranged to solve for P (Substitute in Density for m/v) 12[(0.0821 × 300)/30]= 9.852 atm

Choose the pair of names and formulas that do NOT match. 1. As4O6 : tetraarsenic oxide 2. SO3 : sulfur trioxide 3. NO : nitrogen monoxide 4. Cl2O7 : dichlorine heptoxide 5. N2O5 : dinitrogen pentoxide

1. As4O6 : tetraarsenic oxide correct Cl2O7, N2O5, SO3, and NO are all covalent molecules and are named correctly using the appropriate prefixes. As4O6 is also a covalent compound; appropriate prefixes are used in the name. It should be correctly named tetraarsenic hexoxide to indicate the presence of six oxygen atoms in each molecule.

Which of the following molecules is nonpolar? 1. BF3 2. H2O 3. SO2 4. NF3 5. CH3Br

1. BF3 correct There are no lone pairs on B. This is a symmetrical molecule and non-polar.

Which of the following substances would you predict might evaporate the fastest? 1. C6H14 2. C8H18 3. C12H24 4. C10H22

1. C6H14 correct All the listed molecules are nonpolar hydrocarbons; therefore the dominant intermolecular force that exists in the condensed phase of all listed molecules is dispersion forces. Therefore, the molecule with the least number of atoms and the lowest molecular weight would have the lowest dispersion forces, and therefore would evaporate the easiest.

Carbon tetrachloride (CCl4) and n-octane (C8H18) are both non-polar molecules. At standard pressure, they boil at 345 K and 399 K, respectively. Which answer choice below correctly explains their boiling points? 1. C8H18 has a higher boiling point because its electron cloud is larger and allows it to form more instantaneous dipoles. 2. C8H18 has a higher boiling point because its smaller surface area allows it to form stronger instantaneous dipoles. 3. CCl4 has a lower boiling point because its smaller surface area allows it to form stronger instantaneous dipoles. 4. CCl4 has a lower boiling point because its greater molecular weight enables it to form stronger instantaneous dipoles. 5. C8H18 has a higher boiling point because its greater molecular weight enables it to form stronger instantaneous dipoles.

1. C8H18 has a higher boiling point because its electron cloud is larger and allows it to form more instantaneous dipoles. correct Carbon tetrachloride (CCl4) and n-octane (C8H18 ) both have only instantaneous dipoles. Despite the fact that (CCl4) has a greater molecular weight (153.81 g mol−1) compared to C8H18 (114.23 g mol−1), the latter boils at a substantially higher temperature. In general, dispersion forces are greater in molecules with greater molecular weight, more total electrons and a larger surface area (i.e. the shape of the molecule). Since noctane is a long skinny molecule, more of its electrons are accessible and ready to form instantaneous dipoles.

Which of the following statements is/are true concerning the first law of thermodynamics? I) The internal energy (U) of the universe is conserved. II) The internal energy of a system plus that of its surroundings is conserved. III) The change in internal energy (∆U) of a system and its surroundings can have the same sign. 1. I, II 2. III only 3. II, III 4. I only 5. I, III 6. II only 7. I, II, III

1. I, II correct Statement I and II are true; the first law states that the internal energy of the universe is conserved and since the system plus the surroundings is the universe, their sum is also conserved. Statement III is false; for example, if both the system and its surroundings had a positive change in internal energy, then the internal energy of the universe would have increased or decreasesed - in violation of the first law.

Why is I2 a solid while Cl2 is a gas even though they are both halogens? 1. I2 is more polarizable than Cl2. 2. I2 has H-bonding and Cl2 does not. 3. I2 has a larger dipole than Cl2. 4. I2 has a smaller dipole than Cl2. 5. I2 is less polarizable than Cl2

1. I2 is more polarizable than Cl2. correct I2 has significantly more electrons than Cl2. In addition, those electrons are in orbitals that are significantly farther from the nucleus, making it easier to distort the electron cloud.Higher polarizability leads to higher London dispersion forces. Neither molecule has a dipole nor H-bonding.

Which of the following species has the greatest charge density? 1. Mg2+ 2. K+ 3. Cr+ 4. Ca2+ 5. Al

1. Mg2+ correct charge density is the ratio of magnitude of charge to ionic size (radius). So the ion with the most charge and smallest size would have the highest charge density. Aluminum is small, but has no charge. Mg2+ will have the greatest charge density.

Which of the following compounds would be expected to have the longest N-O bonds? 1. NO3− 2. they will all be the same 3. NO2− 4. NO

1. NO3− correct NO2-and NO3- are both best depicted as resonance structures that have bonds somewhere between single and double bonds. NO has a double bond. The NO3- has a double bond delocalized over three bonds for 1.3 bond. The NO2- is 1.5 bond. Therefore the longest bond will be in NO3-

Rank the compounds CH3CH2OH CH3NH2 CH3OH NaOH in terms of increasing vapor pressure. 1. NaOH < CH3CH2OH < CH3OH < CH3NH2 2. CH3NH2 < CH3OH < CH3CH2OH < NaOH 3. NaOH < CH3NH2 < CH3OH < CH3CH2OH 4. CH3CH2OH < CH3OH < CH3NH2 < NaOH 5. NaOH < CH3OH < CH3NH2 < CH3CH2OH

1. NaOH < CH3CH2OH < CH3OH < CH3NH2 correct

For which of the following chemical equations would ∆H◦rxn = ∆H◦f? 1. O2(g) + H2(g) → H2O2(ℓ) 2. C(s, graphite) + (3/2)O2(g) + H2(g) →CO2(g) + H2O(g) 3. CO(g) + (1/2)O2(g) → CO2(g) 4. N2(ℓ) + 3 F2(g) → 2 NF3(ℓ)

1. O2(g) + H2(g) → H2O2(ℓ) correct For O2(g) + H2(g) → H2O2(ℓ), ∆H◦f of O2(g) and H2(g) are 0. Therefore, ∆H◦rxn =∆H◦f(H2O2(ℓ))

What is the strongest evidence for hydrogen bonding? 1. The boiling points of NH3, H2O, and HF are abnormally high compared with the rest of the hydrides in their respective periods. 2. Hydrogen is able to accept or donate electrons, so it is the most versatile atom in the periodic chart. 3. Hydrogen has an extremely low electronegativity. 4. Hydrogen can be considered either a metal or nonmetal.

1. The boiling points of NH3, H2O, and HF are abnormally high compared with the rest of the hydrides in their respective periods. correct Hydrogen bonds will pose as strong IMFs that will thus make it harder to boil, which is they have such high boiling points

Which of the following statements is true? 1. The magnitude of the heat released by the system will equal the magnitude of the heat absorbed by the surroundings. 2. The standard molar entropy of an element in its standard state is zero. 3. The change in the entropy of the system is always positive for a spontaneous process. 4. The magnitude of the entropy change of the system will always equal the magnitude of the entropy change for the surroundings for all phase changes. 5. The Gibb′s free energy change of the system must be positive for a spontaneous pro

1. The magnitude of the heat released by the system will equal the magnitude of the heat absorbed by the surroundings. correct

Under standard conditions, neopentane is a gas while n-pentane is a liquid. Neopentane has a symmetrical tetrahedral shape N-pentane has a linear straight chain structure Given that both are non-polar and have identical chemical formulas (C5H12), the reason that n-pentane has a higher boiling point is that 1. its straight-chain structure allows the molecules to have more instantaneous dipoles interacting at close distances. 2. it has more C-H bonds that are slightly polar. 3. it has more H-bonding than the neopentane. 4. n-pentane has more valence electrons and therefore can form more instantaneous dipoles. 5. its straight-chain structure allows for a substantial permanent dipole.

1. its straight-chain structure allows the molecules to have more instantaneous dipoles interacting at close distances. correct The more branched a hydrocarbon, the less the effective surface area and therefore less dispersion forces. The more linear (straight chain) a hydrocarbon is, the more surface area and hence more dispersion forces.

Which of the following is not a permitted combination of quantum numbers? 1. n = 2, ℓ = 1, mℓ = −2, ms =1/2 2. n = 4, ℓ = 2, mℓ = 1, ms =1/2 3. n = 4, ℓ = 3, mℓ = 3, ms = −1/2 4. n = 3, ℓ = 0, mℓ = 0, ms = −1/2 5. n = 3, ℓ = 0, mℓ = 0, ms =1/2

1. n = 2, ℓ = 1, mℓ = −2, ms =1/2 correct For n = 2, mℓ can only have values of −1, 0, and +1.

An isolated system allows for the flow of...? 1. none of these 2. sound waves 3. kinetic energy 4. matter 5. heat

1. none of these correct In the natural sciences an isolated system is a physical system without any external exchange- neither matter nor energy (as heat or work) can enter or exit, but can only move around inside.

When a gas phase reaction takes place... 1. some state functions may increase, others may decrease and some may stay constant. correct 2. all state functions must increase or stay constant. 3. all state functions must stay constant. 4. all state functions must change. 5. all state functions must decrease or stay constant.

1. some state functions may increase, others may decrease and some may stay constant. correct

Gas X has a larger value than Gas Y for the van der Waals constant "a". This indicates that 1. the molecules of X have stronger intermolecular attractions for each other than the molecules of Y have for each other. 2. the molecules of gas X have a higher velocity than do the molecules of gas Y. 3. the molecules of X are larger than the molecules of Y. 4. the molecules of gas X repel other X molecules.

1. the molecules of X have stronger intermolecular attractions for each other than the molecules of Y have for each other. correct just think higher a stands for higher "a"ttraction

What is the best explanation for the fact that the ionization energy of boron is lower than that of beryllium? 1. the quantum mechanical stability of the filled s subshell of beryllium 2. boron has a larger effective nuclear charge than beryllium. 3. the increased repulsion experienced by the electrons in beryllium 4. the radius of boron is smaller than that of beryllium making it easy to remove an electron

1. the quantum mechanical stability of the filled s subshell of beryllium correct The additional stability inherent to a filled or a half-filled subshell, such as beryllium's filled s subshell, increases the energy required to strip an electron, i.e. increases beryllium's ionization energy such that boron's is actually lower than beryllium's.

The root mean square speed of nitrogen molecules in air at 20◦C is 511 m/s in a certain container. If the gas is allowed to expand to twice its original volume, the root mean square velocity of nitrogen molecules drops to 325 m/s. Calculate the temperature after the gas has expanded. 1. −154◦C 2. −45.1◦C 3. 154◦C 4. 347◦C 5. −347◦C 6. −261◦C 7. 45.1◦C 8. 261◦C

1. −154◦C correct T1 = 20◦ C + 273.15 = 293.15 K vT1 = 511 m/s vT2 = 325 m/s From kinetic molecular theory, temperature is directly proportional to mean KE. KEmean =(1/2)(MW)(average molecular speed)^2 and the MW is constant (it's the same gas) (vT1)/(vT2)=(√T1)/(√T2) = = 118.581 K , so the final temperature is 118.581 K − 273.15 = −154.569◦C .

A CD player and its battery together do 500 kJ of work, and the battery also releases 250 kJ of energy as heat and the CD player releases 50 kJ as heat due to friction from spinning. What is the change in internal energy of the system, with the system regarded as the battery and CD player together? 1. −800 kJ 2. +200 kJ 3. −200 kJ 4. −700 kJ 5. −750 kJ

1. −800 kJ correct Heat from the CD player is −50 kJ. Heat from the battery is −500 kJ. Work from both together on the surroundings is −250 kJ. This question is testing your ability to see what the system is, and then look at ONLY the energy flow for the system. Here the system is the battery and the CD player together. ∆U = q + w = [−50 kJ + (−250 kJ)] + (−500 kJ) = −800 kJ

When two samples of ideal gases have the same _______ , their molecules must have the same _______ . 1. pressure; average kinetic energy 2. mass; average kinetic energy 3. density; mass 4. density; average kinetic energy 5. volume; average kinetic energy 6. volume; mass 7. pressure; mass 8. mass; density 9. temperature; speed 10. temperature; average kinetic energy

10. temperature; average kinetic energy correct From kinetic molecular theory: For samples of an ideal gas temperature and average kinetic energy are directly proportional.

What is the ground state electron configuration for chromium?

1s2 2s2 2p6 3s2 3p6 4s1 3d5 correct Chromium is an exception to the rule in filling the subshells. There is extra stability afforded to the configuration by having half filled sub-shell. So instead of the 4s subshell filling up completely, the electrons are evenly distributed between the 4s and 3d levels.

If a particle is confined to a one-dimensional box of length 300 pm, for Ψ3 the particle has zero probability of being found at 1. 50 and 250 pm, respectively. 2. 100 and 200 pm, respectively. 3. 75, 125, 175, and 225 pm, respectively. 4. 50, 150, and 250 pm, respectively. 5. 150 pm only.

2. 100 and 200 pm, respectively. correct Box of length 300pm with a Ψ3 means that there are 3 humps and 2 nodes. Zero probability is found at the nodes. Drawing it helps fam.

A 22.4 L vessel contains 0.02 mol H2 gas, 0.02 mol N2 gas, and 0.1 mol NH3 gas. The total pressure is 700 torr. What is the partial pressure of the H2 gas? 1. 28 torr 2. 100 torr correct 3. None of these 4. 14 torr 5. 7 torr

2. 100 torr correct ntotal = 0.14 mol Ptotal = 700 torr nH2 = 0.02 mol XH2 = nH2 / ntotal = 0.02 mol / 0.14 mol = 0.142857 PH2 = XH2 Ptotal = (0.142857) (700 torr) = 100 torr

A vessel contains 0.1 mol H2 gas, 0.1 mol N2 gas, and 0.3 mol NH3 gas. The total pressure is 1000 torr. What is the partial pressure of the H2 gas? 1. 100 torr 2. 200 torr 3. 1000 torr 4. 500 torr 5. 800 torr

2. 200 torr correct P=1000 torr times (0.1mol)/(0.1mol+0.1mol+0.3mol) = 200 torr

Consider the balanced reaction for the combustion of methane below. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) If 100 g of methane react completely with 100 g of molecular oxygen in a sturdy, closed 100 L vessel at 227 ◦C, what is the approximate final pressure in the vessel? 1. 1.92 atm 2. 3.85 atm 3. 0.87 atm 4. 1.75 atm

2. 3.85 atm correct Although we have equal masses of methane and oxygen, their different molar masses and different stoichiometric coefficients result in oxygen being the limiting reagent. 100 g CH4 × (1 mol / 16 g) = 6.25 mol 100 g O2 × (1 mol / 32 g) = 3.125 mol In addition to having fewer moles of oxygen, 2 moles of oxygen are required to react with 1 mole of methane; oxygen is definitely the limiting reagent. From this, we can be certain that 0 moles of oxygen will remain once the reaction has gone to completion. In order to answer the question, we will need to determine the total moles of methane remaining and add to that the number of moles of products. 3.125 mol O2 × 1 CH4 / 2 O2 = 1.5625 mol CH4 6.25 − 1.5625 mol CH4 = 4.6875 mol CH4 3.125 mol O2 × 3/ 2 O2 = 4.6875 mol products Thus, once the reaction has gone to completion, there are a total of 9.375 moles of gas on the container. Under the specified conditions, that produces a pressure of: P = nRT/V= (9.375 × 0.0821 × 500)/100= 3.85 atm

If the interaction energy between a sodium ion and a chloride ion in table salt is 760 kJ/mol, what is the interaction energy between a zinc ion (Zn2+) and a sulfide ion (S2−) in a hypothetical structure in which the inter-ionic distances are the same as that of NaCl? 1. 1.140 kJ/mol 2. 3040 kJ/mol 3. 1520 kJ/mol 4. 1140 kJ/mol 5. 760 kJ/mol

2. 3040 kJ/mol correct qZn = +2 C qS = −2 C qNa = +1 C qCl = −1 C VNaCl = 760 kJ/mol If r is the distance between the two ions (the sum of the ionic radii), the energy interaction between ions is given by Coulomb's Law: VZnS = (qZn qS VNaCl) / (qNa qCl) = (+2 C)(−2 C)(−760 kJ/mol) / (+1 C)(−1 C) = −3040 kJ/mol the energy released by the interaction.

An important reaction that takes place in the atmosphere is NO2(g) −→ NO(g) + O(g) which is brought about by sunlight. Calculate the standard enthalpy of the reaction from the following information reaction. O2(g) → 2 O(g) ∆H◦(kJ)= +498.4 NO(g) + O3(g) → NO2(g) + O2(g) ∆H◦(kJ)=−200.0 (3/2)O2(g) → O3(g) ∆H◦(kJ)= +142.7 1. 449.2 kJ 2. 306.5 kJ 3. 820.5 kJ 4. 963.8 kJ 5. 320.2 kJ 6. 555.7 kJ 7. 106.5 kJ

2. 306.5 kJ correct Using Hess' Law we add the reverse (flip) of reaction 2; the reverse (flip) of reaction 3; and one half of reaction 1: NO2(g) + O2(g) −→ NO(g) + O3(g) ∆H◦(kJ) =200 O3(g) −→(3/2)O2(g) ∆H◦(kJ) =−142.7 (1/2)O2(g) −→ O(g) ∆H◦(kJ)=+249.2 ___________________________________________________________________________________ NO2(g) −→ NO(g) + O(g) ∆H◦(kJ)=306.5

Determine the molecular weight of a compound given the gas phase data: mass of 27.64 g, temperature of 18.4◦C, pressure of 956 torr, volume of 14.32 L. 1. 279 g/mol 2. 36.7 g/mol 3. 0.046 g/mol 4. 24.9 g/mol 5. 191 g/mol

2. 36.7 g/mol correct m = 27.64 g V = 14.32 L T = 18.4◦C + 273.15 = 291.55 K P = 956 torr R = 0.08206 L·atm/mol·K Using the ideal gas law, P V = n R T n = P V / R T plug everything in fam = 0.752909 mol gas If 0.752909 moles of the gas weigh 27.64 g, the molecular weight is MW = 27.64 g / 0.752909 mol = 36.7109 g/mol .

An electron in a hydrogen atom moves from the n = 2 to n = 5 level. What is the wavelength of the photon that corresponds to this transition and is the photon emitted or absorbed during this process? 1. 1875 nm; absorbed 2. 434 nm; absorbed 3. 434 nm; emitted 4. 1875 nm; emitted 5. 276 nm; emitted

2. 434 nm; absorbed correct In order to jump to a higher energy level, the photon must be ABSORBED by the atom. The wavelength can be found using Rydberg's equation

The specific heat of liquid water is 4.184 J/g·◦C, and of steam 2.03 J/g·◦C. The heat of vaporization of water (ℓ) is 2.26 kJ/g and its boiling point is 100◦C. What is the total heat flow when 18 grams of water at 12◦C are heated to become steam at 109◦C? 1. 44.4 kJ 2. 47.6 kJ 3. under 28 kJ 4. 48.9 kJ 5. over 55 kJ 6. 31.7 kJ 7. 40.7 kJ

2. 47.6 kJ correct Just draw out the heat curve man idk

If a particle is confined to a one-dimensional box of length 300 pm, for Ψ3 the particle is most likely to be found at 1. 300 pm. 2. 50, 150, and 250 pm, respectively. 3. 17.3 pm. 4. 100 and 200 pm, respectively. 5. 0 pm.

2. 50, 150, and 250 pm, respectively. correct In a box of length 300pm with Ψ3 means that there will be 3 curves and 2 nodes . The peaks of the curve is where the particle is most likely to be found. Therefore if drawn correctly, 50, 150, and 250 pm are respectively the peaks of a 300pm box with a Ψ3.

What volume will 40.0 L of He at 50.00◦C and 1201 torr occupy at STP? 1. 26.7 L 2. 53.4 L 3. 12.8 L 4. 18.6 L 5. 31.1 L

2. 53.4 L correct P1 = 1201 torr P2 = 760 torr V1 = 40 L T2 = 273.15 K T1 = 50◦ + 273.15 = 323.15 K P1 = 1201 torr P2 = 760 torr V1 = 40 L T2 = 273.15 K T1 = 50◦ + 273.15 = 323.15 K Using the Combined Gas Law, (P1 V1) / (T1) = (P2 V2) / T2 and recalling that STP implies standard temperature (273.15 K) and pressure (1 atm or 760 torr), we have V2 = (P1 V1 T2) / (T1 P2) = =(1201 torr) (40.0 L) (273.15 K) / (323.15 K) (760 torr) = 53.4302 L

Estimate the heat released when ethene (CH2=CH2) reacts with HBr to give CH3CH2Br. 1. 200 kJ/mol 2. 76 kJ/mol correct 3. 1036 kJ/mol 4. 424 kJ/mol 5. 470 kJ/mol

2. 76 kJ/mol correct ∆H = Ebreak − Emake

Calculate the velocity of an oxygen molecule if it has a de Broglie wavelength of 0.0140 nm. 1. 3 × 108 m/s 2. 891 m/s 3. 1780 m/s 4. 445 m/s 5. 8.9 m/s

2. 891 m/s correct λ = 0.0140 nm = 1.4 × 10^−11 m m = 32 g/mol / 6.022 × 1023 = 5.31385 × 10−23 g = 5.31385 × 10−26 kg λ =h / m v v =h / m λ = 890.665 m/s

The wavelength of light with a frequency of 3.30 × 1014 s^−1 is 1. 450 nm. 2. 909 nm. 3. 200 nm. 4. 650 nm.

2. 909 nm. correct c = λν where c= 3 x 10^8 Given frequency (ν), rearrange the equation to find wavelength (λ)

2.26 g of liquid water at 23.5 ◦C was completely converted to ice at 0 ◦C. How much heat was (absorbed/released) by the system during this process? 1. 755 J; released 2. 977 J; released 3. 755 J; absorbed 4. 977 J; absorbed 5. 1478 J; absorbed 6. 1478 J; released

2. 977 J; released correct for 1 gram (cooling + freezing): 23.5(4.184) + 334 = 432.324 J/g scale up to 2.26 g : 432.324(2.26) = 977.052 J = 977 J released

Generally if a liquid has stronger intermolecular attractions it will have A. a higher vapor pressure; B. a higher viscosity; C. a higher boiling point. 1. A only 2. B and C only 3. C only 4. A and C only 5. B only 6. None of the properties 7. A, B, and C 8. A and B only

2. B and C only correct The stronger intermolecular forces will lead to higher viscosity and a higher boiling point. The high boiling point is due to a lower vapor pressure

Acetonitrile has an "a" of17.81 and "b" of 0.117 Butane has an "a" of 14.66 and "b" of 0.123 Freon has an "a" of 10.78 and "b" of 0.0998 Which gas molecule do you expect to be the largest? (a and b are Van der Waals constants.) 1. Acetonitrile 2. Butane 3. Freon

2. Butane correct The Van der Waals constant a represents the role of attractions, so it is relatively large for molecules that attract each other strongly. The Van der Waals constant b represents the role of repulsions and can be thought of as representing the volume per mole of molecules. The larger the b value, the larger the molecules in the gas.

All of the species below have the same bond order except for one of them. Which is it? 1. Ne+2 2. B−2 correct 3. H−2 4. F−2 5. H+2

2. B−2 correct All of the species have a bond order of 0.5 except for B−2, which has a bond order of 1.5.

Which response includes only species that have the electron configuration 1s2, 2s2, 2p6, 3s2, 3p6, and no other species? 1. Na+, K+, Ar 2. Cl−, K+, Ar, P3− 3. Cl−, Na+ 4. Na+, Ar, P3− 5. Cl−, K+, Ar, Mg2+

2. Cl−, K+, Ar, P3− correct For a set of species to have the same electron configuration, they must be isoelectronic. Only one answer choice satisfies this criterion.

Consider a thermodynamic system that is simultaneously releasing heat and doing work. The internal energy of this system will: 1. Increase 2. Decrease 3. Stay exactly the same. 4. Increase, decrease, or stay the same depending on the magnitudes of heat and work

2. Decrease correct The change in internal energy is equal to the sum of the heat absorbed by the system and work done on the system based on the equation: ∆U = q + w. In this case, q and w are both negative. Therefore the internal energy will be decreasing regardless of the magnitudes of heat and work.

Which one shows the substances in the decreasing order of their molar entropy? 1. C(s), H2O(ℓ), H2O(g), H2O(s) 2. H2O(g), H2O(ℓ), H2O(s), C(s) 3. None of these 4. H2O(s), H2O(ℓ), H2O(g), C(s) 5. C(s), H2O(g), H2O(ℓ), H2O(s) 6. C(s), H2O(s), H2O(ℓ), H2O(g)

2. H2O(g), H2O(ℓ), H2O(s), C(s) correct Gases will have a higher entropy than liquids so we expect H2O(ℓ) to have the lowest molar entropy. The gases will increase in entropy in the order Ne(g) < Ar(g) < CO2(g). Ne and Ar are both atoms so they should have less entropy than a molecular substance, which has more complexity. Ar will have a higher entropy than Ne because it has a larger mass and more fundamental particles. The correct order is H2O(ℓ) < Ne(g) < Ar(g) < CO2(g).

Consider the Maxwell-Boltzmann distribution plots for a series of gases with different molar masses but with identical temperatures. Which of the statements below is a correct description of the shape and position of the plot as it pertains to the relative mass of the gas. 1. Lighter molecules tend to have much narrower ranges of velocities and larger average velocities. 2. Heavier molecules tend to have much narrower ranges of velocities and smaller average velocities. 3. Heavier molecules tend to have much broader ranges of velocities and larger average velocities. 4. Lighter molecules tend to have much broader ranges of velocities and smaller average velocities.

2. Heavier molecules tend to have much narrower ranges of velocities and smaller average velocities. correct More massive particles carry the same kinetic energy as smaller ones at much lower velocities because m is much bigger in the equation: Ek = 1/2mv2 Same temperature means the same Ek so the velocities are relatively less than lighter molecules. In addition, the ranges is much more narrow (less spread) in the distribution

Consider four molecules I) CHCl3 II) CH4 III) CH3Cl IV) CCl4 Which of these exhibit permanent dipole-dipole interactions? 1. III only 2. I and III only 3. I only 4. None of these 5. I, III, and IV only

2. I and III only correct Dipole-dipole interactions occur in polar molecules. CHCl3 and CH3Cl are polar because their dipole moments do not cancel. CH4 and CCl4 are symmetric; their dipole moments cancel and the overall molecule is non-polar.

Which of the following statements regarding intermolecular forces (IMF) is/are true? I) Intermolecular forces result from attractive forces between regions of positive and negative charge density in neighboring molecules. II) The stronger the bonds within a molecule are, the stronger the intermolecular forces will be. III) Only non-polar molecules have instantaneous dipoles. 1. II and III 2. I only 3. III only 4. I and II 5. II only 6. I, II, and III 7. I and III

2. I only correct Statement I is true - all IMF result from Coulombic attraction. Statements II and III are both false; the strength of the bonds within a molecule have no bearing on the strength of the bonds between molecules; all molecules have London forces.

Acetic acid (CH3COOH) forms a molecular solid. What type of forces hold it in a solid configuration? I) London forces II) dipole-dipole forces III) hydrogen bonding 1. I only 2. I, II, and III 3. III only 4. II only 5. II and III only 6. I and II only

2. I, II, and III correct Acetic acid will exhibit both London forces and dipole-dipole interactions, in addition to hydrogen bonding.

Which of the following BEST describes the purpose of effective nuclear charge? 1. It is used to rationalize chemical bonding in covalently bonded molecules. 2. It is a method to evaluate how much attraction a given electron "feels" from the nucleus so that periodic trends can be predicted and rationalized. 3. It is used to determine the number of valence electrons of a given species. 4. It exists only to torture foolish CH 301 students who did not study. 5. It is a measure of the effect of filled and half-filled subshells on the stability of atoms and ions. 6. It is a measure of how many protons a given atom has which is useful because of variations from isotope to isotope.

2. It is a method to evaluate how much attraction a given electron "feels" from the nucleus so that periodic trends can be predicted and rationalized. correct Inner shell electrons "shield" the outer shell electrons from the full attraction of the nucleus. An electron in a higher shell, farther from the nucleus, feels much less attraction, for example; in other words the effective nuclear charge it experiences is smaller. This is used to rationalize the periodic trends.

The key to an effective photovoltaic material, is to have a dye that can absorb electromagnetic radiation to cause a promotion of an electron to an excited state with a minimum energy gap. That excited electron is then routed around a circuit to create a current and a voltage. Material A absorbs in the red region of the visible spectrum while material B absorbs in the blue region of the visible spectrum. Assuming the materials cost about the same amount of money to manufacture, which material would be the better choice for the solar cell? 1. Material A would be a better choice because it would have larger HOMO LUMO gap. 2. Material A would be a better choice because it would have smaller HOMO LUMO gap. 3. Material B would be a better choice because it would have smaller HOMO LUMO gap. 4. Material B would be a better choice because it would have larger HOMO LUMO gap

2. Material A would be a better choice because it would have smaller HOMO LUMO gap. correct Since A absorbs in the red region of light which has a lower energy comparing to the blue region, A has a smaller HOMO LUMO gap and thus is a better choice.

Rank the crystal lattice energy of the salts Al2O3, CaCl2, CaO, NaF, Mg3(PO4)2 from least to greatest: 1. NaF < CaO < CaCl2 < Mg3(PO4)2 <Al2O3 2. NaF < CaCl2 < CaO < Mg3(PO4)2 <Al2O3 3. CaO < NaF < CaCl2 < Al2O3 <Mg3(PO4)2 4. Mg3(PO4)2 < NaF < < CaO CaCl2 <Al2O3 5. Mg3(PO4)2 < NaF < CaCl2 < CaO <Al2O3

2. NaF < CaCl2 < CaO < Mg3(PO4)2 <Al2O3 correct Sodium fluoride has the least lattice energy because of its small charges, then calcium chloride and calcium oxide with incrementally increasing charges. Both magnesium phosphate and aluminum oxide have identical magnitudes for their charges, but both of the latter are smaller than than their counterparts in the former and consequently have a greater lattice energy

A particular metal has a work function of 3.05 eV. A light is shined onto this metal with a corresponding wavelength of 524 nm. What is the maximum velocity of the photoelectrons produced? 1. 9.12 × 105 ms−1 2. No photoelectrons are produced. 3. 8.32 × 1011 ms−1 4. 8.72 × 108 ms−1

2. No photoelectrons are produced. correct Using the photoelectric equation: Ek =1/2mv^2 = hν − φ The KE of any photoelectron produced is the remainder after the work function has been subtracted from the energy of the photon. Ephoton = [(6.626 × 10^−34J/s)(3.00 × 10^8 m/s)]/[524 × 10^−9m] Ephoton = 3.794 × 10^−19J Work function = 3.05 eV (1.60 × 10−19 J/eV)= 4.88 × 10^−19J Since the energy of the photon is less than the work function, no photoelectrons are produced.

Which of the following statements about boiling is false? 1. As intermolecular forces increase, boiling point increases as well. 2. The boiling point of a liquid is independent of atmospheric pressure. 3. Boiling occurs when vapor pressure exceeds atmospheric pressure. 4. For a given pressure, the boiling point is always at a higher temperature than melting point.

2. The boiling point of a liquid is independent of atmospheric pressure. correct Boiling point is directly proportional to atmospheric pressure.

The observation and characterization of black body radiation was key to the development of our quantum mechanical view of the atom. Which statement below is correct regarding blackbody radiation? 1. Theoreticians were unable to explain why blackbody radiators emitted so much UV light at room temperature, which became known as the "UV Catastrophe". 2. The peak intensity of the emission curve for a blackbody shifts to shorter wavelengths as the temperature of the blackbody increases. 3. The emission from a blackbody consists of a series of discrete spectral lines, the colors of which are unique to the type of blackbody used. 4. The emission of radiation from an incandescent lightbulb does not resemble the emission of radiation from a blackbody.

2. The peak intensity of the emission curve for a blackbody shifts to shorter wavelengths as the temperature of the blackbody increases. correct A blackbody's radiation emission curve is broad, lopsided and has a peak which shifts to shorter wavelengths as the temperature of the blackbody increases. Theoreticians tried to model this using classical mechanics and never could get their models to behave because they predicted vast amounts of UV, gamma, etc. being produced by a blackbody above absolute zero - the "UV Catastrophe". A plain old incandescent lightbulb with a dimmer switch can be used to reproduce the emission curves of a blackbody - and even vary its temperature by changing the current with the dimmer switch.

Which of the following is NOT a feature of the bomb calorimetry apparatus used to measure the internal energy of a reaction? 1. The heat capacity of the calorimeter should be known to accurately correct for any heat lost to it. 2. The thermometer is inserted directly into the reaction vessel to measure ∆T of the reaction. 3. Large quantities of water surrounding the reaction vessel absorb the majority of the heat loss. 4. The volume of the reaction vessel is held constant to eliminate energy released as work. 5. The large heat capacity of water is beneficial in measuring heat released by combustion reactions.

2. The thermometer is inserted directly into the reaction vessel to measure ∆T of the reaction. correct The thermometer is placed in the water that surrounds the reaction vessel.

Identify the dominant intermolecular force in the following species, respectively: RbCl, C6H6 (benzene), HI, Fe2O3, CH2NH. a) ionic forces b) hydrogen bonding c) dipole-dipole d) instantaneous dipoles 1. a, c, c, d, b 2. a, d, c, a, b 3. a, b, c, b, a 4. a, b, d, a, c 5. c, b, d, c, c 6. b, d, c, d, d 7. c, d, a, a, b

2. a, d, c, a, b correct Rubidium Chloride and Iron(III) oxide are both ion-ion. Benzene is non-polar and thus has only van der Waal's forces. Hydroiodic acid is polar and has dipole-dipole interactions. Methylimine has H-bonding.

Helium has a rms velocity (vrms) that is 4.21 times faster than which of the following gases? 1. neon, Ne 2. chlorine, Cl2 3. xenon, Xe 4. oxygen, O2 5. argon, Ar

2. chlorine, Cl2 correct xplanation: The relationship of rms velocities between different molecules is: rateA / rateB = sqrt (massB / massA) Let massA be helium's mass of 4.0 and 4.21 be heliums rate so that: 4.21 / 1 = sqrt(massB / 4.0) (4.21)^2 = massB / 4.0 70.9 = massB This only matches the chlorine gas, Cl2.

Hund's rule states that 1. it is impossible to determine accurately both the momentum and position of an electron simultaneously. 2. electrons occupy all the orbitals of a given sublevel singly before pairing begins. 3. no two electrons in an atom may have identical sets of four quantum numbers.

2. electrons occupy all the orbitals of a given sublevel singly before pairing begins. correct Literally the definition of hund's rule dawg

Under what conditions is a gas most likely to deviate from ideal behavior? 1. low density 2. high pressure 3. when considering noble gases 4. high temperatures

2. high pressure correct Under high pressure, there are many more collisions between molecules. This increases the likelihood that a non-ideal interaction will occur.

Classify the molecule AsCl3. 1. nonpolar molecule with nonpolar bonds 2. polar molecule with polar bonds 3. polar molecule with nonpolar bonds 4. nonpolar molecule with polar bonds

2. polar molecule with polar bonds correct The difference in electronegativity between As and Cl make polar bonds. When you draw the lewis structure, there is a lone pair on As, implying that the molecule has a dipole moment, meaning it's polar.

Balance the equation Al2(SO4)3+ NaOH → Al(OH)3+ Na2SO4 using the smallest possible integers. What is the sum of the coefficients in the balanced equation? 1. fourteen 2. twelve 3. eight 4. six 5. ten

2. twelve correct Use a balance scale or something, idk fam

Calculate ∆S◦surr at 298 K for the reaction 6 C(s) + 3 H2(g) → C6H6(ℓ) ∆H◦r = +49.0 kJ·mol−1 and ∆S◦r = −253J·K−1·mol−1 1. − 417 J·K−1·mol−1 2. − 164 J·K−1·mol−1 3. +253 J·K−1·mol−1 4. − 253 J·K−1·mol−1 5. +164 J·K−1·mol−1 .

2. − 164 J·K−1·mol−1 correct ∆H◦r = 49000 J · mol−1 T = 298 K ∆S◦surr = qsurr / T= −q / T =−∆H◦r / 298 K = −164.43 J · K−1· mol−1.

Calculate the ∆S surr for the following reaction at 25◦C and 1 atm. Br2(ℓ) → Br2(g) ∆H◦rxn = +31 kJ 1. +93 J/K 2. −104 J/K 3. −93 J/K 4. −124 J/K 5. +124 J/K 6. +104 J/K

2. −104 J/K correct In general for any process: ∆Ssurr =−∆Hsys/Tsurr This is because the heat flow in the surroundings is just the opposite of the heat flow for the system (qsurr = −qsys and at constant pressure the heat is equal to ∆H. therefore ∆Ssurr = −31000/298= −104 J/K

For the combustion reaction of ethylene (C2H4) C2H4 + 3 O2 → 2 CO2 + 2 H2O assume all reactants and products are gases, and calculate the ∆Hrxn using bond energies. 1. 251 kJ/mol 2. −1300 kJ/mol 3. 0 kJ/mol 4. −251 kJ/mol 5. 1300 kJ/mol 6. 680 kJ/mol 7. −680 kJ/mol

2. −1300 kJ/mol correct ∆H0rxn =BE reactants −BE products= =[ (C=C) + 4 (C-H) +3 (O=O)] − [4 (C=O) + 4 (H-O) ] = [ (602kJ/mol)+ 4 (413 kJ/mol) + 3(498 kJ/mol)] − [4(799 kJ/mol)+ 4(463 kJ/mol)] = −1300kJ/mol

Calculate the standard reaction enthalpy (∆H◦rxn) for the final stage in the production of nitric acid, when nitrogen dioxide dissolves in and reacts with water: 3NO2(g) + H2O(ℓ) → 2HNO3(aq) + NO(g) 1. −104 kJ 2. −137 kJ 3. +136 kJ 4. −370 kJ 5. −304 kJ 6. +70 kJ

2. −137 kJ correct Values for ∆H◦f from external table are in order (from reaction) +33, -286, -207, and+90 ∆H◦rxn =(n ∆H◦jproducts)−(n ∆H◦jreactants)= (2∆H◦f, HNO3(aq) + ∆H◦f, O(g)) - (3∆H◦f, NO2(g) + ∆H◦f, H2O(ℓ))= (2(−207) + 90) − (3(33) + (−286) = −137 kJ

Calculate the standard reaction enthalpy for the reaction N2H4(ℓ) + H2(g) → 2 NH3(g) given N2H4(ℓ) + O2(g) → N2(g) + 2H2O(g) ∆H◦ = −543 kJ · mol−1 2 H2(g) + O2(g) → 2 H2O(g) ∆H◦ = −484 kJ · mol−1 N2(g) + 3 H2(g) → 2 NH3(g) ∆H◦ = −92.2 kJ · mol−1 1. −1119 kJ · mol−1 2. −151 kJ · mol−1 3. −935 kJ · mol−1 4. −243 kJ · mol−1 5. −59 kJ · mol−1

2. −151 kJ · mol−1 correct Reverse the 2nd reaction and add them

Calculate the standard enthalpy of combustion of butane (C4H10(g)) at 298 K from standard enthalpy of formation data. 1. −2056.49 kJ · mol−1 2. −2877.04 kJ · mol−1 3. −895.49 kJ · mol−1 4. −2843.5 kJ · mol−1 5. −2342.32 kJ · mol−1

2. −2877.04 kJ · mol−1 correct The balanced equation is C4H10(g) + 13/2O2 → 4 CO2(g) + 5 H2O(ℓ) Products - reactants

Consider a reaction that is non-spontaneous at all temperatures. What would be the signs of ∆Gsys, ∆Hsurr, and ∆Suniv respectively for such a reaction? 1. +, -, + 2. -, +, - 3. +, -, - 4. -, +, + 5. +, +, +

3. +, -, - correct For a reaction that is non-spontaneous at all temperatures, the free energy of the system will increase and the entropy of the universe will decrease. Such a reaction must be endothermic, and the heat it gains will be lost from the surroundings.

Methyl tert-butyl ether or MTBE is an octane booster for gasoline. The combustion of 0.9211 grams of MTBE (C5H12O(ℓ), 88.15g/mol) is carried out in a bomb calorimeter. The calorimeter's hardware has a heat capacity of 1.540 kJ/◦C and is filled with exactly 2.022 L of water. The initial temperature was 26.336◦C. After the combustion, the temperature was 29.849◦C. Analyze this calorimeter data and determine the molar internal energy of combustion (∆U) for this octane booster. 1. -2286 kJ/mol 2. -1957 kJ/mol 3. -3362 kJ/mol 4. -3120 kJ/mol 5. -4293 kJ/mol 6. -3560 kJ/mol 7. -2748 kJ/mol

3. -3362 kJ/mol correct ∆T = 29.849 − 26.336 = 3.513◦ Ccal = [2022(4.184) + 1540 ]/1000 = 10.00 kJ/◦C qcal = 10(3.513) = 35.130 kJ moles = 0.9211/88.15 = 0.01045 mol ∆U = -35.130 kJ / 0.01045 mol = -3362 kJ/mol

There are two glass bulbs (A & B) containing gas connected through a valve. The volume for gas A is 48L and the volume for gas B is 16L. The gases also happen to be at the same temperature (337K) and pressure (420 torr). After the valve is opened, the two gases mix completely. What is the partial pressure of gas B in this new (opened valve) state? 1. 525 torr 2. 420 torr 3. 105 torr 4. 168 torr 5. 84 torr 6. 280 torr 7. 140 torr 8. 210 torr

3. 105 torr correct When the valve is opened the total available volume for both gases is now 64 L but the total pressure remains 420 torr. This means (Boyle's Law) that the gas B pressure will drop by a factor of 16/64 or 1/4 or 0.25. 0.25(420) = 105 torr for gas B. Gas A has a partial pressure of 420-105 = 315 torr. Or you could know that volume ratio (before to after) for A is 48/64 or 3/4 or 0.75. 0.75 of 420 is 315 torr.

What is the entropy at T = 0 K for one mole of chloroform (CHCl3)? 1. 1.9 × 10−23 J K−1 2. 1.38 J K−1 3. 11.5 J K−1 4. −11.5 J K−1 5. 0 J K−1

3. 11.5 J K−1 correct Since the question is concerned with the residual entropy of a mole of chloroform, which as 4 orientations, we can use the Boltzmann equation to calculate the residual entropy. S = Na k ln W = R ln 4 = 11.5 J K−1

Which transition between energy levels in a hydrogen atom corresponds to the shortest wavelength of light. 1. 5 → 6 2. 2 → 4 3. 2 → 5 4. 3 → 4 5. 3 → 5 6. 2 → 3

3. 2 → 5 correct The gap between 2 to 5 gives off the highest energy out of the other choices. The energy (E)and wavelength (λ)of light are inversely proportional. So, the shortest wavelength would correspond to the highest energy [and the highest frequency (ν), though that isn't germane to this particular question]. Therefore the transition between n=2 and n=5 will have the highest energy and shortest wavelength.

What is the final volume if 20 L methane reacts completely with 20 L oxygen CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(ℓ) at 100◦C and 2 atmospheres? 1. 10 L 2. 30 L 3. 20 L 4. Cannot be determined from the information given. 5. 15 L

3. 20 L correct V1 = 20 L CH4 V2 = 20 L O2 Avogadro's Principle tells us that because P and T are the same, V ∝ n, so we can work with the volume instead of the number of moles. From the equation above, we need 2 times more L of O2 than that of CH4, but we only have 20 L, not 40 L of O2! This means O2 is the limiting reactant. Find out how much CO2 is made based on the L of O2: LCO2 = 20 L O2 × (1 L CO2 / 2 L O2) = 10 L CO2 We now find out how much CH4 was used reacting with all the L of O2: LCH4 = 20 L O2 × (1 L CH4 / 2 L O2) = 10 L CH4 This means 20 L − 10 L = 10 L CH4 are unused and still present. The final mixture is 10 L unreacted CH4 + 10 L CO2 = 20 L of gas. We assume the volume of the water is insignificant.

What is the total non-vibrational internal energy of 10 nitrous oxide (N2O) molecules? 1. 15 k T 2. 10 k T 3. 25 k T 4. 10 R T 5. 25 R T 6. 15 R T

3. 25 k T correct N2O is a linear molecule, therefore it has 5 modes of freedom. 5 x 1/2RT then multiply it by how many molecules/moles there are however in this case you use kT because it's referring to molecules. If it was in moles, you would use RT

Which of the reactions below is a formation reaction? 1. H2(g) + Cgraphite(s) + 3/2O2(g)→ CO2(g) + H2O(g) 2. 2H2(g) + 2Cgraphite(s) + O2(g)→ 2CH2O(ℓ) 3. 2Fe(s) + 3/2O2(g) → Fe2O3(s) 4. N2(ℓ) + 2H2(g) → N2H4(ℓ)

3. 2Fe(s) + 3/2O2(g) → Fe2O3(s) correct A formation reaction produces exactly one mole of one product from elements in their standard states.

The two reactions shown below are both endothermic. For which reaction is ∆H < ∆U? N2(g) + O2(g) → 2NO(g) 2NO(g) + O2(g) → 2NO2(g) 1. Neither reaction has ∆H < ∆U. 2. N2(g) + O2(g) → 2NO(g) 3. 2NO(g) + O2(g) → 2NO2(g) 4. Both reactions have ∆H < ∆U.

3. 2NO(g) + O2(g) → 2NO2(g) correct ∆U= q + w, q= ∆H w= -P∆V or -∆nRT If no work is done, then ∆H=∆U The reaction N2(g) + O2(g) → 2NO(g) has no change in moles and thus has zero work: (Final gas mol- Initial gas mol)=(2-2)=0 w=-(0)RT= 0 thus ∆H=∆U However, the reaction 2NO(g) + O2(g) → 2NO2(g) has a mol change of (-1): (Final gas mol- Initial gas mol)=(2-3)=-1 w=-(-1)RT= a value > 0 Therefore ∆U= q + value > 0, where q is smaller than ∆U

2.0 g of H2 and 8.0 g of He are put in a 22.4 liter container at 0◦C. The total pressure is 1. 1.0 atm. 2. 10.0 atm. 3. 3.0 atm. 4. 5.0 atm

3. 3.0 atm. correct V = 22.4 L nH2 = 2 g · mol / 2 g = 1 mol nHe = 8 g · mol / 4 g = 2 mol ntotal (add them up) = 3 mol T = 0◦C + 273 = 273 K PV = NRT Solve

What is the change in entropy (∆S) for the heating of 20.0 grams of methanol (CH3OH, liquid) from 34◦C to 62◦C? 1. -30.42 J/K 2. 0 J/K 3. 4.42 J/K 4. 168.81 J/K 5. 1418 J/K 6. 0.22 J/K 7. 30.42 J/K

3. 4.42 J/K correct The specific heat capacity of methanol is equal to 2.533J/g◦C via table of data. Use the equation: ∆S = m Cs ln (T2/T1) ∆S = 20(2.533) ln(335/307) = 4.42

Consider the molecular orbital diagram for oxygen, O2. What is the total number of electrons that occupy the π and π∗ molecular orbitals? How many of those electrons are unpaired? 1. 4 ; 2 2. 4 ; 0 3. 6 ; 2 4. 6 ; 0 5. 0 ; 0 6. 6 ; 4 7. 8 ; 2

3. 6 ; 2 correct There are 6 total in the π and π∗ orbitals. The two in the π∗ orbitals are unpaired.

When a certain element is excited with electricity, we see three main lines in its emission spectrum: Two red lines and one orange line. What would the absorption spectrum of this element look like? 1. Similar to the emission spectrum with two red lines and one orange line, but with each of those shifted to lower wavelengths 2. Identical to the emission spectrum, with two red lines and one orange line at the same wavelengths 3. A continuous spectrum broken by thin black lines at the same wavelengths as the red and orange lines in the emission spectrum 4. It would have lines complementary to the emission spectrum colors, so it would have two green lines and one blue line.

3. A continuous spectrum broken by thin black lines at the same wavelengths as the red and orange lines in the emission spectrum correct Literally what an absorption spectrum looks like fam

All of the species below have the same bond order except for one of them. Which is it? 1. H+2 2. F−2 3. B−2 4. H−2 5. Ne+2

3. B−2 correct All of the species have a bond order of 0.5 except for B−2, which has a bond order of 1.5.

Arrange the following atoms in order of increasing radius: Li, Au, C, Cs. 1. Li < C < Cs < Au 2. Cs < Au < Li < C 3. C < Li < Au < Cs 4. Au < Cs < C < Li

3. C < Li < Au < Cs correct Explanation: Atomic radius decreases across a period due to increasing effective nuclear charge (ENC) and increases down a group due to decreasing ENC and the increase in total electrons.

Which of the reactions below will likely have the largest increase in entropy (∆Srxn)? 1. 2 CH4(g) + 2O3(g) → 4 H2O(g) + 2 CO(g) 2. N2H4(g) + H2(g) → 2 NH3(g) 3. C5H12(ℓ) + 8 O2(g) → 6 H2O(g) + 5 CO2(g) 4. Na+(g) + Cl−(g) → NaCl(s) 5. S3(g) + 9 F2(g) → 3 SF6(g)

3. C5H12(ℓ) + 8 O2(g) → 6 H2O(g) + 5 CO2(g) correct The reaction with the greatest positive value for ∆ngas will have the greatest value of ∆Srxn.

The lattice energy of calcium bromide is the energy change for the reaction 1. CaBr2(s) → Ca(g) + Br2(g) 2. CaBr2(s) → Ca(g) + 2 Br(g) 3. CaBr2(s) → Ca2+(g) + 2 Br−(g) 4. Ca(g) + 2 Br(g) → CaBr2(g) 5. Ca(s) + Br2(ℓ) → CaBr2(s)

3. CaBr2(s) → Ca2+(g) + 2 Br−(g) correct Lattice energy is the energy change at constant pressure when a 1 mol of an ionic solid is broken into ions of its component elements in gaseous form at 298 K, 1 atm.

Write an equation that represents the second ionization energy of copper. 1. Cu+(g) −→ Cu2+(g) + 2 e− 2. Cu(g) −→ Cu+(g) + e− 3. Cu+(g) −→ Cu2+(g) + e− 4. Cu(g) −→ Cu2+(g) + e− 5. Cu(g) −→ Cu2+(g) + 2 e−

3. Cu+(g) −→ Cu2+(g) + e− correct it just is fam

C(s, graphite) has a ∆G◦f of 0. C(s, diamond) has a ∆G◦f of 2.9. C60(s, buckminsterfullerene) has a ∆G◦f of 2.4 Which of the following statements is supported by these data? 1. C60 is thermodynamically more stable than graphite under standard conditions. 2. Graphite could spontaneously form C60 under standard conditions. 3. Formation of graphite from C60 would be exergonic under standard conditions. 4. Diamond is the least thermodynamically stable allotrope of carbon under standard conditions.

3. Formation of graphite from C60 would be exergonic under standard conditions. correct A lower ∆G◦f means it's more stable. Therefore 1 and 4 are wrong. 2 is wrong because a spontaneous change means there's a negative ∆G but going from Graphite to C60 is a positive ∆G. 3 is correct because going from C60 to Graphite is a negative ∆G (products-reactants)

Which of the following statements concerning the Schrodinger equation and its solutions is true? I) Its solutions are wave functions. II) It can be used to determine an electron's exact position. III) Both attractive and repulsive V (r) terms are used when solving the Schr¨odinger equation for the hydrogen atom. 1. II only 2. II, III 3. I only 4. I, II 5. I, III 6. I, II, III 7. III only

3. I only correct Solutions to the Schrodinger equation are wave functions, which when squared express the probable location of electrons; but, the exact location cannot be known. Attractive potential energy terms are found in all solutions for all atoms. Repulsive potential energy terms are found only in atoms that have more than one electron, i.e. everything beyond hydrogen. Polar coordinates are preferred for 3-D solutions because they simplify the math.

Which of the following statements concerning hybrid orbitals is/are true? I) Hybrid orbitals are energetically degenerate. II) Any element can form sp3d2 hybrid orbitals. III) Hybridizing a 2s and a 2p orbital would produce one single sp hybrid orbital. 1. III only 2. II, III 3. I only 4. II only 5. I, II 6. I, III 7. I, II, III

3. I only correct Statement I is true; hybridization was developed as a theoretical framework to explain the energetic degeneracy of bonds in molecules. Statement II is false; hybridization involving d orbitals requires access to empty d orbitals, and thus begins in period 3. Statement III is false; the number of orbitals used to hybridize is always equal to the number of hybridized orbitals, so using a 2s and a 2p orbital would result in two sp hybrid orbitals.

Consider the electron filling diagram 3p ↑↓ ↑ 3s ↑ 2p ↑↓ ↑↓ ↑↓ 2s ↑↓ 1s ↑↓ for a ground state atom. Which of the following does it violate? I) The Aufbau principle II) Hund's rule III) The Pauli exclusion principle 1. III only 2. I, II, III 3. I, II 4. I only 5. II only 6. II, III 7. I, III

3. I, II correct Aufbau's principle says you must fill the orbitals in order from lowest energy to highest. Putting electrons in the 3p orbitals before filling the 3s orbital violates that. Hund's rule says that orbitals in the same subshell must each have an unpaired electron before any of them can have a pair. This is violated in the 3p subshell. The Pauli exclusion principle says that no two electrons can have the same four quantum numbers, which the given configuration does not violate.

Rank the following species in terms of increasing electron affinity: Sulfur (S), Rubidium (Rb), Germanium (Ge), Krypton (Kr), Floruine (F) 1. Kr < Ge < Rb < S < F 2. Ge < Rb < S < F < Kr 3. Kr < Rb < Ge < S < F 4. Not enough information 5. Rb < Ge < S < F < Kr 6. F < Ge < S < Rb < Kr

3. Kr < Rb < Ge < S < F correct Elements' electron affinities increase across a given period and up and given group. Noble gases (i.e. Kr) have an electron affinity of essentially zero. Rb is greater than zero, Ge is greater than Rb, S is greater than Ge, and F is greater than P.

Which of the following individual atoms has a paramagnetic electronic structure? 1. Be 2. Mg 3. Li 4. He 5. Ne

3. Li correct Look for the element with unpaired electrons Diamagnetic means that they have no unpaired electrons

If the following crystallize in the same type of structure, which has the highest lattice energy? 1. KCl 2. LiCl 3. LiF 4. KBr 5. KF

3. LiF correct Highest charge densities due to small size.

In the 20th century, quantum mechanics addressed the failures of classical mechanics by introducing the concept of wave-particle duality. Why did classical mechanics able to explain the world just fine up until then? 1. Only at very high velocity, rarely traveled by macroscopic objects, does the wavelength of particles become large enough to influence its behavior. 2. The wave nature of particles does not affect any macroscopic phenomena and is only important at the atomic scale. 3. Macroscopic objects can be modeled purely as particles because their wavelength is so small compared to their scale that it can be neglected for most purposes and still give a good description of their behavior. 4. Planck's constant is proportional to the size of the object it describes, so that the wavelength of that particle is only significant for small objects. 5. Macroscopic objects have no wave-like properties

3. Macroscopic objects can be modeled purely as particles because their wavelength is so small compared to their scale that it can be neglected for most purposes and still give a good description of their behavior. correct When Newton was laying the foundation for classical mechanics, the implicit assumption throughout was that macroscopic objects do not have a wavelength. Although de Broglie later demonstrated that they do, he also demonstrated that de Broglie wavelengths for macroscopic objects are so small that they can be neglected without impacting our calculations and predictions about behavior. So, although Newton and his successors had no idea that everything has a wavelength, the framework they developed (classical mechanics) still worked quite well.

Which of the following have standard Gibbs free energy of formation values equal to zero? N2(g) O2(ℓ) Ar(ℓ) CO2(g) He(g) 1. N2(g), CO2(g), and He(g) 2. Ar(ℓ) and He(g) 3. N2(g) and He(g) 4. N2(g), O2(ℓ), Ar(ℓ) , and He(g) 5. O2(ℓ) and Ar(ℓ)

3. N2(g) and He(g) correct Standard state for all of these should be gas state. CO2 is not an element. Only elements in their standard states will have ∆G◦f equal to zero. Only N2(g) and He(g) match this criteria

Which of the following is planar? 1. NH3 2. SO2−3 3. NO−3 4. H3O+ 5. PF3

3. NO−3 correct All except the nitrate ion have trigonal pyramidal molecular geometries; the nitrate ion has trigonal planer geometry.

Which one of the processes listed below (if any) have a positive value for ∆S ? 1. None of the choices here have a positive ∆S. 2. The formation of ice crystals from water in a freezer compartment. 3. Rubbing alcohol (isopropanol) evaporating from your skin. 4. The condensation of water droplets on an ice cold drink.

3. Rubbing alcohol (isopropanol) evaporating from your skin. correct Evaporation is liquid to gas which has a +∆S value. Freezing and condensation have negative values for ∆S.

All of the following statements, except one, are important postulates of the kineticmolecular theory of ideal gases. Which one is not a part of this kinetic molecular theory? 1. The time during which a collision between two molecules occurs is negligibly short compared to the time between collisions. 2. There are no attractive nor repulsive forces between the individual molecules. 3. The average kinetic energy of the molecules is inversely proportional to the absolute temperature. 4. The volume of the molecules of a gas is very small compared to the total volume in which the gas is contained. 5. Gases consist of large numbers of particles in rapid random motion.

3. The average kinetic energy of the molecules is inversely proportional to the absolute temperature. correct The average kinetic energy of gas molecules is DIRECTLY (not indirectly) proportional to the absolute temperature. As temperature increases, so does kinetic energy.

What is responsible for the solubility of substances that dissolve endothermically? 1. The negative value of qsys for the dissolution process. 2. The decrease in the entropy of the system. 3. The increase in entropy of the system. 4. The large amount of heat absorbed by the surroundings for the process.

3. The increase in entropy of the system. An example of a substance that dissolves endothermically is a cube of sugar in a cup of hot tea. The sugar cube will dissolve because heat is entering the system, thus endothermic. A phase change from a solid to liquid correlates with an increase in entropy

Which of the following is NOT true about gases? 1. Gases can expand without limit. 2. Gases exert pressure on their surroundings. 3. The volume a gas occupies is directly proportional to its molecular weight. 4. The gas is at STP if it is at 273 K and 1 atm. 5. The density of a gas can be increased by applying increased pressure.

3. The volume a gas occupies is directly proportional to its molecular weight. correct P V = n R T Volume is directly proportional to the number of moles of a gas present, not its molecular weight.

Consider the reaction C6H14(ℓ) + 9.5 O2(g) → 6 CO2(g) + 7 H2O(ℓ) at constant pressure. Which response is true? 1. No work is done as the reaction occurs. 2. Work is done by the system as the reaction occurs. 3. Work is done on the system as the reaction occurs. 4. Work may be done on or by the system as the reaction occurs, depending upon the temperature.

3. Work is done on the system as the reaction occurs. correct ni = 9.5 mol gas nf = 6 mol gas For P = const, w = −P ∆V = −(∆n) R T , ∆n = nf − ni = −3.5 mol gas. For −∆n, w will be positive, which indicates that work was done on the system, which progresses as the reaction progresses.

NO gas exhibits a density of 12 g/L in a closed container at STP. What would be the density of N2 gas under the same conditions? 1. about half 12 g/L 2. a little more than 12 g/L 3. a little less than 12 g/L 4. about double 12 g/L

3. a little less than 12 g/L correct Nitrogen gas has a molecular weight of 28 g/mole which is slightly less than NO which has a molecular weight of 30 g/mole. Because there is a direct relationship between density and molecular weight, the density of nitrogen gas under the same conditions will be a little less than 12 g/L.

What is the correct name for (NH4)2CO3? 1. ammonium bicarbonate 2. biammonium carbonate 3. ammonium carbonate 4. ammonium carboxide

3. ammonium carbonate correct Note the individual ions are NH4+ and CO2^3-

Which of the following is an intensive property? 1. weight 2. volume 3. density 4. mass 5. number of moles of molecules

3. density correct Intensive properties are independent of the amount of matter present. The density of a substance will be the same regardless of the size (large or small) of the sample.

Dispersion (London) forces result from 1. the balance of attractive and repulsive forces between two polar molecules. 2. the formation of a loose covalent linkage between a hydrogen atom connected to a very electronegative atom in one molecule and another very electronegative atom in a neighboring molecule. 3. distortion of the electron cloud of an atom or molecule by the presence of nearby atoms or molecules. 4. attraction between molecules in a liquid and molecules or atoms in a solid surface with which the liquid is in contact. 5. attractive forces between a molecule at the surface of a liquid and those beneath it which are not balanced by corresponding forces from above

3. distortion of the electron cloud of an atom or molecule by the presence of nearby atoms or molecules. correct Just remember dat fam

What kind of carbon-carbon bond is in a molecule of ethylene (C2H4)? 1. ionic 2. single 3. double 4. triple

3. double correct To draw the dot structure for C2H4 we must first calculate the number of valence electrons available from the atoms: A = 4 × 2 (C atom) + 1 × 4 (H atoms) = 12 e− Hydrogen can form only one bond, so carbon atoms must serve as the central atoms. We place the most symmetrical arrangement possible. The correct dot structure for the molecule should show a complete octet (8 electrons) around the carbon atoms, two electrons around each hydrogen atom, and a total of 12 valence electrons for the entire structure:the carbon carbon bond is a double bond. (If using the S = N − A rule to determine the dot structure, N = (8 × 2) + (2 × 4) = 24 e− and S = 24 − 12 = 12 e−. This would indicate 6 bonds

A substance has a melting point of 1200 K, and it conducts electricity in the melted state (liquid) but not in the solid state. What is the name of the major attractive force that holds this substance together? 1. dispersion forces 2. hydrogen bonds 3. ionic bonds 4. dipole-dipole attractions 5. metallic bonds

3. ionic bonds correct Only an ionic compound has these characteristics: 1. high melting point 2. Non-conducting in the solid state 3. Conducting in the liquid state

Consider the following specific heats: copper, 0.384 J/g·◦C; lead, 0.159 J/g·◦C; water, 4.18 J/g·◦C; glass, 0.502 J/g·◦C. If the same amount of heat is added to identical masses of each of these substances, which substance attains the highest temperature? (Assume that they all have the same initial temperature.) 1. copper 2. water 3. lead 4. glass

3. lead correct Use q=Mcat and compare them. There will be a higher temperature for the one with the lowest specific heat.

Antibonding orbitals 1. are higher in energy than bonding orbitals and are therefore populated with electrons prior to bonding orbitals. 2. are responsible for high ionization energies in atoms. 3. lend instability to a molecule when populated with electrons. 4. are lower in energy than bonding orbitals and are therefore populated with electrons prior to bonding orbitals. 5. are responsible for dipole moments in molecules.

3. lend instability to a molecule when populated with electrons. correct Antibonding orbitals have nothing to do with dipole moments or ionization energies. They form from the overlap of atomic orbitals, are higher in energy than bonding orbitals, are populated with electrons after the corresponding bonding orbitals are populated and lend instability to the molecule when populated with electrons

What is the correct order of increasing energy? 1. microwaves, ultraviolet light, visible light, x-rays, γ-rays 2. microwaves, x-rays, γ-rays, visible light, ultraviolet light 3. microwaves, visible light, ultraviolet light, x-rays, γ-rays 4. ultraviolet light, visible light, microwaves, x-rays, γ-rays 5. x-rays, γ-rays, microwaves, visible light, ultraviolet light

3. microwaves, visible light, ultraviolet light, x-rays, γ-rays correct microwaves < visible light < ultraviolet light < x-rays < γ-rays

An electron in a hydrogen atom could undergo any of the transitions listed below, by emitting light. Which transition would give light of the shortest wavelength? 1. n = 4 to n = 2 2. n = 2 to n = 1 3. n = 4 to n = 1 4. n = 4 to n = 3 5. n = 3 to n = 1

3. n = 4 to n = 1 correct Shorter wavelengths have more energy than longer wavelengths. n = 4 is a very excited energy level, shorter than n = 3, 2 or 1. A transition from n = 4 to n = 1 will release more energy than any other option given.

What is the shape (molecular geometry) of IF+4? 1. square planar 2. tetrahedral 3. seesaw 4. trigonal bipyramidal 5. T-shaped

3. seesaw correct There are five regions of electron density (including one lone pair) around the central atom:

Compared to a 320 nm photon, a 280 nm photon has: 1. shorter wavelength, higher frequency, lower energy 2. longer wavelength, lower frequency, higher energy 3. shorter wavelength, higher frequency, higher energy 4. longer wavelength, lower frequency, lower energy 5. longer wavelength, higher frequency, higher energy 6. shorter wavelength, lower frequency, lower energy

3. shorter wavelength, higher frequency, higher energy correct As wavelength increases, frequency and energy decrease As wavelength decrease, frequency and energy increase

What is the hybridization of carbon in CH2O? C is the central atom. 1. sp3d2 2. sp3d 3. sp2 4. sp3 5. sp

3. sp2 correct There are 3 regions of HED. This corresponds to sp2 hybridization.

If a molecule has square planar molecular geometry, what must be its hybridization? 1. sp3d 2. sp 3. sp3d2 4. sp 5. sp3

3. sp3d2 correct A molecule with square planar molecular geometry has octahedral electronic geometry which cooresponds to sp3d2 hybridization.

For an isoelectronic series of ions, the ion that is the smallest is always 1. the ion with the most electrons. 2. the ion with the most neutrons. 3. the ion with the highest atomic number. 4. the least positively (or most negatively) charged ion. 5. the ion with the fewest protons.

3. the ion with the highest atomic number. correct Isoelectronic ions all have the same number of electrons. However, the one with the highest atomic number will have the largest effective nuclear charge.

ICl3 is sp3d hybridized. What is the electronic and molecular geometry? 1. octahedral; T-shaped 2. trigonal bipyramidal, seesaw 3. trigonal bipyramidal; T-shaped 4. tetrahedral; pyramidal 5. trigonal planar; trigonal planar

3. trigonal bipyramidal; T-shaped correct The hybridization tells us that there are 5 regions of high electron density. Three of those regions are the bonded Cl atoms. The other two regions must be lone pairs of electrons on the central I atom. This corresponds to trigonal bipyramidal electronic geometry and T-shaped molecular geometry.

Consider the polyatomic ion PCl4- and its three-dimensional structure. What is the electronic geometry and the molecular geometry for this ion? 1. trigonal bipyramidal; tetrahedral 2. trigonal bipyramidal; T-shaped 3. trigonal bipyramidal; seesaw 4. tetrahedral; tetrahedral 5. octahedral; square planar 6. octahedral; square pyramidal

3. trigonal bipyramidal; seesaw correct The correct line structure has the central P atom with 4 single covalent bonds to the Cl's and one lone pair. This means there are 5 electron regions which corresponds to a trigonal bipyramidal electronic geometry. Having one lone pair on this geometry gives a seesaw molecular geometry.

Some of the following terms characterize both the bonding within a molecule (intramolecular) and that between atoms and molecules (intermolecular). Which of the following is normally considered only when characterizing intermolecular forces? 1. ionic forces 2. covalent bonding 3. van der Waals forces 4. polar covalent bonding 5. electrostatic forces

3. van der Waals forces correct Only van der Waals forces occur between separate molecules; i.e., are intermolecular forces.

Consider the compound ethene, C2H4. The bond between the two carbons that is formed above and below the internuclear axis is a ________ bond. The atomic orbitals that combine to form this bond are _________ orbitals. 1. π; sp2 2. π; 1p 3. π; 2p 4. σ; sp3 5. σ; sp2

3. π; 2p correct π bonds form above and below the internuclear axis (off-axis), σ bonds form along the axis (on-axis). Two 2p atomic orbitals overlap side to side to form a π bond.

Calculate ∆S◦surr at 298 K for the reaction 6 C(s) + 3 H2(g) → C6H6(ℓ) ∆H◦r = +49.0 kJ·mol−1 and ∆S◦r = −253J·K−1·mol−1 1. − 417 J·K−1·mol−1 2. +253 J·K−1·mol−1 3. − 164 J·K−1·mol−1 4. +164 J·K−1·mol−1 5. − 253 J·K−1·mol−1 .

3. − 164 J·K−1·mol−1 correct ∆S◦surr = qsurr / T = −q / T= −∆H◦r / 298 K

Calculate the change in entropy when 3.28 moles of an ideal gas are compressed isothermally such that the volume changes from 14.4 L to 3.6 L. 1. −56.7 J/K 2. +24.7 J/K 3. −37.8 J/K 4. +38.1 J/K 5. −11.5 J/K 6. −16.4 J/K 7. −32.2 J/K

3. −37.8 J/K correct ∆S = n R ln (V2/V1) = (3.28)(8.314) ln(3.614.4) = −37.8041 J/K

Calculate the ratio of the rate of effusion of He to that of CO2 (at the same temperatures). 1. 1 : 11 2. 1 : 112 3. √11 : 1 4. 11 : 1 5. 1 : 1 6. 121: 1 7. 1 : √11

3. √11 : 1 correct EffHC / EffCO2 = √MWCO2 /√MWHC = √(44/4) =√11

Consider a flea of mass 4.5×10^−4g moving at 1.0 m/s midway through its jump. What is its de Broglie wavelength? 1. 1.4725 × 10^−30 m 2. 2.9818 × 10^−40 m 3. 2.9818 × 10^−37 m 4. 1.47244 × 10^−27 m

4. 1.47244 × 10^−27 m correct λ = h / p =h / (m · v) m = 0.00045 g × 0.001 kg / 1 g = 4.5 × 10^−7kg [6.626 × 10−34 kg·m^2/s] / [(4.5 × 10−7 kg)(1 m/s)]= = 1.47244 × 10^−27 m

Consider a situation in which two solid reactants are mixed together to generate an unknown gaseous product. The vapor from the gas effuses at a rate that is 1.77 times slower than the same amount of carbon dioxide (CO2) under the same temperatures and pressures. What is the molar mass of this unknown gas? 1. 156 g/mol 2. 102 g/mol 3. 87.7 g/mol 4. 138 g/mol

4. 138 g/mol correct Because the unknown gas took 1.77 times as long to effuse, the rate of effusion of CO2 must be 1.77 times that of the unknown gas: EffCO2 = 1.77 Effunknown MWCO2 = 12.01 g/mol + 2(16 g/mol) = 44.01 g/mol Effunknown/ Eff CO2 = √(MWunkown/MW CO2) =137.879 g/mol .

You are caught in a radar trap and hope to show that the speed measured by the radar gun is in error due to the uncertainty principle. If you assume that the uncertainty in your position is large, say about 10 m, and that the car has a mass of 2150 kg, what is the uncertainty in the velocity? 1. 4 × 1038 m/s 2. 1 × 10−34 m/s 3. 5.0 × 10−42 m/s 4. 2.5 × 10−39 m/s 5. 1 × 1033 m

4. 2.5 × 10−39 m/s correct ∆x = 10 m m = 2150 kg ∆v = h / 2 m ∆x = 2.45349 × 10^−39 m/s

How many different types of sigma (σ) bonds are found in ethanoic acid, CH3COOH? In other words, how many different combinations of atomic orbitals are used when forming the σ bonds in ethanoic acid. 1. 5 2. 4 3. 7 4. 3 5. none of the above

4. 3 correct All of the σ bonds found in ethanoic acid are sp3 − 1s, sp3 − sp2, or sp2 − sp2 .

How much energy does a photon with a wavelength equal to 350 nm have? 1. 1.89 × 10^−27 J 2. 2.32 × 10^−31 J 3. 3.50 × 10^−7J 4. 5.68 × 10^−19 J 5. 1.29 × 10^−48 J

4. 5.68 × 10^−19 J correct λ = 350 nm c = 3 × 108 m/s h = 6.63 × 10−34 J · s. For a photon c = λ ν , so E = h ν = (h c) / λ plug in the numbers homes = 5.68 × 10−19 J

What is the energy, in Joules, of a photon of wavelength 200 nm? What bond energy would this correspond to, in kJ · mol−1? 1. 1.32 × 10−40 J; 7.95 × 10−20 kJ · mol−1 2. 1.32 × 10−21 J; 795 kJ · mol−1 3. 9.94 × 10−21 J; 5.99 kJ · mol−1 4. 9.94 × 10−19 J; 599 kJ · mol−1 5. 9.94 × 10−17 J; 1.65 × 10−43 kJ · mol−1 6. 1.32 × 10−31 J; 7.95 × 10−11 kJ · mol−1

4. 9.94 × 10−19 J; 599 kJ · mol−1 correct λ = 200 nm c = 3 × 108 m/s h = 6.626 × 10−34 J · s. For a photon c = λ ν , so E = h ν = h c / λ where c is the speed of light and h is Planck's constant. E =h c / λ

Which of the following is expected to boil at the highest temperature? 1. C4H10 2. C3H8 3. C2H6 4. C5H12 5. CH4

4. C5H12 correct The boiling point tends to increase with increasing molecular weight. C5H12 has the biggest MW, so it should boil at the highest temperature.

Which of the following substances would you predict might evaporate the fastest? 1. C8H18 2. C10H22 3. C12H24 4. C6H14

4. C6H14 correct All the listed molecules are nonpolar hydrocarbons; therefore the dominant intermolecular force that exists in the condensed phase of all listed molecules is dispersion forces. Therefore, the molecule with the least number of atoms and the lowest molecular weight would have the lowest dispersion forces, and therefore would evaporate the easiest.

Which of the following can be expected to exhibit the strongest hydrogen bonding in the liquid state? 1. CH4 2. CH3CH3 3. CH3COCH3 4. CH3OH 5. CH3OCH3

4. CH3OH correct H-bonds are a special case of very strong dipole-dipole interactions. They only occur when H is bonded to small, highly electronegative atoms - F, O or N only. Methyl alcohol is the only molecule given that has H bonded to N, O, or F (O in this case).

When wood is burning (i.e. a combustion process is occurring), which of the following quantities is positive? 1. Change in enthalpy. 2. Work. 3. Change in Gibbs' free energy. 4. Change in entropy.

4. Change in entropy. correct A burning piece of wood produces a lot of gas and thus does expansion work on the surroundings so work is negative, not positive. It is an exothermic reaction (producing heat) so enthalpy change is negative, not positive. It happens spontaneously so change in free energy is negative, not positive. Finally, all the gas produced yields an increase in entropy so the change is indeed positive.

Which of the following has bond angles slightly less than 109.5◦? 1. I−3 2. O3 3. CH+3 4. HOCl 5. NO−2

4. HOCl correct Only HOCl has four regions of electron density around the central atom; the two lone pairs repel the bonds more, making the bond angle less than 109.5◦ .

Consider a potential energy diagram for the interaction of a sodium ion (Na+) with a chloride ion (Cl−). Which of the following statements is/are true? I) Repulsive forces predominate at very small internuclear distances. II) The minimum potential energy occurs when attractive forces are greatest. III) Attractive forces are linearly dependent on the internuclear distance. 1. I and II 2. III only 3. I and III 4. I only correctIII 6. I, II and III 7. II only

4. I only correct It is true that repulsive forces predominate at very small internuclear distances; this is the result of the positively charged nuclei electrostatically repelling each other, which occurs despite that fact that sodium ion and chloride ion have opposite charges overall. The attractive forces actually continue to increase as you force the ions closer together, but the repulsive forces increase more, thus offsetting the attractive forces and producing a net increase in potential energy. Thus, the minimum potential energy does not occur at maximum attractive force, but rather at a "sweet spot" where the sum of attractive and repulsive forces is minimized. Attractive forces are a rectangular hyperbola that asymptotically approach zero as the internuclear distance approaches infinity.

What types of intermolecular interactions does ammonia (NH3) exhibit? I) dispersion forces II) dipole-dipole interaction III) hydrogen bonding IV) covalent bonding 1. II and III only 2. I only 3. I and II only 4. I, II, and III only 5. II and IV only 6. II only

4. I, II, and III only correct All molecules, because they have electron clouds, experience dispersion forces. NH3 is polar and thus experiences dipole-dipole interactions. NH3 contains N H bonds and therefore experiences hydrogen bonding interactions. Covalent bonding is an intramolecular interaction.

Which of the following statement(s) is/are true? I) The failure of classical mechanics to predict the absorptions/emission spectra of gases is called the ultraviolet catastrophe. II) Quantum mechanics accurately predicted the behavior of blackbody radiators. III) The emission spectra of gases are discrete rather than continuous. IV) Any frequency of light will eject an electron from a metal surface as long as the intensity is sufficient. 1. I and III 2. I, II and IV 3. II, III, and IV 4. II and III 5. III and IV

4. II and III correct Classical mechanics predicted that the power radiated by a blackbody radiator would be proportional to the square of the frequency at which it emitted radiation, and thus approach infinity as the frequency increased. This was false, since at higher frequencies blackbody radiators emit less, not more power. This was termed the ultraviolet catastrophe. Classical mechanics also predicted that the energy (velocity) of electrons emitted from a metal surface is proportional to the intensity of light. In reality, the energy (velocity) is only dependent upon the frequency of light. Once the threshold frequency is reached, however, the number of emitted electrons is proportional to the intensity of light. Classical mechanics also fails in explaining the discrete lines in absorption/emission spectrum, which are due to discrete energy levels of electrons in atoms.

Which of the following species exhibit resonance/delocalization? I) HCN II) O3 III) CO2−3 1. I, III 2. III only 3. I only 4. II, III 5. I, II, III 6. II only 7. I, II

4. II, III correct Both ozone and carbonate have a single pair of resonant electrons and are famous examples of resonant molecules. Cyanide cannot have resonance since hydrogen can only form a single bond.

Which of the following is diamagnetic? 1. N2+ 2. N2^(2-) 3. N2- 4. N2

4. N2 correct Explanation: All but N2 are paramagentic (possess unpaired electrons).

Arrange Al2O3, Nb, I2, C(s) (diamond) in the order metallic solid, covalent network, molecular solid, ionic solid. 1. C(s) (diamond); Nb, Al2O3; I2 2. Nb, I2; C(s) (diamond); Al2O3 3. Al2O3; C(s) (diamond); I2; Nb 4. Nb; C(s) (diamond); I2; Al2O3 correct

4. Nb; C(s) (diamond); I2; Al2O3 correct

Which elements are correctly listed in order of INCREASING ionization energy? 1. C < N < O 2. N < O < F 3. N < P < As 4. O < F < Ne 5. O < S < Se

4. O < F < Ne correct Ionization energy corresponds to how much energy it takes to ionize an element. Elements with high ionization energies do not want to give up their electrons, low ionization energies mean they do not want to receive an electron, but would rather give one up. In general, ionization energies increase as you move right along a period and up a group. Some atoms (like B, O, Al and S) can achieve more stable electronic configuration in filled or half-filled shells.

Which of the following is true of a general thermodynamic state function? 1. The change in the value of a state function is always negative for a spontaneous reaction. 2. The value of the state function remains constant. 3. The value of a state function does NOT change with a change in temperature of a process. 4. The change of the value of a state function is independent of the path of a process.

4. The change of the value of a state function is independent of the path of a process. correct The change in the value of the state function is always positive for endothermic processes. A change in a state function describes a difference between the two states. It is independent of the process or pathway by which the change occurs.

The following three statements refer to the Bohr theory of the atom. Z1) An electron can remain in a particular orbit as long as it continually absorbs radiation of a definite frequency. Z2) The lowest energy orbits are those closest to the nucleus. Z3) An electron can jump from an inner orbit to an outer orbit by emitting radiation of a definite frequency. Which response contains all the statements that are consistent with the Bohr theory of the atom and no others? 1. Z1, Z2 and Z3 2. Z2 and Z3 only 3. Z1 and Z2 only 4. Z2 only 5. Z3 only

4. Z2 only correct Z1 is wrong because it's never going to absorb radiation forever, that just sounds stupid Z3 is wrong because a jump from an inner orbit to an outer orbit is actually absorption NOT emission.

For the reaction 2 C(s) + 2 H2(g) → C2H4(g) ∆H◦r = +52.3 kJ · mol−1 and ∆S◦r = −53.07 J · K−1· mol−1 at 298 K. The reverse reaction will be spontaneous at 1. no temperatures. 2. temperatures below 1015 K. 3. temperatures above 985 K. 4. all temperatures. 5. temperatures below 985 K

4. all temperatures. correct ∆G = ∆H − T∆S is used to predict spontaneity. (∆G is negative for a spontaneous reaction.) T is always positive; for the reverse reaction, we reverse the sign of ∆H and ∆S. We thus have ∆G = (−) − T(+) for the reverse reaction, so ∆G will be negative for any physically possible value of T.

You have two liquids of identical mass, and both with initial temperatures of 15◦C. One is ethanol, C2H5OH, with a specific heat of 2.46 J/g◦C and the other is benzene, C6H6, with a specific heat of 1.74 J/g◦C. If both liquids absorb the same amount of heat, which one will have the highest final temperature? Assume that neither liquid reaches its boiling point. 1. ethanol 2. Both liquids will have the same final temperature. 3. Cannot tell without more information given. 4. benzene

4. benzene correct Temperature rise (∆T) is inversely proportional to the heat capacity. ∆T =qmCs Therefore, because benzene has a smaller heat capacity, Cs, it will have the larger temperature rise.

Dispersion (London) forces result from 1. the balance of attractive and repulsive forces between two polar molecules. 2. attraction between molecules in a liquid and molecules or atoms in a solid surface with which the liquid is in contact. 3. the formation of a loose covalent linkage between a hydrogen atom connected to a very electronegative atom in one molecule and another very electronegative atom in a neighboring molecule. 4. distortion of the electron cloud of an atom or molecule by the presence of nearby atoms or molecules. 5. attractive forces between a molecule at the surface of a liquid and those beneath it which are not balanced by corresponding forces from above.

4. distortion of the electron cloud of an atom or molecule by the presence of nearby atoms or molecules. correct literally what it is fam

What kind of radiation causes molecules to vibrate? 1. visible 2. microwave 3. x-rays 4. infrared 5. ultraviolet

4. infrared correct They just do ,ok fam

In reactions to form ionic compounds, metals generally 1. become non-metals. 2. gain electrons. 3. do not react. 4. lose electrons.

4. lose electrons. correct Ionic bonds occur between metals and nonmetals. Metals will generally lose electrons, as the nonmetals will generally receive electrons.

Consider two balloons filled with gas and arranged so that P, V , T are the same in both. The number of molecules in each balloon 1. would be the same only if the filling gases are the same. 2. must be different. 3. could be different if the filling gases are different. 4. must be the same.

4. must be the same. correct PV=nRT can be rearranged into n=PV/TR so, if PVT is constant in both balloons, then the number of moles( and thus molecules) must be the same.

What are the electronic and molecular geometries of the molecule BrF5? 1. octahedral, trigonal bipyramidal 2. octahedral, octahedral 3. trigonal bipyramidal, trigonal bipyramidal 4. octahedral, square pyramidal correct 5. trigonal bipyramidal, square pyramidal

4. octahedral, square pyramidal correct In the molecule bromine pentafluoride, the central atom bromine has five bonded atoms and one non-bonding pair. The molecule is thus octahedral, square pyramidal.

Advertising claims sometimes state that adding something mechanical to a car's engine will allow it to recover 100% of the energy that comes from burning gasoline. You should be skeptical of such claims because they violate the 1. first law of thermodynamics. 2. activation energy requirements of all chemical reactions. 3. law of conservation of matter. 4. second law of thermodynamics.

4. second law of thermodynamics. correct If you burn gasoline in an engine to move a car, you are ultimately converting chemical potential energy into kinetic energy. But much of this energy is lost as heat. There is NO way to make any energy conversion 100% efficient.

When a given molecule or ion is shown via resonance structures, the numerous structures 1. show the various possible geometries that a molecule or ion can assume. 2. show the various stable isomers (or kinds) of a compound. 3. show how an isotope of one or more of the atoms is distributed within a molecule. 4. show the set of bonding extremes of which the average will better represent the actual bonding. 5. show the nature of the various vibrations possible based on the nature of the kinds of atoms and bonds.

4. show the set of bonding extremes of which the average will better represent the actual bonding. correct Resonance structures are an attempt to clearly show the extremes of where the electrons could be. The average of all the structure is more like the true bonding.

Surface tension describes 1. the resistance to flow of a liquid. 2. the forces of attraction between the surface of a liquid and the air above it. 3. adhesive forces between molecules. 4. the inward forces that must be overcome in order to expand the surface area of a liquid. 5. capillary action. 6. the forces of attraction between surface molecules of a solvent and the solute molecules.

4. the inward forces that must be overcome in order to expand the surface area of a liquid. correct Molecules in the interior of a liquid interact with molecules all around them, whereas molecules at the surface of a liquid can only be affected by those beneath the surface layer. This phenomenon leads to a net inward force of attraction on the surface molecules, contracting the surface and making the liquid behave as though it had a skin. Surface tension is a measure of the inward forces that must be overcome to expand the surface area of a liquid.

Which of the following experiments provided evidence that the electrons in atoms are arranged in distinct energy levels? 1. the results of the Millikan oil-drop experiment 2. the scattering of α particles by a metal foil 3. the existence of elements with noninteger values for atomic weights 4. the observation of line spectra from gas discharge tubes 5. the deflection of ions in a mass spectrometer

4. the observation of line spectra from gas discharge tubes correct The fact that gases emitted only specific wavelengths of energy suggested that electron energy states are quantized.

1.95 mol of an ideal gas at 300 K and 3.00 atm expands from 16 L to 28 L and a final pressure of 1.20 atm in two steps: (1) the gas is cooled at constant volume until its pressure has fallen to 1.20 atm, and (2) it is heated and allowed to expand against a constant pressure of 1.20 atm until its volume reaches 28 L. Which of the following is CORRECT? 1. w = 0 for the overall process 2. w = −6.03 kJ for the overall process 3. w = −4.57 kJ for (1) and w = −1.46 kJ for (2) 4. w = 0 for (1) and w = −1.46 kJ for (2) 5. w = −4.57 kJ for the overall process

4. w = 0 for (1) and w = −1.46 kJ for (2) correct For step (1): If there is no change in volume, w = 0. For step (2): For expansion against a constant external pressure, w = −Pext ∆V = (−1.2 atm)(18 L − 6 L) × (101.325 J · L−1· atm−1) = −1.45908 kJ . The total work done by the system would be the sum of the work for each step.

H2S has a lower boiling point than H2O or H2Se. Which of the following is NOT part of the explanation for this observation? 1. ∆EN for the O H bond is larger than ∆EN for the S H bond. 2. The strength of London forces is greater for H2Se than for H2S. 3. Hydrogen bonding is most significant for compounds containing electronegative atoms in the second row. 4. ∆EN for the Se H bond is larger than ∆EN for the S H bond.

4. ∆EN for the Se H bond is larger than ∆EN for the S H bond. correct All three compounds are polar molecules which posess polar bonds but the ∆EN (electronegatively difference) for O-H is much greater than for S-H or Se-H. H2O exhibits H-bonding, so it has the highest boiling point of the three compounds. The relative polarities (∆EN) of the H2Se and H2S bonds are similar and one would predict slightly smaller for H-Se based on general trends, so dipoledipole interactions do not explain the trend. However, London interactions become longer as the size of the molecule's electron cloud increases; since Se is bigger than S, H2Se has stronger London forces and thus a higher boiling point than H2S.

Consider a system where 2.50 L of ideal gas expands to 6.25 L against a constant external pressure of 330 torr. Calculate the work (w) for this system. 1. −1238 J 2. +1238 J 3. +1.63 J 4. −165 J 5. −1.63 J 6. +165 J

4. −165 J correct Convert torr to atm, and then convert answer in L·atm to joules. The answer will be negative due to the expansion of the gas. w = −P ∆V = −(330/760)(3.75L) w = −1.628 L atm ×101.325 J/(L atm) = −165 J

What is ∆G◦r for the combustion of liquid n-pentane? 1. 3389 kJ/mol 2. −383 kJ/mol 3. 383 kJ/mol 4. −3389 kJ/mol 5. −451 kJ/mol 6. 451 kJ/mol

4. −3389 kJ/mol correct The combustion of liquid n-pentane occurs according to the following reaction. C5H12(ℓ) + 8 O2(g) → 5 CO2(g) + 6 H2O(ℓ) Do products minus reactants for ∆H Do products minus reactants for ∆S then plug everything in ∆G = ∆H − T ∆S −3512 kJ/mol − (298 K) (−0.413 kJ/mol · K) = −3389 kJ/mol

A 1.00 g sample of n-hexane (C6H14) undergoes complete combustion with excess O2 in a bomb calorimeter. The temperature of the 1502 g of water surrounding the bomb rises from 22.64◦C to 29.30◦C. The heat capacity of the hardware component of the calorimeter (everything that is not water) is 4042 J/◦C. What is ∆U for the combustion of n-C6H14? One mole of n-C6H14 is 86.1 g. The specific heat of water is 4.184 J/g·◦C. 1. −1.15 × 10^4kJ/mol 2. −4.52 × 10^3kJ/mol 3. −7.40 × 10^4kJ/mol 4. −5.92 × 10^3kJ/mol 5. −9.96 × 10^3kJ/mol

4. −5.92 × 10^3kJ/mol correct Just do MCAT of water + CAT of hardware That answer will be over (in this case) 1g. Now multiply by molar mass to get the answer in mol.

Based on the enthalpy of sublimation (∆Hsub = 393.5 kJ · mol−1) and entropy of sublimation (∆Ssub = 2.023 kJ · mol−1· K−1) of carbon dioxide, at what temperature does this phase transition occur? 1. 78.5◦C 2. −78.5 K 3. 0.2◦C 4. −78.5◦C 5. 0.2 K

4. −78.5◦C correct Tsub = ∆Hsub / ∆Ssub

A 50/50 mix (by mass) of nitrogen gas and carbon dioxide is made. What is the mole fraction of nitrogen in this mixture? 1. 0.27 2. 0.50 3. 0.73 4. 0.56 5. 0.61 6. 0.44 7. 0.39

5. 0.61 correct 50/50 mix means the same mass for both gases. So let that mass be 44 g each. 44g is exactly 1 mole of the CO2. 44/28 = 1.57 mol of N2 gas XN2 = 1.57/(1+1.57) = 0.61

What is the effective nuclear charge experienced by the 2p and 3p electrons of an Chlorine atom (Cl), respectively? 1. 7, 2 2. 17, 7 3. 17, 10 4. 11, 7 5. 15, 7

5. 15, 7 correct Electrons within the same energy level do not shield one another. Therefore the 2s electrons are not shielding the 2p electrons. To calculate effective nuclear charge the number of shielding electrons (all electrons in lower energy levels) are subtracted from the actual nuclear charge (number of protons in the nucleus). Chlorine, atomic number 17, has 17 protons. All of its electrons in principal energy level 2 are shielded only by the electrons in principal energy level 1, i.e. the 2 1s electrons. 17 - 2 = 15 and for 3p 17 - 10 = 7.

How many sigma (σ) and pi (π) bonds are in the Lewis structure for C(COOH)4? 1. 8 σ, 4 π 2. 12 σ, 4 π 3. 16 σ, 0 π 4. 12 σ, 0 π 5. 16 σ, 4 π

5. 16 σ, 4 π correct

How many sigma (σ) and pi (π) bonds are in the Lewis structure for C(COOH)4? 1. 12 σ, 0 π 2. 8 σ, 4 π 3. 16 σ, 0 π 4. 12 σ, 4 π 5. 16 σ, 4 π

5. 16 σ, 4 π correct

Calculate the mass of NH3 that can be produced from 30.0 g N2 and 5.0 g H2 in the following reaction. N2(g) + 3 H2(g) → 2 NH3(g) 1. 36.4 g 2. 63.8 g 3. 42.5 g 4. 18.2 g 5. 28.3 g

5. 28.3 g correct mN2 = 30.0 g mH2 = 5.0 g First, we need to find the limiting reagent. We can do this by finding the number of moles that each amount of nitrogen gas and hydrogen gas can make: ? moles NH3 (from 30.0 g N2) = 30 g N2 ×(1 mol N2 / 28 g N2) × (2 mol NH3 / 1 mol N2) = 2.1 mol NH3 ? mol NH3 (from 5.0 g H2) = 5.0 g H2 ×(1 mol H2 / 2 g H2) × (2 mol NH3 / 3 mol H2) = 1.7 mol NH3 Since we can make fewer moles of the product with the H2 that we have, it is our limiting reagent. That means we base the following equations on the amount of H2 that we start with: ? g NH3 = 5.0 g H2 × (1 mol H2 / 2 g H2) × (2 mol NH3 / 3 mol H2) × (17 g NH3 / 1 mol NH3) = 28.3 g NH3

Draw the Lewis structure of xenon difluoride and give the number of lone pairs of electrons around the central atom. 1. 5 2. 1 3. 4 4. 2 5. 3

5. 3 correct Just draw it fam

A 6.35 L sample of carbon monoxide is collected at 55.0◦C and 0.892 atm. What volume will the gas occupy at 1.05 atm and 59.0◦C? 1. 6.68 L 2. 4.82 L 3. 1.96 L 4. 6.10 L 5. 5.46 L

5. 5.46 L correct V1 = 6.35 L P1 = 0.892 atm T1 = 55◦C = 328 K P2 = 1.05 atm T2 = 59◦C = 332 K We can use the combined gas law and solve for V2: (P1 V1/T1) = (P2 V2)/T2 V2 =(P1 V1 T2)/(T1 P2) = (6.35 L) (0.892 atm) (332 K) / (328 K) (1.05 atm) = 5.46 L

A 6.00 L sample of C2H4(g) at 2.00 atm and 293 K is burned in 6.00 L of oxygen gas at the same temperature and pressure to form carbon dioxide gas and liquid water. If the reaction goes to completion, what is the final volume of all gases at 2.00 atm and 293 K? 1. 4.00 L 2. 2.00 L 3. 12.00 L 4. 6.00 L 5. 8.00 L 6. 2.66 L

5. 8.00 L correct The balanced equation is C2H4(g) + 3 O2(g) → 2 H2O(ℓ) + 2 CO2(g) Avogadro's Principle tells us that when P and T are constant (which they are in this problem), V ∝ n, so we can work with the volume of the gases instead of the number of moles. From the equation above, we need 3 times more L of O2 than of C2H4 (18 L, not the given 6 L!) so O2 is the limiting reagent. Find out how much CO2 is made based on the L of O2: LCO2 = (6 L O2) × (2 L CO2 / 3 L O2) = 4 L CO2 We now find out how much C2H4 was used reacting with all the L of O2 LC2H4 = (6 L O2) × (1 L C2H4 / 3 L O2) = 2 L C2H4 This means 6 L − 2 L = 4 L C2H4 are unused and still present. The final mixture is 4 L unreacted C2H4 + 4 L CO2 = 8 L of gas. We assume the volume of the water is insignificant.

A sample of nitrous oxide gas (NO) has a density of 12 g L−1. What pressure does the sample exert at 27 ◦C? 1. 1.0 atm 2. 61.6 atm 3. not enough information 4. 997.9 atm 5. 9.9 atm

5. 9.9 atm correct Explanation: By thoughtful substitutions and rearrangement, the ideal gas law can be used to relate the mass density of a gas to its pressure. PV = nRT Recalling that a number of moles (n) is equal to a mass (m) divided by a molecular weight (M.W.), we can substitute into the ideal gas law.

Which of the following would you expect to have the highest heat of vaporization? 1. C8H18 2. CH4 3. C3H6 4. C5H12 5. C12H26

5. C12H26 correct C12H26 has the highest IMF (dispersion forces) and thus it takes a lot more heat and energy to vaporize it. Therefore it has the highest heat of vaporization

For which of the following reactions at room temperature (25◦C) would there be 5.0 kJ of work done on the system? 1. 2 H2O(ℓ) + O2(g) → 2 H2O2(ℓ) 2. CO2(g) + 2 H2O(g) → CH4(g) + 2 O2(g) 3. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g) 4. N2H2(g) + CH3OH(g) → CH2O(g) + N2(g) + 2 H2(g) 5. CH2O(g) + N2(g) + 2 H2(g) → N2H2(g) + CH3OH(g) 6. 2 H2O2(ℓ) → 2 H2O(ℓ) + O2(g)

5. CH2O(g) + N2(g) + 2 H2(g) → N2H2(g) + CH3OH(g) correct At room temperature (298 K), the product of the gas constant (R =8.314 J · mol−1· K−1) and T is very closeto 2.5 kJ mol−1. Based on 5.0 kJ =−∆ngas2.5 kJ · mol−1 , the reaction forwhich ∆ngas = −2 will be the correct answer.

Put the following compounds LiF, HF, F2, NF3 in order of increasing melting points. 1. NF3, HF, F2, LiF 2. NF3, F2, HF, LiF 3. LiF, NF3, HF, F2 4. F2, HF, NF3, LiF 5. F2, NF3, HF, LiF correct 6. LiF, HF, NF3, F2 7. LiF, F2, HF, NF3 8. LiF, HF, F2, NF3

5. F2, NF3, HF, LiF correct

Which of the following statements concerning the laws of thermodynamics is not true? 1. S = 0 for a perfect crystal at absolute zero. 2. ∆Uuniv = 0 3. Entropy always increases in an isolated system. 4. ∆Suniv > 0 5. Free energy is conserved in a closed system.

5. Free energy is conserved in a closed system. correct Free energy is conserved in an isolated system, but not in a closed system.

Which of the following is not correctly paired with its dominant type of intermolecular forces? 1. NH3, hydrogen bonding 2. SiH4, instantaneous dipoles 3. C6H6 (benzene), instantaneous dipoles 4. CaO, ionic forces 5. HBr, hydrogen bonding

5. HBr, hydrogen bonding correct London forces, dispersion forces, van der Waals' forces, instantaneous or induced dipoles all describe the same intermolecular force. London forces are induced, short-lived, and very weak. Molecules and atoms can experience London forces because they have electron clouds. London forces result from the distortion of the electron cloud of an atom or molecule by the presence of nearby atoms or molecules. Permanent dipole-dipole interactions are stronger than London forces and occur between polar covalent molecules due to charge separation. H-bonds are a special case of very strong dipole-dipole interactions. They only occur when H is bonded to small, highly electronegative atoms - F, O or N only. Ion-ion interactions are the strongest due to extreme charge separation and occur between ions (including polyatomic ions). They can be thought of as both inter- and intramolecular bonding. HBr is a polar molecule that does not contain H bonds; therefore, dipole-dipole forces will be the most significant type of intermolecular forces present.

Consider the following statements: I) Real gases act more like ideal gases as the temperature increases. II) When n and T are constant, a decrease in P results in a decrease in V . III) At 1 atm and 273 K, every molecule in a sample of a gas has the same speed. IV) At constant T, CO2 molecules at 1 atm and H2 molecules at 5 atm have the same average kinetic energy. Which of these statements is true? 1. III and IV only 2. I and II only 3. II and III only 4. II and IV only 5. I and IV only

5. I and IV only correct Real gases act more like ideal gases at higher temperatures and lower pressures. From Boyle's Law, if n and T are constant, P and V are inversely related. Molecules in a sample of gas at a given temperature are moving at different speeds; there is a distribution of speeds and a mean speed, and a mean kinetic energy. The mean kinetic energy is directly proportional to the temperature of the gas sample.

Consider the following processes. (Treat all gases as ideal.) I) The pressure of one mole of oxygen gas is allowed to double isothermally. II) Carbon dioxide is allowed to expand isothermally to 10 times its original volume. III) The temperature of one mole of helium is increased 25◦C at constant pressure. IV) Nitrogen gas is compressed isothermally to one half its original volume. V) A glass of water loses 100 J of energy reversibly at 30◦C. Which of these processes leads to an increase in entropy? 1. III and V 2. I and II 3. I and IV 4. V 5. II and III

5. II and III correct Entropy is directly proportional to volume Volume is inversely proportional to volume Temperature is proportional to entropy Using Q/T, losing 100J of energy actually decreases entropy

Which of the following are true consequences of the uncertainty principle? I) The uncertainty in an electron's momentum can never be less than h/2; II) An electron can be measured in twoplaces at once; III) Electrons and other particles do not have a well-defined position or momentum like particles in classical mechanics do. 1. I only 2. II and III 3. I and II 4. II only 5. III only 6. I and III

5. III only correct I is false because ∆p may be less than h/2 provided that ∆x is greater than unity (and vice versa). II is false because an electron can only be observed in one place at one time, although there may be an equal probability of observing it at two different places. III is true because quantum mechanics can only give probabilities of a particle having a certain position or momentum and not an exact value.

The oxidation of sugar to carbon dioxide and water is a spontaneous chemical reaction. Since we know that reactions that occur spontaneously in one direction cannot occur spontaneously in the reverse direction, how can we understand photosynthesis? 1. Thermodynamics deals only with closed systems; photosynthesis is an open system. 2. Thermodynamics does not apply to living systems. 3. Thermodynamics does not apply to photochemical reactions. 4. This reaction is characterized by an energy change so close to zero that it is essentially reversible. 5. It is not a spontaneous chemical reaction; it is driven by an external source of energy - light

5. It is not a spontaneous chemical reaction; it is driven by an external source of energy - light correct Spontaneous occur naturally in one direction. A ball will spontaneous(naturally) roll down a hill. It cannot roll back UP a hill because that is un-natural, nonspontaneous, and unfavorable. The only way it goes back up the hill is with an external force such as someone rolling it back up. Photosynthesis does not spontaneously occur and requires an external force, (energy in this case) which is light.

Which of the following statements about gas laws is/are true? I) Boyle's law says that above the boiling point, the pressure and volume of a gas are directly proportional. II) Jacques Charles measured an inverse relationship between volume and temperature. III) The ideal gas law is only accurate at very high concentrations. 1. II, III 2. I, II, III 3. II only 4. I, II 5. None are true 6. I only 7. III only 8. I, III

5. None are true correct All three statements are false. Boyle's law describes an inverse proportionality between P and V , and says nothing about the boiling point. Charles' law describes a direct proportionality between V and T. Only gases that closely approximate the assumptions of kinetic molecular theory are well described by the ideal gas law - at high concentrations, intermolecular forces become significant.

We conduct an experiment by shining 500 nm light on potassium metal. This causes electrons to be emitted from the surface via the photoelectric effect. Now we change our source light to 450 nm at the same intensity level. Which of the following is the result from the 450 nm light source compared to the 500 nm source? 1. No electrons would be emitted from the surface. 2. Fewer electrons would be emitted from the surface. 3. More electrons would be emitted from the surface. 4. The same number of electrons would be emitted, but they would have a lower velocity 5. The same number of electrons would be emitted, but they would have a higher velocity

5. The same number of electrons would be emitted, but they would have a higher velocity correct 500 nm light has more energy than than the work function of potassium due to the fact that electrons were emitted. Therefore 450 nm light, which is higher in energy than the 500 nm light, will also emit electrons. The number of electrons emitted must be the same because the intensities (photons/s) are the same. However, the higher energy photons from the 450 nm light would yield electrons with a higher kinetic energy and therefore a higher velocity. 1/2mv^2 = hν − Φ

Write the ground-state electron configuration of a europium atom. 1. [Xe] 4f5 5d2 6s2 2. [Xe] 4f9 3. [Xe] 5d7 6s2 4. [Xe] 4f2 5d5 6s2 5. [Xe] 4f7 6s2

5. [Xe] 4f7 6s2 correct The Aufbau order of electron filling is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, etc. s orbitals can hold 2 electrons, p orbitals 6 electrons, and d orbitals 10 electrons. Note some exceptions do occur in the electron configuration of atoms because of the stability of either a full or half-full outermost d-orbital, so you may need to account for this by 'shuffling' an electron from the (n − 1) s orbital. Finally use noble gas shorthand to get the answer: [Xe] 4f7 6s2 .

Fill in the blanks: potassium is one of the most well-known elements in the alkali metal___________. It is in the ___________ which makes it a ___________ element. Its single valence electron is in the ______subshell of the ________shell, making it very reactive. It reacts readily with non-metals to form _______ 1. row; d block; main group; ℓ = 1; n = 4; salts 2. row; d block; non-metal; ℓ = 1; n = 3; alloys 3. series; s block; common; ℓ = 2; n = 4; networks 4. family; s block; reactive; ℓ = 0; n = 3; alloys 5. family; s block; main group; ℓ = 0; n = 4; salts

5. family; s block; main group; ℓ = 0; n = 4; salts correct Family refers to the common name of a group or groups of similar elements, e.g., rare earth, coinage metal, halogen. The number (1-18) of the column of an element is the group. All elements in rows 3-12 are called d-block elements, while the rest of the rows are called main group elements. Potassium is on row 4, but the principal quantum number n always refers to an electron shell. The 4s electron of K is its entire valence. K+ is isoelectronic to group 18 which are called the noble gases. The reaction of a metal and non-metal usually produces a salt.

For the evaporation of water from an open pan at 25◦C, the values of ∆S for the water, the surroundings, and the universe must be, respectively, 1. positive, positive, positive. 2. positive, negative, zero. 3. None of these is correct. 4. negative, negative, negative. 5. positive, negative, positive.

5. positive, negative, positive. correct The process is spontaneous, which means ∆Suniverse > 0 according to the Second Law of Thermodynamics. Entropy (S) is high for systems with high degrees of freedom, disorder or randomness and low for systems with low degrees of freedom, disorder or randomness. S(g) > S(ℓ) > S(s). The system is H2O(ℓ) → H2O(g) so ∆Ssystem > 0. The entropy of the surroundings must be negative. Energy is removed from the surroundings to get the water to evaporate.

The organic compound CH3CH2CH3 is called 1. pentane. 2. ethane. 3. methane. 4. butane. 5. propane.

5. propane. correct Propane is c3H8 Tip to remembering your first four alkanes: Monkeys Methane Eat Ethane Peanut Propane Butter Butane

An antibonding orbital is formed when 1. a px-orbital overlaps a pz-orbital. 2. a free electron is present in the molecule. 3. an s-orbital overlaps a p-orbital. 4. None of these is correct. 5. the overlap of the corresponding atomic orbitals leads to destructive interference.

5. the overlap of the corresponding atomic orbitals leads to destructive interference. correct In-phase overlap of two atomic orbitals results in molecular bonding orbitals; out-ofphase overlap of two atomic orbitals results in molecular antibonding orbitals.

For the reaction 2 SO3(g) → 2 SO2(g) + O2(g) ∆H◦r = +198 kJ·mol−1 at 298 K. Which statement is true for this reaction? 1. The reaction is driven by the enthalpy. 2. The reaction will not be spontaneous at any temperature. 3. The reaction will not be spontaneous at high temperatures. 4. ∆G◦r will be positive at high temperatures. 5. ∆G◦r will be negative at high temperatures.

5. ∆G◦r will be negative at high temperatures. correct ∆G = ∆H − T∆S is used to predict spontaneity. (∆G is negative for a spontaneous reaction.) ∆H is positive and T is always positive. For the reaction 2 mol gas → 3 mol gas. The more moles of gas, the higher the disorder, so ∆S is positive and ∆G = (+) − T(+). For ∆G to be negative, T must be large.

If we set up a bomb calorimetry experiment to determine the molar internal energy of combustion of ethene (C2H4) using 1 L of water as our heat sink, 2.805 g of ethene, and measure an initial and final temperature of 25.20◦C and 58.92◦C, respectively, what will be the experimentally determined molar internal energy of combustion of ethene? Assume the density of water is 1.00 g · mL−1 and the calorimeter itself absorbs no heat. The specific heat capacity of water is 4.184 J · g−1· K−1. 1. −14.11 kJ · mol−1 2. −141.1 kJ · mol−1 3. −14, 110 kJ · mol−1 4. −141, 100 kJ · mol−1 5. −1, 411 kJ · mol−1

5. −1, 411 kJ · mol−1 correct In bomb calorimetry, ∆Urxn = ∆H, and ∆H= -mCAT (combustion reaction)

The density of a gas is 1.96 g/L at STP. What would the mass of 2.00 moles of the gas be at STP? 1. 0.0875 g 2. 43.9 g 3. There is not enough information to solve. 4. 0.510 g 5. 44.8 g 6. 87.8 g

6. 87.8 g correct d = 1.96 g/L P = 1 atm T = 0◦C = 273 K R = 0.0821 L·atm/mol·K d =m/V M =(mRT) / (PV) = (m/V)(RT/P) =(d R T)/(P) = [(1.96 g/L)(0.0821 L·atm/mol·K)(273 K)]/(1 atm) = 43.9301 g/mol Two moles of the gas will weigh (2 mol) x 43.9301 g/mol = 87.8601 g .

The absolute entropy of a system (S measured in J/K) is related to the number of microstates in that system. Consider the three processes listed below. Which one(s) will result in an increase in the number of microstates in the system? I) The temperature of a gas is raised by 3◦C. II) A fixed amount of gas is allowed to expand to a slightly larger volume. III) The total number of gas molecules in a system is reduced to a smaller number. 1. II and III only 2. III only 3. I and III only 4. I only 5. I, II, and III 6. I and II only 7. II only

6. I and II only correct Raising the temperature will always add to the number of available energy states in a system. More volume allows more states as well. Reducing the number of molecules however, will lower the number of microstates.

Consider a chemical reaction that is endothermic and has a negative change in entropy. Which of the following is/are true? I) ∆Suniv is negative at all temperatures. II) This reaction will reach equilibrium when T = ∆H/∆S . III) The reaction is spontaneous only at relatively high temperatures. IV) ∆G is positive at all temperatures. 1. II and III only 2. I, II, III, and IV 3. I, III, and IV only 4. III only 5. I and II only 6. I and IV

6. I and IV only correct A reaction that is endothermic and has a negative change in entropy is nonspontaneous at all temperatures. This means that ∆Suniv is negative and ∆G is positive. Because it is never spontaneous, there will never be a point in which ∆G = 0.

Which of the following electronic transitions for a hydrogen atom would correspond to the highest energy emission found in the Balmer series? 1. n=2 to n=4 2. n=2 to n=1 3. n=3 to n=1 4. n=1 to n=2 5. n=3 to n=2 6. n=4 to n=2

6. n=4 to n=2 correct The Balmer series is electronics transitions the involve n=2 and some higher principle energy level (n greater than 2) for the hydrogen atom. An emission occurs when an electron moves from some energy level to a lower energy level. Combining these constraints, n=4 to n=2 will be the highest energy emission in the Balmer series, of the available answer choices.

Which of the following does not affect the ideality of gases? I) the temperature of the gas II) the density of the gas III) the size of the gas molecules 1. III only 2. I only 3. II only 4. I, II, and III 5. II and III 6. none of the above 7. I and III 8. I and II

6. none of the above correct All of the listed factors influence gas ideality

Using the provided bond enthalpy data, calculate the change in enthalpy for the following reaction: CH4(g) + O2(g) ←→ CH2O(g) + H2O(g) 1. −577 kJ · mol−1 2. 710 kJ · mol−1 3. 577 kJ · mol−1 4. −710 kJ · mol−1 5. 349 kJ · mol−1 6. −349 kJ · mol−1

6. −349 kJ · mol−1 correct Reactants - Products

150 grams of iron is heated from 25◦C to 150◦C. What is ∆S for this change? The specific heat capacity of iron is 0.450 J/g K. 1. −8438 J/K 2. 0 J/K 3. +121 J/K 4. −23.6 J/K 5. −121 J/K 6. +8438 J/K 7. +23.6 J/K

7. +23.6 J/K correct ∆S = mC ln (T2/T1)

Which of the following species is/are paramagnetic? I) Li−2 II) O2 III) H+2 1. I only 2. I and III 3. III only 4. II only 5. II and III 6. I and II 7. I, II and III

7. I, II and III correct Li−2 and H+2 both have an odd number of electrons and therefore must be paramagnetic. O2 has 16 total electrons, the last two of which must go into separate degenerate π∗ anti-bonding orbitals.

A chemist prepares a sample of helium gas at a certain pressure, temperature, and volume and then removes all but a fourth of the gas molecules (only a fourth remain). How must the temperature be changed (as a multiple of T1) to keep the pressure and the volume the same? 1. T2 =1/2T1 2. T2 = 16 T1 3. T2 = 2 T1 4. T2 =1/16 T1 5. None of these 6. T2 =1/4T1 7. T2 = 4 T1

7. T2 = 4 T1 correct Because we want P and V to be constant, we can use the relationship n1 · T1 = n2 · T2 . The fraction of the moles is the same as the fraction of the molecules, so n1 · T1 =(n1/4)T2 T2 = 4 T1 For V and P to remain constant, the temperature must be four times.

Calculate ∆G◦ for the following reaction at 298 K. NH4NǑ(s) → N2O(g) + 2H2O(g) 1. −113 kJ 2. +130 kJ 3. −1.33 × 105 kJ 4. −130 kJ 5. +169 kJ 6. +97.2 kJ 7. −169 kJ c

7. −169 kJ correct ∆G = ∆H − T∆S ∆S = [220 + 2(189)] − 151 = 447 J/K ∆H = [82 + 2(−242)] − (−366) = −36 kJ ∆G = −36000 − 298(447) = −169206 J ∆G = −169 kJ

An ideal gas occupies 45.0 L at 963 torr and 31.2◦C. What volume would it occupy at STP? 1. 0.0195 L 2. 45.7 L 3. 22.4 L 4. 52.3 L 5. 62.2 L 6. 71.2 L 7. 63.5 L 8. 51.2 L 9. 0.0157 L

8. 51.2 L correct P1 = 963 torr P2 = 760 torr V1 = 45 L T1 = 31.2◦C + 273.15 = 304.35 K Using the Combined Gas Law (P1 V1)/(T1) = (P2 V2)/(T2), and recalling that STP implies standard temperature (0◦C or 273.15 K) and pressure (1 atm or 760 torr), we have V2 = (P1 V1 T2)/(T1 P2) = (963 torr) (45 L) (273.15 K) / (304.35 K) (760 torr) = 51.1744 L

Which of the following sets of quantum numbers are invalid, i.e. violate one or more boundary conditions? I) n = 3, ℓ = 2, mℓ = −2, ms = +1/2 II) n = 9, ℓ = 5, mℓ = 6, ms = +1/2 III) n = 2, ℓ = 1, mℓ = 0, ms = +1 IV) n = 2, ℓ = 0, mℓ = 0, ms = +1/2 V) n = 1, ℓ = 0, mℓ = 0, ms = −1/2 1. III only 2. I, III, IV 3. IV only 4. I, IV 5. I, II, IV 6. II only 7. I only 8. II, III

8. II, III correct Set II and III are invalid. For II, ml = 6 is disallowed because ℓ = 5. For III, ms = +1 is disallowed because ms may only be +1/2 or −1/2 .

Under which of the following conditions is a real gas most likely to deviate from ideal behavior? 1. high volume 2. low density 3. if it is a noble gas 4. Tuesdays and Thursdays 5. zero pressure 6. low pressure 7. new moon 8. low temperature

8. low temperature correct Deviations from ideality occur due to molecular attractions or repulsions. More attractions or repulsions can occur when the molecules are closer together. Low pressure, high volume, and low density all correspond to molecules being far apart. Low temperature often corresponds to molecules being close together. It also corresponds to low kinetic energy which allows molecules to 'stick together' easier.

What is the expected ground state electron configuration for Zr2+ ? 1. [Xe] 6s2 2. [Xe] 3. [Xe] 5d2 4. [Xe] 6s25d4 5. [Kr] 5s14d5 6. [Kr] 5s2 7. [Kr] 8. [Kr] 5s24d2 9. [Kr] 4d2 10. [Kr] 5s24d4

9. [Kr] 4d2 correct Valence shell electrons are the first electrons lost when the atom ionizes. The 5s electrons are both lost first, leaving the two 4d electrons.

If 250 mL of a gas at STP weighs 2 g, what is the molar mass of the gas? 1. 179 g · mol−1 2. 44.8 g · mol−1 3. 56.0 g · mol−1 4. 28.0 g · mol−1 5. 8.00 g · mol−1

1. 179 g · mol−1 correct V = 250 mL P = 1 atm T = 0◦C = 273.15 K m = 2 g The density of the sample is ρ =m / V= 2 g / 0.25 L = = 8 g/L The ideal gas law is P V = n R T with unit of measure mol/L on each side. Multiplying each by molar mass (MM) gives n/V · MM = P / R T · MM = ρ with units of g/L MM = ρ R T / P =179.318 g/mol

What volume will 40.0 L of He at 50.00◦C and 1201 torr occupy at STP? 1. 53.4 L 2. 26.7 L 3. 31.1 L 4. 12.8 L 5. 18.6 L

1. 53.4 L correct P1 V1 / T1 = P2 V2 / T2 Convert Torr to ATM

Rank the following species from least to greatest electron affinity: F, Ge, S, As, Se. 1. As < Ge < Se < S < F 2. Ge < As < Se < S < F 3. F < Se < S < Ge < As 4. F < S < Se < Ge < As 5. As < Ge < S < Se < F

1. As < Ge < Se < S < F correct The electron affinity trend increase as one moves from the lower left corner of the periodic table to the upper right corner of the periodic table, with exceptions occurring at filled and half-filled subshells, wich are inherently stable and this have a lower-than-expected electron affinity.

Which of the following has the highest lattice energy? 1. MgO 2. BaO 3. KI 4. NaCl 5. CaO

1. MgO correct Mg2+ and O2− have the highest charge densities.

Consider a solid that has a molar mass of 180.2 g/mol and a melting point of 423K. This solid is a terrible electrical conductor, even when fully dissolved in an aqueous medium. What type of solid is this compound? 1. Molecular 2. Network 3. Ionic 4. Metallic

1. Molecular correct Since the solid has a relatively small molar mass and low melting point, and it is not a conductor in solution, it is most likely molecular compound.

Consider two separate 1 L gas samples, both at the same temperature and pressure. The two gases have different molar masses. Which is true? 1. The particles in both gas samples have the same average kinetic energies. 2. You would need more information to be able to compare the average kinetic energies of these two gas samples. 3. The particles in both gas samples have different average kinetic energies.

1. The particles in both gas samples have the same average kinetic energies. correct Average kinetic energies of gases are determined by temperature.

What is the ground state electron configuration expected for germanium, Ge? 1. [Ar] 4s23d104p2 2. [Kr] 4s13d103p3 3. [Kr] 4s23d104p2 4. [Ar] 4s13d104p3 5. [Ar] 4s23d103p2 6. [Kr] 7. [Ar] 4s24d104p2 8. [Ar] 4s23f104p2 9. [Ar]

1. [Ar] 4s23d104p2 correct Ge is in the fourth row and Group IV.

The dominant forces between molecules (intermolecular forces) are in origin. 1. electrostatic 2. electrodynamic 3. electromagnetic 4. gravitational 5. magnetic

1. electrostatic correct Intermolecular forces are all electrostatic in origin.

The speed of light in air 1. is independent of the wavelength and frequency of light. 2. depends only on the wavelength of light. 3. depends on both the wavelength and the frequency of light. 4. depends only on the frequency of the light.

1. is independent of the wavelength and frequency of light. correct The speed of light in air is constant (3.00 × 108 meters/second) regardless of the wavelength or frequency.

Which of the following elements would have the lowest first ionization energy? 1. O 2. Na 3. Rb 4. Cl 5. I

3. Rb correct Ionization energy generally increases from left to right and decreases going down the Periodic Table.

Which set of quantum numbers does NOT provide a satisfactory solution to the wave equation? 1. n = 2, ℓ = 0, mℓ = −1 2. n = 1, ℓ = 0, mℓ = 0 3. n = 4, ℓ = 2, mℓ = +2 4. n = 5, ℓ = 3, mℓ = −3 5. n = 3, ℓ = 2, mℓ = −1

1. n = 2, ℓ = 0, mℓ = −1 correct ℓ is (n-1) and mℓ is any number equal to or between +ℓ or -ℓ

Which of the following compounds would you expect to have the highest S? 1. CH4(g) 2. CH3F(g) 3. Ar(g) 4. H2O2(ℓ) 5. C6H14(ℓ)

You should compare first based on the phase and second based on the complexity of the molecule. The most complex molecule in the gas phase at standard conditions is CH3F(g)