UCI General Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

Midterm 2 Review Packet

Evaluation Link: https://forms.gle/X68ZYBmyQAaCB5EN9

Week 4 Material: Dalton’s Law, Effusion, K-M Theory, Ideal vs Real Gases ( ) UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

Week 5 Material: Liquids, Solids and Intermolecular Forces (Raychelle)

1. What kind of attractive forces must be overcome to: a. melt ice Only a change in state is occurring, so only intermolecular forces must be broken. Between molecules, there are bonding, dipole-dipole forces, and London dispersion forces. You can think of hydrogen bonding as a type of specialized dipole-dipole force, however. b. boil molecular bromine Like the previous question, there is no chemical change; only a change in state is occurring. Molecular bromine has no dipole and is not capable of hydrogen bonding, so the only force to overcome is London dispersion. c. melt solid iodine Iodine is similar to bromine in that its only is London dispersion. Therefore, in a change of state, only London dispersion forces must be overcome.

d. dissociate F2 into F Unlike the previous problems, this is a chemical change, not a physical change. The molecules are being dissociated, meaning a bond is broken. The force to be

overcome is the between the two fluorine atoms in F2.

Take Away: When changing phases (g,l,s), you are dealing with IMFs. When changing compounds, you are dealing with a chemical change where intramolecular forces are broken and new ones are formed.

2. List the following substances in order of increasing boiling point. UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

Answer by increasing boiling point: D

Start first with the strongest IMF. Which molecules are capable of H-bonding? Dipole-Dipole? All atoms have London dispersion, so start with the strongest, largest differences, then get into the smaller differences (i.e. which has strong LDF).

Molecules A and B have bonded to hydrogens. This fulfills the requirement for an atom to be capable of hydrogen bonding: F, O, N directly bonded to a H. Thus, molecules A and B have stronger IMF’s than C and D which are not capable of hydrogen bonding. (Note: If an atom is capable of hydrogen bonding, it is also capable of dipole-dipole)

Examining A and B: Substances A and B can form hydrogen bonds, but B has one more possible site of bonding than A (has 2 OH rather than only 1 OH). So B has stronger intermolecular forces than A.

Examining C and D: Substances C and D are made of only and are therefore nonpolar and cannot , so we can only look at London dispersion forces. Larger molecules have stronger London dispersion forces, so C has stronger intermolecular forces than D.

If time allows, question H-bond donor, acceptor and whether an atom can be both. UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

3. When the liquid metal mercury, Hg, is placed in a small tube, the meniscus actually curves upward (see figure below), just the opposite of water. The reason for this is that:

a. The cohesive force is greater than the adhesive force. b. The adhesive force is greater than the cohesive force. c. The density of mercury is much larger than that of water. d. The density of mercury is much greater where it is in constant with the glass. e. Mercury is less volatile than water. Answer: A The cohesive force between mercury atoms is much stronger than the adhesive forces of mercury and glass. This creates a strong surface tension in mercury as well as the upward bulge in the test tube. In water, the adhesive forces are greater, which causes the molecules on the edge of the test tube to cling to the glass while the rest hang down from the sides. This is why water curves in the opposite direction. 4. The structures of two isomers of heptane are shown here. Which of these two compounds would you expect to have the greater viscosity? UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

Answer: Compound A

a. Compound A will have a greater viscosity than compound B. The factors that affect viscosity are temperature and intermolecular forces i. Temperature: a higher temperature decreases viscosity since there is more motion and , the intermolecular forces are easier to overcome to allow for the liquid to flow. ii. Intermolecular forces: the stronger the intermolecular forces the greater the viscosity. 1. Compounds A and B only have London dispersion forces as they are nonpolar molecules. Compound A has stronger forces than compound B because a greater surface area allows for more interactions. Thereby since compound A has more/stronger London dispersion forces it will have the higher viscosity. iii. (side note: surface tension, the energy needed to stretch or increase the surface of a liquid by unit area, is similar to viscosity in that it increases with greater intermolecular forces and decreases with temperature). 5. How will the time it takes to hard-boil an egg compare at higher altitudes? Answer: It will take longer. Recall the scientific definition of boiling: vapor pressure of the liquid is equal to the pressure of the surrounding environment. There is less atmospheric pressure at higher altitudes than at lower ones. Thus, at higher altitudes, water boils at a lower temperature than at sea level. 6. One way our body is cooled is by evaporation of the water in sweat. In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); ∆퐻 = 43.46 kJ/mol at 37 °C. 푣푎푝 Start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:

1000 푔 1 푚표푙 43.36 푘퐽 3 1. 5 퐿 푥 1 퐿 푥 18 푔 푥 1 푚표푙 = 3. 6 푥 10 푘퐽 Thus, 3600 kJ of heat is removed by the evaporation of 1.5 L of water.

Week 6 Material: Phase Diagrams and Structures of Solids ( ) 1. How much heat is required to convert 200 grams of -25 °C ice into 150 °C steam? Given: Specific heat of ice = 2.06 J/g°C Specific heat of water = 4.19 J/g°C Specific heat of steam = 2.03 J/g°C

Heat of fusion of water ΔHf = 334 J/g

Heat of vaporization of water ΔHv = 2257 J/g

Heating cold ice to hot steam requires five distinct steps, which when you add all the heats together, you get the total heat you need to do this process. 1. Heat -25 °C ice to 0 °C ice a. Q1 = mcΔT

b. ΔT = Tf - Ti = 0 - (-25C) = +25C c. Q1 = (200 g) x (2.06 J/g°C) x (25 °C) = 10300 J 2. Melt 0 °C solid ice into 0 °C liquid water

a. Q2 = m · ΔHf b. Q2 = 200 · 334 J/g c. Q2 = 66800 J 3. Heat 0 °C water to 100 °C water a. Q3 = mcΔT

b. ΔT = Tf - Ti = 100 - (0) = +100C c. Q3 = (200 g) · (4.19 J/g°C) · (100 °C) d. Q3 = 83800 J 4. Boil 100 °C liquid water into 100 °C gaseous steam

a. Q4 = m · ΔHv b. Q4 = 200 · 2257 J/g UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

c. Q4 = 451400 J 5. Heat 100 °C steam to 150 °C steam a. Q5 = mcΔT b. ΔT = (150 °C – 100 °C) c. Q5 = (200 g) · (2.03 J/g°C) · (50 °C) d. Q5 = 20300 J 6. Find the Total Heat a. Qtotal = Q1 + Q2 + Q3 + Q4 + Q5 b. Qtotal = 10300 J + 66800 J + 83800 J + 4514400 J + 20300 J c. Qtotal = 632600 J = 632.6 kJ

2. Using the phase diagram for water given in the figure below, determine the state of water at the following temperatures and pressures: a. −10 °C and 50 kPa i. Solid ii. This point is in the solid region, so it is a solid. b. 25 °C and 90 kPa i. Liquid ii. This point is in the liquid region, so it is a liquid. c. 50 °C and 40 kPa i. Liquid ii. This point is in the liquid region, so it is a liquid. d. 80 °C and 5 kPa i. Gas ii. This point is in the gas region, so it is a gas. UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

3. Fill in this table

Unit Cells

Image of Unit Name of # of Atoms Coordination Number Edge Stacking Cell Unit Cell per Unit Length of Atoms Cell Formula (if if applicable) applicable (ex: abcabc, abab)

Simple 1 atom/unit 6 n/a Cubic cell l=2r

There are 8 corner atoms in a simple cubic unit cell. If an atom is Imagine you are the located in a black point on the 3D corner, it is plane and the other considered colored points are as ⅛ th of UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

an atom. immediately next to Since there you. (closer together are 8, this than in the picture so adds up to 1 that all the points are full touching) There are a atom/unit total of 6 different cell. atoms touching you, so the coordination # is 6.

Body 2 atom/unit 8 n/a Centered cell Look at the provided 푙 = 4푟 Cubic image to the left. 3 Imagine you are the full atom in the center of the unit cell. There are 8 total atoms touching you, so the coordination # is 8.

FaceCentere 4 atom/unit 12 abcabc d/Cubic cell l=r 8 Close Packed Cubic

Imagine you are the center atom. There are a total of 12 atoms touching you, so the coordination UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

Hexagonal 2 atom/unit 12 n/a abab Closed cell Packed (hcp)

Imagine you are the center atom. There are a total of 12 atoms touching you, so the coordination number is 12.

4.

Lattice Type Hint (images) Ratio of Size comparisons between cation and anion (blue = anion) Cations (red = cation) to Anion

Cesium 1:1 Similar Sized Chloride

Rock 1:1 Medium difference in size (NaCl) UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

Zinc Blende 1:1 Very different sized ions

Fluorite 1:2 N/A (for this class)

Antifluorite 2:1 N/A (for this class)

5. Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm. a. What is the atomic radius of Ag in this structure? b. Calculate the density of Ag. UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin

6. An element has a simple body-centred cubic unit cell. The corresponding density is 4.253 g/cm3 and the atomic radius is 178 pm. Calculate the atomic mass (in g/mol) for this element and identify what this element could be based on the atomic mass. UCI General Chemistry Peer Tutoring Chemistry 1B Professor Holton Website: https://sites.uci.edu/gcptutoring/ Tutors: Nile Luu, Raychelle Benito, Craig Sutkin