Questions and Problems Involving Content from Chapters 1-3

1. Public bowling alleys generally stock bowling balls from 8 to 16 lb, where the mass is given in whole numbers. Given that regulation bowling balls have a diameter of 8.6 in, which (if any) of these bowling balls would you expect to float in water?

The volume of the bowling ball is given by

3 33 3 34 4 8.6 in 2.54 cm

r = × = 5.5 × 10 cm3 3 2 1 inπ π

Starting with an 8 lb bowling ball and assuming two significant figures in the mass, converting pounds to grams

gives

3453.6 g8 lb × = 3.6 × 10 g

1 lb

So the density of an 8 lb bowling ball would be

33

3 3

mass 3.6 × 10 gdensity = = = 0.65 g/ cm

volume 5.5 × 10 cm

Carrying out analogous calculations for the higher-weight bowling balls gives

ball weight (lb) mass (g) density (g/cm3) 9 4.1 103 0.75 10 4.5 103 0.82 11 5.0 103 0.91 12 5.4 103 0.98 13 5.9 103 1.1

Therefore we would expect bowling balls that are 11 lb or lighter to float because they are less dense than

water. Bowling balls that are 13 lb or heavier would be expected to sink because they are denser than water. The 12 lb bowling ball is borderline, but it would probably float.

Note that the above calculations were carried out by rounding off the intermediate answers, as discussed in

Section 1.8 of Change 12e. If we carry an additional digit past the number of significant figures to minimize errors from rounding, the following densities are obtained:

ball weight (lb) density (g/cm3) 10 0.83 11 0.91 12 1.01

The differences in the densities obtained are slight, but the value obtained for the 12 lb ball now suggests that

it might sink. This problem illustrates the difference rounding off intermediate answers can make in the final answers for some calculations.

practice problems adapted, in part, from previous 1045/1050 exams – page 2

2. (Spring 2016) One of the most exciting areas of chemistry today is in the field of nanoscience and its ambitious cousin, nanotechnology. A major player in the new science and applications of nanoscience is the quantum dot. Quantum dots are nanoparticles consisting of a small number of atoms, and they look to have great potential in number of areas, including diagnostic and therapeutic medicine, for example, the treatment of cancer. Nanoparticles formed from metals like gold or silver raise concerns about heavy-metal toxicity. Recent work in several different labs has shown promise for treating cancer with nanoparticles based on calcium carbonate, CaCO3(s).

(See www.sciencedaily.com/releases/2016/02/160202143936.htm.) A. The CaCO3 particles were described as being “100-nanometer-sized.” Let’s see if that holds up.

A typical CaCO3 nanoparticle weighs 4.8 1015 g. Assuming a cubic particle, determine the length on one side of the cube in nm. The density of CaCO3 is 2.17 g/cm3.

4.8 1015 g CaCO3 x 1 cm3/2.17 g = 2.21 x 10-15 cm3 x (109 nm/100 cm)3 = 2.21 x 106 nm3 take the cube root => 130 nm = one side of the cube

(So “100-nanometer-sized” is about right.) B. Calcium carbonate is the active ingredient of some antacids; for example, Tums. The acid-base

chemistry of CaCO3 plays an important role in its mechanism as an antitumor agent, and we will look at that chemistry as we work through Chapter 4 in our text.

One Tums “Extra Strength” tablet contains 750 mg of CaCO3. How many 4.8 1015 g CaCO3 nanoparticles could you make from the calcium carbonate in one Tums “Extra Strength” tablet?

750 mg CaCO3 x (1 g/1000 mg) = 0.75 g CaCO3 0.75 g CaCO3 / 4.8 1015 g/nanoparticle = 1.6 x 1014 nanoparticles

3. (Fall 2012) Because of the very small size of quantum dots, their physical properties are often different than the bulk materials. For example, we think of color as being an intensive property (it does not depend on the amount of material), but the color of quantum dots varies with size. (We will consider quantum effects in Chapter 7; see page 314 for more information about quantum dots. Geoff Strouse and Hedi Mattoussi in our department are leaders in this field.)

A lot of work has been done with quantum dots based on the compound CdSe. A. Orange CdSe quantum dots have a diameter of 7 107 cm. (That would be 7 nm, hence the term

“nanoparticle.”) Assuming quantum dots to be spherical, calculate the volume of an orange CdSe quantum dot in cm3.

3-7

3 -19 34 4 7 × 10 cmV = r = = 2 × 10 cm

3 3 2π π

B. A blue CdSe quantum dot has a volume of 4 1021 cm3. Assuming the density of a CdSe quantum dot is the same as that of the bulk compound (4.82 g/cm3), determine the mass of a blue CdSe quantum dot.

-21 3 -2093

4.82 g 4 × 10 cm = 1. × 10 g

cm

practice problems adapted, in part, from previous 1045/1050 exams – page 3

C. On average, how many CdSe formula units would be contained in a blue CdSe quantum dot. One CdSe formula unit has a mass of 112.41 + 78.96 = 191.37 amu.

191.37 amu x 1.66 x 1024 g/amu = 3.18 x 1022 g

2019

221. 10 g / blue dot

6 x 10 CdSe formula units per blue quantum dot3.18 10 g / CdSe formula unit

In other words, only around 60 Cd2+ ions and 60 Se2 ions – a very small number of ions!

4. Fusing “nanofibers” with diameters of 100–300 nm gives junctures with very small volumes that would potentially allow the study of reactions involving only a few molecules. Estimate the volume in liters of the junction formed between two such fibers with internal diameters of 200 nm.

The volume could be approximated in several ways. One approach would be to model the junction as a cylinder

with an internal diameter of 200 nm and a height of 200 nm. The volume of the cylinder would be

32 22 -18

9 3

200 nm 10 cm 1 mL 1 Lr h = × 200 nm × × × = 6 × 10 L

2 10 nm 1 cm 1000 mLπ π

To get a sense of that volume relative to a molecule, consider that the volume of a water molecule in liquid

water is roughly 3 1026 L. Therefore a volume of 6 1018 L could still potentially contain (6 1018 L)/( 3 1026 L) ≈ 2 108 water molecules! Water molecules are very small, and they pack tightly

in the liquid state due to strong intermolecular forces (Chapter 11). For larger molecules such as those involved in biological processes, it might be possible to trap a much smaller number in a nanofiber juncture.

5. In water conservation, chemists spread a thin film of a certain inert material over the surface of water to cut

down on the rate of evaporation of water in reservoirs. This technique was pioneered by Benjamin Franklin three centuries ago. Franklin found that 0.10 mL of oil could spread over the surface of water about 40 m2 in area. Assuming that the oil forms a monolayer, that is, a layer that is only one molecule thick, estimate the length of each oil molecule in nanometers.

We assume that the thickness of the oil layer is equivalent to the length of one oil molecule. We can calculate

the thickness of the oil layer from the volume and surface area. The thickness of the oil layer corresponds to the length of one molecule. Therefore, the volume of oil can be calculated from the

equation: length of molecule surface area = volume of oil

22 5 21 cm

40 m × = 4.0 × 10 cm0.01 m

Given that 0.10 mL = 0.10 cm3, we write volume = surface area thickness. Rearranging that equation gives

3

-75 2

volume 0.10 cmthickness = = = 2.5 × 10 cm

surface area 4.0 × 10 cm

Converting to nm:

–7–9

0.01 m 1 nm2.5× 10 cm× × = 2.5 nm

1 cm 1× 10 m

practice problems adapted, in part, from previous 1045/1050 exams – page 4

6. (Spring 2016) The International Union of Pure and Applied Chemistry (IUPAC) recently announced that synthesis of the four elements needed to completing the seventh row of the periodic table had been confirmed. The new elements will be named and given atomic symbols later this year, but in the meantime a systematic naming system will be used for the four new elements: ununtrium, Uut; ununpentium, Uup; ununseptium, Uus; and ununoctium, Uuo.

(See: www.rsc.org/chemistryworld/2016/01/new-elements-periodic-table-seventh-row-iupac.)

The specific isotopes synthesized are listed below. For each one give the number of protons and neutrons.

1. 279113 Uut 113 protons 279-113 = 166 neutrons 2. 288

115 Uup 115 protons 288-115 = 173 neutrons

3. 293117 Uus 117 protons 293-117 = 176 neutrons 4. 294

118 Uuo 118 protons 294-118 = 176 neutrons

Which of these elements do you expect to be the least reactive? ununoctium, Uuo

Explain your reasoning. It is in the same column as the noble gases. Write the formula of the compound that you would expect to form between

ununseptium (Uus) and barium (Ba). BaUus2

7. (Spring 2016) Work just published in the journal Science (1/29/2016) suggests that our moon was formed by a head-on collision with another object almost 4.5 billion years ago, not a glancing blow as was once thought. They reached this conclusion by comparing the concentrations of oxygen-16 and oxygen-17 found on Earth and the moon.

What is the difference between oxygen-16 and oxygen-17? (Be specific without using a lot of words.) An oxygen-17 nucleus has one more neutron than an oxygen-16 atom. Earth, Mars, and other planetary bodies in our solar system each has a unique ratio of oxygen-16 to oxygen-

17. The distribution of isotopes would be different for other elements, as well. How and why would the periodic table look different for different planets?

Different distribution of isotopes => different average atomic masses for those elements in the periodic table.

8. (Fall 2015) Breaking News: At 11 am this morning NASA announced that they have evidence liquid water still flows on Mars. Water can stay liquid at such low temperature because of the high concentration of salt dissolved in the water, and some scientists believe perchlorate ions play a major role.

Write a formula for the perchlorate ion. Chlorate is -3ClO , so perchlorate is - -

3 4ClO + O => ClO .

9. (Spring 2016) Balance the following chemical equations and answer the associated questions.

A. 2NH3 + 3CuO 3Cu + N2 + 3H2O Name CuO. copper(II) oxide

B. 3NaClO + NH3 3NaOH + NCl3 Name NaClO. sodium hypochlorite Name NaOH. sodium hydroxide Name NCl3. nitrogen trichloride

practice problems adapted, in part, from previous 1045/1050 exams – page 5

C. 2NaOH + H2SO4 Na2SO4 + 2H2O Determine the molar mass of Na2SO4. 2(22.99) + 32.07 + 4(16.00) = 142.00 g/mol D. Write a balance chemical equation for the combustion of propane pentane, C5H12. C5H12 + 8O2 5CO2 + 6H2O

10. (Fall 2015) Boron and Fluorine. A. Give the number of protons and neutrons in the following isotopes: 1. boron-10 5 protons 10 – 5 = 5 neutrons 2. boron-11 5 protons 11 – 5 = 6 neutrons 3. fluorine-19 9 protons 19 – 9 = 10 neutrons B. Balance the following chemical equations involving boron and/or fluorine: 1. B + Br2 → BBr3

B + 3/2Br2 → BBr3 (accept 2B + 3Br2 → 2BBr3) 2. B2O3 + C → B4C + CO

2B2O3 + 7C → B4C + 6CO 3. F2 + H2O → O2 + HF

F2 + H2O → 1/2O2 + 2HF (accept 2F2 + 2H2O → O2 + 4HF) C. Name the following compounds:

1. CaF2 calcium fluoride 2. CoF2 cobalt(II) fluoride 3. BF3 boron trifluoride

11. (Fall 2012) Cadmium and selenium are involved in some very interesting and important chemistry outside of nanoscience.

A. How many protons and neutrons would you expect to find in the most common isotope of selenium?

Because the (average) atomic mass of selenium is 78.96, it is reasonable to assume that the most common isotope is selenium-79, so that the #p+ + #n = 79. The atomic number of selenium is 34, so the nucleus would contain 34 p+ and 79 34 = 45 n.

_______ protons _______ neutrons Why would it be harder to answer this question for cadmium?

Because the (average) atomic mass of cadmium is 112.41, which is midway between two whole numbers, suggesting significant contributions from at least two isotopes.

34 45

practice problems adapted, in part, from previous 1045/1050 exams – page 6

B. Cadmium is a very important element in the manufacturing of batteries, the most important example being nickel-cadmium (“NiCad”) batteries. Name the following compounds involved in the chemistry of NiCad batteries:

NiO ______________________________ Ni(OH)2 ______________________________

C. One could argue that selenium is even more important cadmium, given the role it played in saving the

planet from extraterrestrials, as we all learned in the 2001 documentary Evolution. Name the following selenium compounds:

SeO2 ______________________________ K2Se ______________________________

Given that the ion 24SO is named sulfate, we can infer that the ion 2

4SeO would be named selenate. Name the following compounds containing the selenate ion: Cr2(SeO4)3 __________________________________________________ MgSeO4 4H2O __________________________________________________ Name the following ions:

23SeO ________________________________________

4HSeO ________________________________________

12. (Spring 2016) Remember Problem 2? So, where do we get calcium carbonate? Calcium carbonate is ubiquitous in nature, so it can be mined and purified, but it can also be prepared from

calcium oxide and carbon dioxide. It’s a two-step process, but for our purposed we can combine the steps to get the following overall reaction:

CaO(s) + CO2(g) CaCO3(s) A. Suppose you wanted to make 25 g of CaCO3. Turns out you also have 25 g of CaO. Is that enough to

make 25 g of CaCO3? Explain you reasoning using words or numbers from Part B, but be sure to actually answer the question: Do you have enough CaO to make 25 g of CaCO3?

Molar mass of CaO is less than the molar mass of CaCO3, so 25 g of CaO will contain more moles of CaO

than the moles of CaCO3 in 25 g. Therefore, 25 g of CaO will make more than 25 g of CaCO3.

- or-

Only one product is formed, so this reaction simply adds mass to the 25 g of CaO to give more than 25 g of CaCO3.

-or- Use the numbers calculated below in Part B.

nickel(II) hydroxide

nickel(II) oxide

selenium dioxide

potassium selenide

chromium(III) selenate

magnesium selenate tetrahydrate

selenite

hydrogen selenate

practice problems adapted, in part, from previous 1045/1050 exams – page 7

B. Determine the amount of CaCO3 you can make from 25 g of CaO, assuming CaO is the limiting reagent.

CaO => 40.08 + 16.00 = 56.08 g/mol CaCO3 => 40.08 + 12.01 + 3(16.00) = 100.09 g/mol 25 g CaO x (1 mol/56.08 g) x (1 mol CaCO3/1 mol CaO) x (100.09 g/mol) = 47 g CaCO3 Does the amount you calculate agree with your answer to Part A? Yes C. Quality control told you to send a sample of the CaCO3 out to have it analyzed for percent calcium by

mass. If the sample is pure, what should you get for %Ca? Lots of ways to approach this calculation. Here’s one... Let’s assume we have 1 mole of CaCO3. It would weigh 100.09 g. One mole of CaCO3 would contain one mole of Ca with atomic mass 40.08 g/mol. (40.08 g Ca / 100.09 g CaCO3) x 100% = 40% (2 sf) Ca Suppose the %Ca is too high. What might that mean? The sample includes an impurity. Suppose the %Ca is too low. What might that mean? The sample includes unreacted CaO.

13. (Fall 2009) Some unusually heavy isotopes of magnesium, aluminum, and silicon were reported a couple of

years ago (Nature, 2007, 449, 1022). Give the number of electrons, protons, and neutrons in the following species:

A. a neutral 44Si atom _______ electrons _______ protons _______ neutrons the most common 40Mg ion _______ electrons _______ protons _______ neutrons B. How many more neutrons does an 43Al atom contain compared to an average aluminum atom?

14. (Fall 2009) Concerns about vitamin D deficiency have received a great deal of attention over the last few

years as it was discovered that a significant number of people are deficient in this important vitamin. Adults identified to have insufficient amounts of vitamin D are often recommended to take a daily supplement containing 2000 IU (International Units) of vitamin D, which corresponds to 50 μg. (The IU to mass conversion varies with different drugs and vitamins since the IU reflects the potency of the drug.)

The annual production of vitamin D is roughly 10 ton per year. Assuming all of this vitamin D was used to manufacture pills containing 50 μg of the vitamin per pill, and that these pills are sold in bottles containing 100 pills per bottle, how many bottles could be produced in one year?

1414

10

30

12 28

practice problems adapted, in part, from previous 1045/1050 exams – page 8

15. (Fall 2009) Phosphorous Chemistry A. Balance the following chemical equations involving phosphorous compounds: 1. _____P4(l) + _____Cl2(g) _____PCl3(g) 2. _____P4O10(s) + _____H2O(l) _____H3PO4(aq) 3. _____PCl5(s) + _____H2O(l) _____H3PO4(aq) + _____HCl(aq) B. Name the following compounds that appeared in the above reactions. 1. PCl3(g) ___________________________________________________ 2. P4O10(s) ___________________________________________________ 3. H3PO4(aq) ___________________________________________________ C. The following reaction is balanced for you:

P4(s) + 3NaOH(aq) + 3H2O(l) 3NaH2PO2(aq) + PH3(g)

I had never encountered the compound NaH2PO2 before writing this exam, so I typed it into the Google and sure enough, not only does it exist, but I discovered that it is used to treat nervous disorders and phthisis (an archaic term for atrophy of the body).

Suppose you wanted to prepare 25.0 g of NaH2PO2. How many grams of P4 would be needed, assuming complete conversion of P4 to NaH2PO2? The molar mass of NaH2PO2 is 87.978 g; i.e. the molecular weight is 87.978 g/mol.

D. Now suppose that you wanted to prepare 25.0 g of KH2PO2 from P4 and KOH. Would you need more,

less, or the same amount of P4 compared to what you calculated in part C?

Explain your reasoning, briefly, but also in sufficient detail; i.e. do not just say how NaH2PO2 and KH2PO2 are different, but also explain how this difference led to your answer.

E. Name the compound NaH2PO2. sodium dihydrogen hypophosphite F. When NaH2PO2 is isolated as a solid, it is actually obtained as NaH2PO2 H2O. Name this compound.

sodium dihydrogen hypophosphite hydrate

6

6 4

4

4 5

tetraphosphorous decaoxide

phosphorous trichloride

phosphoric acid

practice problems adapted, in part, from previous 1045/1050 exams – page 9

16. (Fall 2009) More of the name game.... A. Predict the formula for the compound formed between aluminum and nitrogen ____________________ B. Name the following compounds: 1. Cr(NO2)3 _____________________________________________________ 2. Ba(IO4)2 _____________________________________________________ 3. NH4C2H3O2 _____________________________________________________ C. Give formulas for the following compounds: 1. calcium hydroxide ___________________________________ 2. platinum(II) chloride dihydrate ___________________________________

D. Name the acid formed from the following anions in water: 1. S2 __________________________________________ acid 2. ClO3

__________________________________________ acid E. The naming of compounds containing metalloids is a bit tricky. Name the compound Sb2S3 as... 1. a molecular compound. __________________________________________ 2. an ionic compound. __________________________________________

practice problems adapted, in part, from previous 1045/1050 exams – page 10

17. (Fall 2009) Prior to 1982, pennies were 95% copper, but since then pennies are made mostly of zinc covered by a thin layer of copper. Which will contain more pennies: a pound of pennies minted in 1970, or a pound of pennies minted in 1990? (The density of Cu is 8.95 g/cm3; the density of Zn is 7.14 g/cm3.)

a pound of pennies minted in 1990

Explain your reasoning. (Do not just say “because copper is more dense”; explain how the difference in density led to your answer.)

18. (Summer 2010) Fluorescent light bulbs use considerably less energy than incandescent light bulbs, which

saves money and reduces the amount of carbon dioxide released into the atmosphere, but fluorescent bulbs also contain mercury (Hg) which is a toxic heavy metal of some concern. The Environmental Protection Agency has estimated that approximately 670 million fluorescent light bulbs are discarded each year in the United States. The amount of mercury in a fluorescent light bulb typically ranges between 3.5 mg to 15 mg, so it is reasonable to assume an average of 9 mg Hg per bulb.

(Source: http://www.epa.gov/wastes/hazard/wastetypes/universal/lamps/index.htm) Some additional information about our favorite liquid metal: The density of mercury is 13.534 g/cm3. The current market value of mercury is around $650 per flask, where 1 flask of mercury = 76 lbs of

mercury. A. What is the total amount of mercury (in g) discarded each year in fluorescent light bulbs?

What is that value in tons?

practice problems adapted, in part, from previous 1045/1050 exams – page 11

B. What is the total volume of mercury (in L) discarded each year in fluorescent light bulbs?

C. What would be the total value of the mercury (in dollars) discarded each year in fluorescent light bulbs?

19. (Summer 2010) Chemical Name Dropping.

A. Name the following anions: 1. HCO3

_____________________________________________________ 2. IO _____________________________________________________

3. MnO3 _____________________________________________________

B. Name the following: 1. CrBr3 7H2O _____________________________________________________ 2. Ba(HCrO4)2 _____________________________________________________ 3. BrF5 _____________________________________________________ C. Give formulas for the following compounds: 1. calcium chlorate ______________________________ 2. barium phosphite ______________________________ 3. tetraphosphorous hexaoxide ______________________________ 4. copper(II) nitrate decahydrate ______________________________

practice problems adapted, in part, from previous 1045/1050 exams – page 12

20. (Summer 2010) Balance the following chemical equations:

A. Fe2O3 + CO Fe + CO2 (important in the conversion of iron ore to cast iron) B. C3H8 + O2 CO2 + H2O (the reaction that powered many a backyard barbeque) C. SF4 + H2O H2SO3 + HF Now name the two acids produced in the above reaction. H2SO3 ________________________________________ acid HF ________________________________________ acid

practice problems adapted, in part, from previous 1045/1050 exams – page 13

21. (Summer 2010) A compound isolated from a marine organism showed promising antibacterial properties. A subsequent analysis showed the compound to be an organic thiol consisting of carbon, hydrogen, and sulfur. A small sample of the purified compound was sent off for elemental analysis which showed that the compound is 38.66% C and 9.73% H by mass. What is the empirical formula of this compound?

22. (Fall 2008) Common gases used in laboratories are generally obtained from pressurized metal gas

cylinders, but for small amounts of occasionally-used gases, it is sometimes easier just to prepare them chemically. For example, oxygen gas can be prepared by heating KMnO4(s) according to the following chemical reaction:

2KMnO4(s) K2MnO4(s) + MnO2(s) + O2(g)

A. How many grams of KMnO4 would you need to produce 0.27 moles of O2, assuming 100% conversion?

The molar mass of KMnO4 is 158.034 g/mol. B. The above procedure was carried out, and it was later determined that all of the KMnO4 reacted

according to the above equation except 11.7 g. What was the percent yield for the reaction?

practice problems adapted, in part, from previous 1045/1050 exams – page 14

C. Name the following compounds: KMnO4 ____________________________________________________________ MnO2 ____________________________________________________________ KMnO3 ____________________________________________________________ D. Here are some other chemical equations for reactions that can be used to prepare gases, but they are not

balanced. Balance the chemical equations and answer any other questions. 1. _______HCl(aq) + _______Zn(s) _______H2(g) + _______ZnCl2(aq)

2. _______Na2SO3(s) + _______HCl(aq) _______SO2(g) + _______H2O(l) + _______NaCl(aq) Name Na2SO3 and determine its molar mass

practice problems adapted, in part, from previous 1045/1050 exams – page 15

23. (Fall 2008) The oxygen gas you prepared in the previous question could be used to carry out a number of different reactions. For example, it could be reacted with a hydrocarbon compound (in the presence of a catalyst) to give any number of oxyhydrocarbons of the general formula CxHyOz. One such a reaction gave a product that was purified and sent off for elemental analysis giving the following mass percents: 68.85% C and 4.95% H.

Determine the empirical formula of this compound.

24. (Fall 2007) In a moment of weakness I purchased the “Big Bag” of Nacho Cheese Flavor Bugles from the

vending machine in Dittmer Laboratory. Besides giving me 52% of my daily value of saturated fat (in 1 serving!), it also contained sodium phosphate and disodium phosphate.

Why doesn’t “disodium phosphate” make sense as a systematic name? Give the formula and systematic name for the substance they probably intended to name. formula __________________ name ______________________________________________

25. (Fall 2007) For many people, their first time giving blood is on a college campus. The standard volume of blood donated is one pint. Some of you will use blood obtained in this manner as medical professionals; perhaps others of you will require blood for your research. All of us will benefit directly or indirectly from the simple act of one person giving blood.

A. After a brief and confidential survey to access your health and suitability for giving blood, the

phlebotomist will check the iron (hemoglobin) content of your blood to make sure you can spare a pint. One simple way to do this is to add a drop of the blood to a solution of copper sulfate of a certain concentration to make sure the drop sinks, indicating that the blood is above a minimum density required for donation. Assuming a typical blood density of 1.06 g/mL, how much would 1.0 pint of blood weigh? Give your answer in pounds.

practice problems adapted, in part, from previous 1045/1050 exams – page 16

First, estimate an answer based on your knowledge of what a gallon of milk weights, and briefly

explain your reasoning. Now, calculate the weight (in pounds) of 1.0 pint of blood.

B. Average blood flow rates in the human body vary considerably depending on the vein or artery, but

taking blood by needle from the antecubital vein in the arm typically gives a flow rate of about 1 mL/s. How long should it take to deliver a pint of blood once the needle is inserted in the vein? Give your answer in minutes.

C. You were told to go to the lab to get a bottle of copper sulfate so you could prepare a solution like the

one described in part A above. You know how much solid you need to prepare the correct concentration of solution, but the person in the stockroom tells you that he cannot tell what you need based on the name you gave him. What’s the problem?

practice problems adapted, in part, from previous 1045/1050 exams – page 17

26. (Fall 2010) We considered quantum dots at last night’s colloquium. Do not worry if you were not there – we did not discuss the chemistry to make these materials. We are going to do that now. The best know quantum dots are based on cadmium selenide, CdSe, which is prepared by reacting dimethyl cadmium, (CH3)2Cd, with an appropriate selenium compound. A possible reaction for making (CH3)2Cd is shown below.

CdBr2 + 2KCH3 (CH3)2Cd + 2KBr

A. Starting with 55 g CdBr2 and 25 g KCH3, what is the theoretical yield (in grams) of (CH3)2Cd? (Molar masses: CdBr2, 272.4 g/mol; KCH3, 54.13 g/mol; (CH3)2Cd, 142.5 g/mol.

B. When you carried out this reaction you obtained a yield of 19.3 g. What was your percent yield?