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Chem 1211 Yearlong Notes

by: Aminah Notetaker

Chem 1211 Yearlong Notes Chem1211

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These notes cover all of the lectures in Chem 1211 with helpful hints!
Chemistry 1211
Dr. Caughran
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This 114 page Bundle was uploaded by Aminah Notetaker on Sunday July 31, 2016. The Bundle belongs to Chem1211 at University of Georgia taught by Dr. Caughran in Summer 2016. Since its upload, it has received 11 views. For similar materials see Chemistry 1211 in Chemistry at University of Georgia.


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Date Created: 07/31/16
Ability to influence people. Credit to success Freshman Chemistry Lecture Notes Date: Tuesday, January 26, 2016 Chapter 3, Day 1 Put Into Perspective: A mole is a HUGE number! The Mole One mole is greater than 23 ' Mole= 6.022×10 atoms(Avogadr o s Number) the number of stars in the universe! A mole is just a unit of measurement like a dozen. You can have a mole of *It is important that you think about what the answer means! Don’t just memorize calculations. Problem 1 Question: Is it possible that a mole of fire ants occupies the Earth? Mass of Ant: 5 mg 24 Mass of Earth: 6×10 kg Work: 23 ,005 1mile 1.609k 1000m 6.022×10 m 1 ant 5280ft 1mile 1km Answer: There is not enough information to solve this problem because you only know mass. You need to know how much surface area an ant would take up. This problem is an example of you actually understanding the information and using it to rationalize the answer Problem 2 Question: How many moles are in a 500 mL bottle of water? Helpful Hints: Use density of water to convert from the Work: 500 ml to grams (D= Helpful Hints: 1g/ml) 500 mL 1g 1mol Molar Mass (g/mol) of Water: 1mL 18.02g H2O Step 1: Identify Elements There are 2 hydrogen (atomic mass: 1.01g) and 1 oxygen (atomic mass: 16g) Step 2: Add atomic masses 2(1.01g) + 1(16g) = 18.02g of Answer: 27.75mols Problem 3 Question: A mole of H O2molecules is distributed to all humans. Estimate the volume of H 2 each person would receive. Would volume be enough to quench your thirst? 9 Human Population: 7 ×10 Work: Step 1: One mole of H 2 divided by the number of people in the world = How many molecules of H O 2ne person gets 23 1 6.022×10 Step 2: molecul es 13 7×10 9 = 9×10 molecules per person people 13 Convert the number of molecules per person 9×10 1 mol 18g of H 2 into grams by using Avogadro’s number to molecules convert it back into mols 23 6.022×10 1 mol −9 = 3×10 g molecules −9 Answer: 3×10 g , This is not enough water to quench your thirst. Problem 4 Question: How many moles of nitrogen atoms are there in a 64g sample of hydrazine H N2H or2H N (4o2sn’t matter how you write the hydrazine formula) Helpful Hints: This is the stoichiometry conversion factor. You base this number off of how many moles of H N4th2re Work: are compared to the moles of Nitrogen 64 g H4N 2 1 mol H 4 2 2 mols N =4 mols of N 32.06 g H 4 2 1 mol Answer: 4 moles of Nitrogen H 4 2 Exam 1 - No curve - You need to assess what you got on the first progress check and Exam 1 to determine if you should stay in this class or drop to Chem 1210 prep course. - You need to study and work problems every day to do well in this class - Exam 1 answers will be posted by this weekend Test 2 Countdown: TWO WEEKS!!! Freshman Chemistry Lecture Notes Date: Tuesday, January 28, 2016 Chapter 3, Day 2 Problem 1 Question: How many moles of gold are in the Temple of Doom golden idol? How many moles of sand (SiO ) 2re required to balance the idol? Work: Part 1: Helpful Hints: 197 g/mol in Au (gold) This is a conceptual problem. Gold 60.1 g/mol in SiO 2 weighs more than sand. Therefore, you need more moles of sand to replace the gold (equally) Answer: mol Au< mol SiO 2 Part 2: Recall: Solid gold idol= 19.3 kg Density of gold: 19.3g/cm 3 Helpful Hints: Density of sand: 1.6 g/cm 3 Density is irrelevant. They just put this information there to distract you. = 98 mols of Au 19.3kg Au 1 g 1 mol Au 193 g (molar 1kg mass) = 321 mol 19.3kgSiO 2 1 g SiO2 1 mol SiO 2 1kg 60.1 g (molar mass) Answer: 98.0 mol Au, 321 mol SiO 2 Problem 2 Question: Calculate mass in grams of a single iron atom? Work: 55.85 g Fe 1 mol 9×10−23 23 = atoms 1 mol 6.022×10 −23 atoms Answer: 9×10 atoms Problem 3 Question: How many grams of oxygen are in 28.3g of Copper(II) Sulfate Pentahydrate? Work: 28.3g 1 mol 9 mols of O in 16.0 g O CuSO 45H 2 CuSO 45H 2 CuS O 45 H2O 249.67 1 mol 1 mol O g CuSO .5H O CuSO 45H 2 4 2 =16.3 g OR 144gO(molar massof Oxygen) ×100 = 57.7% 249.6gCuSO4.5H 2O(molar massof Copper (I)SulfatePentahydrate) O (The percent of oxygen that is in Copper(II) Sulfate Pentahydrate) 57.7% of 28.3g =16.3g (Take that percentage and multiply it by the amount ofCuSO4.5H 2O given to get the amount of Oxygen in the total sample) 16.3gof Oxygen Answer: in Copper(II) Sulfate Pentahydrate Example: Empirical Formula Molecular Formula: H4N2or H 2NH 2 - Simple, reduced formula - Smallest whole number ratio of atoms present Molecular Formula - Actual ratio of elements in the compound - Integer multiple of the empirical formula Problem 4 Question: Pseudoephedrine is %72.69 carbon, 9.15% hydrogen, 8.48% nitrogen, and 9.68% oxygen by mass. What is the empirical formula of pseudoephedrine? Work: Step 1: Use the percentages to convert each element into moles Step 2: Divide the mols of each element by the smallest number of mols -In this case, divide each element by .6052 mols Step 3 72.69C 1 mol C The numbers you calculated in Step 2 are the subscripts for 12.01g each element = 6.025 mols 9.15 H 1 mol H = 9.05 mols 1.01g 8.48N 1 mol N 9.68O 1 mol O =.6052 mols 14.01g 16.00g =.6052 mols 6.025molsC =10 .6052mols 9.05mols H=15 .6052 mols .6052molsN =1 .6052mols .6052 molsO .6052 mols=1 C10 15 Answer: C H10O15 Problem 5 Question: There are many environmentally important oxides of nitrogen. You analyze a sample and find 25.9% nitrogen and 74.1% oxygen. What is the empirical formula of your Sample? Work Step 1: Use the percentages to convert each element into moles Helpful Hints: You cannot have a fraction of an element. Therefore Step 2: you need to multiply both Divide the mols of each element by the smallest number of molsls of N and O by 2 to make 2.5 mols of O a whole -In this case, divide each element by 1.85 mols number Step 3 The numbers you calculated in Step 2 are the subscripts for 25.9N 1 mol N each element 14.01g = 1.85 mols N = 4.63 mols O 74.1O 1 mol O 16.00g 1.85mols N 1.85mols =1×2=2 4.63molsO =2.5×2=5 1.85mols N2O 5 Answer: N O2 5 Stoichiometry Uses relationships between reactants and/or products in a chemical reaction to determine desired quantitative data Example: You want to make pizza! 1 pizza consist of 1 crust, and 2 cups of cheese, 5oz tomato sauce. You have 4 crust, 10 cups of cheese, and 15oz of tomato sauce. How many pizza can you make? 4 crust= 4 pizzas 10 cups of cheese= 5 pizzas 15oz of tomato sauce= 3 pizzas You can only make 3 pizzas because you don’t have Balancing enough tomato sauce (limiting reagent). A chemical equation is balanced when there are the same number of each kind of atom on both sides of the reaction arrow. To balance the equation 1. Only coefficients (numbers in front of compounds) can be changed. Changing the subscript would change the compound entirely 2. Begin with atoms that appear in only one compound on the left and right 3. Balance atoms that appear as free elements last 4. When possible, balance polyatomic ions as if they were elements (keep it as a whole, don’t separate it) Problem 6 Question: If 9 ethane molecules, C H2,6are reacted with twenty one oxygen molecule, how many carbon dioxide molecules could be produced? Step 1: Balance the equation On the left (reactants): 2 C, 6H, and 2O On the right (products): 1 C, 2H, and 3O *The odd number of oxygen on the right, lets you know that you will have to multiply the O2on the right by an odd number 2 C H (g) + 7 O (g) → 4 CO (g) + 6 H O(g) 2 6 2 2 2 Now there are 4 C, 12H, and 14O on left and 4C, 12H, and 14O on right so the equation is balanced Step 2: Take the 9 molecules of ethane and 21 molecules of oxygen given and use a ratio from the balanced equation to convert to CO 2 9 C2H 6 4 CO 24 carbon =18 CO 2 dioxides in equation) 21 4 CO 24 carbon dioxides in =1 2 C2H 62 ethane in 2 equation) equation) O2 7 CO 2 O2(7 oxygen in equation) Answer: You can only make 12 molecules of CO beca2se you don’t have enough O (limiting reagent) 2 Announcements - Professor Norton said that lab starts next week. Remember that safety is very important! Come to class with your lab book and goggles. (It is not advised to share goggles due to the spread of pink eye) th - Exam 2 will be held on the 6 floor of the Chemistry Learning Center. You need to register for your testing time slot starting Tuesday at 11am (Schedule your slot appropriately. There will be no excused absences if you miss your other classes due to the test) Freshman Chemistry Lecture Notes Date: Tuesday, February 2, 2016 Chapter 4, Day 1 Types of Reactions Combinati 2Ca s +O (g)→2CaO s) 2 or more reactants combine to on 2 make 1 product Reactions CaO (s)+H 2 (l→Ca(O H) (s) CaO (s)+C O 2g )→CaC O 3s) Decomposi CaC O (s→CaO (s)+C O (g) 1 reactant breaks down into 2 or tion 3 2 more products Reactions 2 Na N 3s →2 Na (s)+3N (2) Displacem The number of reactants = the ent number of products Reactions 1. Fe s)+CuSO (a4)→Cu (s+FeS O (4q) 1. AB+X→ XB+A 1. Single 2. Double 2. 2. AB+XY →XY +BX K 2O 4aq +Ba(N O ) 3 2)→2KN O 3aq +BaS O (4) Precipitati K 2O 4aq +Ba(N O ) 3 2)→2KN O 3aq +BaS O (4) Precipitate forms when the product on is insoluble (Solubility Rules) Acid- Base/ HCl aq +NaOH(aq)→NaCl aq +H O2l) Uses Acids and Base (Strong and Neutralizat Weak Rules) ion Reaction Gas- 2 Na (s+2H O2l)→2 NaOH (aq +H (2) Forms a gas as a product Forming Reaction Redox H 2(g)+F 2g)→2 HF(g) Oxidized- Reducing Agent: The Reaction species loses an electron Reduced- Oxidizing Agent: The species gains an electron Chemical Questions Formula Unit Equation NaCl(aq)+AgNO (aq)→NaN O +AgCl(s) 3 3 Total Ionic Equation −¿(aq)+AgCl(s) ¿ +¿ (aq)+N O 3 −¿ ( )→Na ¿ ¿ +¿ ( )+NO 3 ¿ −¿ (aq)+Ag +¿ (aq)+Cl ¿ ¿ Na Net Ionic Equation Helpful Hint: −¿ (aq → AgCl(s) Formula Unit: Both compounds written out as normal +¿(aq +Cl ¿ ¿ Total Ionic: The soluble compounds dissociate into Ag individual ions (solubility rules) Net Ionic: The spectator ions (ions that are on both sides of the equation) are taken out of the equation Solubility Rules You must memorize all the Solubility Rules!!! Example: Ex: NaCl solution mixed with AgNO s3lution At first, both the NaCl and AgNO br3ak up into + separate ions (Na , Cl ,Ag ,NO 3 because they’re both soluble. Then the AgCl combine to make an insoluble precipitate! Acid- Base Rules Strong Acid: Ionizes completely in H O 2 −¿(s) ¿ +¿(aq)+Cl ¿ HCl(aq)→H O 3 Weak Acid: Does not completely ionize +¿ aq +CH C O (aq) 3 2 ¿ C H 3O H2 aq →H O2 Strong Base: Ionizes completely in H O2 −¿ (aq) ¿ +¿(aq)+OH ¿ NaOH s)+H 2(l)→Na Weak Base: Does not completely ionize −¿ aq ) ¿ +¿ aq +OH N H (g)+H O(l)→N H ¿ 3 2 4 Strong Acids Weak Acid Strong Base Weak Base HI HCOOH NaOH NH 3 HBr CH 3OOH KOH N(CH 3 3 HClO CCl COOH LiOH C H N 4 3 5 5 HCl HF RbOH NH 4H HClO 3 HCN CsOH H 2O 4 H 2 Ca(OH) 2 HNO 3 Ba(OH) 2 Sr(OH) 2 Oxidation Numbers The oxidation number or state is the charge an atom has or appears to have in a compound 1. Atoms in their elemental state= 0 Ex: The element Iron (Fe) has a 0 oxidation number 2. Monoatomic ions2+charge of ion Ex: The ion Br has a +2 oxidation number 3. In a compound, fluorine, chlorine, bromine, and iodine= -1 Ex: In NaCl, Clhas a -1 oxidation number 4. Oxygen= -2 (except in H O 2 2 In AgNO 3 O has a 5. Hydrogen = -1 (when paired with a metal= +1) The combination of oxidation numbers in a neutral compound must equal 0. If it’s a polyatomic ion, then the total sum of the oxidation numbers must equal the charge of the ion. Find the oxidation number of Al in 1. Set up an equation Al O 2. Al O is a neutral compound so the 2 3 2 3 2(Al) + 3(O)= 0 total oxidation number equals 0 2(Al) + 3(-2)= 0 3. You know that O always has an oxidation number of -2 in 2(Al) - 6= 0 2(Al)= 6 compounds Al= 3 4. Solve for the oxidation number of Al Find the oxidation number of S in Do the same process except the total Sulfide oxidation number will equal -2 because 2- SO 3 that is the charge of the polyatomic (S) + 3(O) = -2 number S + 3(-2)= -2 S – 6 = -2 S = 4 Problem 1 Question: What species is reduced in the reaction of iron (III) oxide with carbon monoxide to produce iron metal and carbon dioxide? What is the oxidizing agent? Helpful Hint: Work: Fe O +CO→Fe+CO Fe2O 3 started with a 3+ 2 3 2 oxidation number (oxidation of Fe in a compound is equal to its charge. If O has a -2 charge Answer: Fe 2 3 is the answer to both questions and there are 3 of them, then both Fe must have a -3 charge in order to be equivalent On the opposite side of the reaction, Fe becomes a solid. Elemental Fe has a 0 oxidation number. Therefore, Fe went from +3 to 0, meaning it gained electrons. Freshman Chemistry Lecture Notes th th Date: March 12 and 14 , 2016 Chapter 7 Problem 1 Question: In the “egg in the bottle” demonstration, was the egg pushed into the bottle or pulled into the bottle? Answer: Pushed by the pressure on the outside Problem 2 Question: When volume is increased… A. Gas molecules speed up B. Gas molecules slow down C. No change Answer: C Problem 3 A sample of CO 2as has a pressure of 85.3 mmHg in a 125 mL flask. The sample is transferred to a new flask, where it has a pressure of 0.0820 atm at the same temperature. Part 1 Question: The new flask is… the old flask? A. Smaller than B. Same size C. Larger than Work: 0.0820 atm 760 mmHg =62.3 mmHg 1 atm Answer: C Part 2 Question: What is the volume of the new flask? Work: P1V 1P V2 2 (85.3 mmHg)(125 mL) = (62.3 mmHg)(V ) 2 V2=171 Answer: 171 mL Problem 4 o A 10.0-mL sample of CO g2s is enclosed in a gastight syringe at 22.0 C. The syringe is immersed in a dry ice/ acetone bath (-78.0 C). Part 1 Question: The gas will occupy a… volume? A. Smaller B. Same size C. Larger Answer: A Part 2 Question: What is the new volume, assuming pressure is constant? Work: V1 V 2 = T1 T 2 10.0 mL V = 2 295K 195K V2=6.61 Answer: 6.61 mL Problem 5 Question: A balloon is filled with a quantity of nitrogen, N 2 at room temperature, 20 C. At what temperature would the volume of the balloon have doubled? o A. -78 C B. 5 Co C. 10 Co o D. 20 C E. 40 Co o F. 80 C o G. 310 C H. Can’t be determined Work: V1 V2 = T1 T2 1 2 = 293K T2 T =586 K -273.15K = 313 2 Answer: G Problem 6 Question: A sample of hydrogen, H , ex2rts a pressure of 732 torr in a 1.25 L rigid vessel at a temperature of 24.6 C. A sample of carbon dioxide, CO , a2so exerts a o pressure of 732 torr in a 1.25 L rigid vessel at a temperature of 24.6 C. A. There must be more moles of hydrogen in the first vessel than carbon dioxide in the second vessel B. There must be more moles of carbon dioxide in the second vessel than hydrogen in the first vessel C. There must be a larger mass of hydrogen in the first vessel than carbon dioxide in the second vessel D. There must be a larger mass of carbon dioxide in the second vessel than hydrogen in the first vessel Answer: D Problem 7 Question: A weather balloon with a volume of 268 ft is released from the ground o where the pressure is 1.00 atm and the temperature is 25.0 . What will be the volume of the balloon when it reaches Earth’s lower stratosphere, about 10 km above the ground, where the pressure is 0.0750 at and the temperature -65.0 C? o Work: P1V 1 P 2 2 = T 1 T 2 1atm )(268 ft ) (0.0750atm)(V 2 = 298K 208K T2=2494 Answer: 2490 ft 3 Problem 8 Question: As molar mass increases the density of a gas… at constant T, P, and V. A. Increases B. Decreases C. Doesn’t Change D. Molar mass and density are unrelated Answer: A Problem 9 Question: if the density of air is 1.29 g/L at STP, which of the following hydrocarbons would have been safe to use in the demonstration? A. Methane, CH 4 B. Ethane, C H2 6 C. Propane, C H3 8 D. Butane, C H3 10 E. Pentane, C H 5 12 Work: 1 mol of an ideal gas occupies 22.4 L at STP 16.0g =0.714g/L Density of CH =4 22.4L 30.1g =1.34g/L Density of C 2 =6 22.4L Answer: A Freshman Chemistry Lecture Notes Date: Tuesday, February 9, 2016 Chapter 3.5, Day 1 Problem 1 Question: A 24.92 g sample of cobalt is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 49.02g. Determine the empirical formula of the metal fluoride. Work: Step 1: Subtract the mass of metal 49.02 g metal fluoride – 24.92 g Co = 24.10 g F fluoride from cobalt to get the mass of fluoride. Step 2: Calculate the moles of F and Co = 1.268 mol F 1 mol 24.10 gF 19.00g = .423 mols Co 24.92gCo 1 mol 58.93g Step 3: 1.268mols F=3 Divide the mols of each .423mols element by the smallest number of mols -In this case, divide each .423molsCo =1 element by .423 mols .423mols CoF3 Answer: CoF 3 Problem 2 Question: An unknown compound has the formula C H O .xAy0z0956 g sample of the compound is burned, and 0.1356 g of carbon dioxide and 0.0833 g of water are isolated. What is the percent composition of oxygen in the unknown compound? Work: Part 1 Step 1: 0.1356 g 1 mol 1 mol C 12.011 Covert grams of CO to grams of Co g C 2 2 C 44.01 g 1 mol 1 mol CO 2 CO 2 =.0379g of C Step 2: 0.0833 g 1 mol 2 mol H 1.088 Convert grams of H O to grams H O g H 2 2 of H 18.02 g 1 mol 1 mol H 2 H 2 =.00933 g of H Step 3: Subtract the grams of C and 0.0956 g sample – 0.0379 g C – 0.00933 g H = grams of H from the total 0.0493 g O sample to find the grams of O Step 4: Divide the grams of O by the .0493gO total grams in the sample and .0956gtotal×100=51.6 O multiply by 100 51.6 O Answer: Part 2 Question: If the molar mass of the unknown compound is 62.1 g/mol, what is the molecular formula? Work: Step 1: 0.1356 g 1 mol 1 mol C Use the information in part 1 to Co = 3.08 x 10- 2 3 mols of C convert to moles of C, H, and O 44.01 g 1 mol CO 2 CO 2 0.0833 g 1 mol 2 mol H -3 H2O = 9.26 x 10 mols of H 18.02 g 1 mol H 2 H2O 0.0493 g 1 mol -3 O = 3.08 x 10 16 g O mols of O Step 2: Divide the mols of each element 3.08x10 molsC by the smallest number of mols =1 3.08x10 mols -In this case, divide each element 3.08x10−3 by mols −3 9.26x10 molsH −3 =3 3.08x10 mols −3 3.08x10 molsN =1 3.08x10 mols Step 3: Write the empirical formula CH 3 Step 4: Divide the grams of the sample by 62.1gsample the molar mass of the empirical 31.0 gC H O =2 formula 3 Step 5: Multiply the ratio from Step 4 by C H O C H O 2 x 3 = 2 6 2 each subscript in the empirical formula to get the molecular formula Answer: C 2 O6 2 Problem 3 Question: A 1.959 g sample of a compound composed of iron and carbon monoxide, Fe (CO) , is burned in pure oxygen and 0.799 g iron(III) oxide and 2.200 g x y of carbon dioxide are recovered. What is the empirical formula of the compound? Work: Step 1: Use grams of Fe2O 3 to find 0.799gFe O2 3 1 mol 2 mol Fe moles of Fe 159.7 g 1 mol Fe2O3 =0.0100 mol Fe Step 2: Use grams of CO t2 find moles 2.200gCO 2 mol 1 mol of CO CO 44.01 g 1 mol CO 2 CO 2 =0.0500 mol CO Step 3: Divide the mols of each element 0.0100molsC by the smallest number of mols 0.0100mols =1 -In this case, divide each element by 0.0100 mols 0.0100mols H =5 0.0500mols Step 4: Write empirical formula Fe(CO) 5 Answer: Fe(CO) 5 Problem 4 Question: A discarded gas cylinder contains a pure compound of only carbon and hydrogen. To determine the identity of the contents, a sample is burned. The products of the combustion are separated and analyzed. 5.28 g of CO a2d 2.88 g of H2O are isolated. What is the empirical formula for the compound in the gas cylinder? Work: Step 1: 1 mol 1 mol C Use grams of CO 2 to find 5.28gCO 2 moles of C 44.01 g 1 mol CO 2 =0.120 mol C Step 2: Use grams of H 2 to find moles 2.88gH O2 1 mol 1 mol H of H 18.0 g 1 mol H2O =0.320 mol H Step 3: Divide the mols of each element 0.120molsC by the smallest number of mols =1 -In this case, divide each 0.120mols element by 0.120 mols 0.320molsH =2.67 0.120mols Step 4: Multiply the ratio by 3 in order to1 x 3= 3 C make 2.67 a whole number 2.67 x 3 = 8 H Step 4: Write empirical formula C 3 8 Answer: C 3 8 Problem 5 Question: A sample of copper (II) sulfate pentahydrate ore is analyzed to determine its value. A 3.365 g sample is heated to remove all the water from the sample and 2.940 g are recovered after cooling. Assuming that the only source of water in the ore is from the copper (II) sulfate pentahydrate and all of the copper can be recovered, what is the value of the ore in dollars/ton? Work: Helpful Hints: Step 1: Copper (II) sulfate Subtract the grams of the 3.365 g – 2.940 g = 0.425 g H O2 sample from the grams left after pentahydrate is heating CuSO 4H O2 Step 2: Use the grams of water that was 0.425gH O 1 mol 1 mol 1 mol 63.55 released from Step 1 to solve for 2 CSP Cu g grams of Cu 18.0 g 5 mol 1 mol 1 mol H 2 CSP Cu =0.300 g Cu Step 3: Divide the grams of the Cu .300g Cu/ 3.365 ore = 8.92% Cu found from Step 2 by the total grams of the ore sample and multiply by 100 Step 4: Convert 1 ton of ore to tons of 1tonore 8.424 2000 lb $1.90 copper then solve for the price tons Cu 100 ton 1 ton 1 1b ore Cu =$340 Answer: $340 Freshman Chemistry Lecture Notes Date: Tuesday, February 17, 2016 Chapter 5, Day 1 Problem 1 (Review from Unit 2) Question: Dr. Norton discovered a bottle of HCl solution in the back of the chemistry prep room, but it is not labeled with a concentration. So she takes 10.00 mL of solution and dilutes it to 250.00 mL. She then takes a 25.00 mL sample of this solution and titrates it with 0.100 M sodium hydroxide. 80.00 mL of the solution hydroxide solution are needed to neutralize the sample. What is the concentration of HCL solution Dr. Norton found in the prep room? Work: Step 1: Write a balanced equation HCl+NaOH →NaCl+H O 2 Step 2: Convert the mL of NaOH to mols =.00800 mol HCl 80.00mL NaOH .100 1 mol mol HCl NaOH Step 3: 1000mL 1 mol .00800molHCl NaOH =0.320M Determine the concentration of .02500L HCl in 250 mL Step 4: Use M 1 1M V2t2 determine (.320 M)(250.00 mL)= M (20.00mL) dilution Answer: M =28.00 M Problem 2 (Start of Unit 3) Question: 8 oz of Gatorade contains 50 Calories. How much energy is that expressed in joules? Work: Helpful Hints: 50 Cal 1000 cal 4.184 J = 209,200 J There is a lower case (c)alories and an Cal 1 cal upper case (C)alories. The upper case Answer: 200,000 Joules (only 1 significant figure) Calories is used for food (can also be written as kcal). To find joules you need to convert C to c. Energy System: The specific object you are studying/ observing Surroundings: Everything else around/ effected by that object Endothermic: Exothermic: Requires the input of HEAT Requires the release of -Heat from the surroundings go into the system, heat making the surroundings cooler - Heat from the system leaves into the surroundings, making the system cooler Examples: Examples 1. Liquid  Gas (evaporation) 1. Gas  Liquid When you work out, your body gets hot and (Condensation) sweats (sweat is the system). Your body (body 2. Liquid  Solid is the surroundings) then uses its heat to (Freezing) evaporate your sweat which makes your body 3. Combustion cooler. 2. Solid  Liquid (melting) 3. Food  Cooked Food 4. Decomposition of Compounds Energy us put into the molecules to break apart the bonds Problem 3 Question: Which of the following are Exothermic? A. Heating a gallon of water to boil pasta B. Combustion of gasoline in a car engine C. Whisking/ beating an egg D. Heating calcium carbonate so that it decomposes to form calcium oxide and carbon dioxide E. Cooling a can of soda from room temperature to 35 F. F. None of the Above Answer: B and E Problem 4 Question: Aluminum pans are great for sautéing foods quickly because the pans heats up very quickly. Cast iron pans are great for searing because they heat up slowly and don’t change temperature as much as an aluminum pan when a cold steak is placed in the pan. Which statement do you know to be true? A. Chefs prefer aluminum pans to cast iron Helpful Hints: B. Cast iron has a higher specific heat than aluminum C. Aluminum pans are more expensive than cast iron paThe higher the specific heat, the D. Cast iron pans are less environmentally friendly more heat energy it takes to E. Aluminum has a higher specific hear than cast iron increase its temperature Answer: B Problem 5 Question: A 3.50 g sample of copper that is initially at room temperature, 22.0 C, is dropped into a beaker that contains 100 grams of water that is at a rolling boil and kept at a rolling boil. How many joules of heat will the copper absorb to reach thermal equilibrium? Helpful Hints: The specific heat of copper is 0.385 J/g K First you need to know that the boiling point of water is The specific heat of water is 4.184 100 degrees. Therefore you use 100 as your final temperature. Then you needed to know that the temperature of water didn’t change from 100 degrees therefore it is irrelevant Work: to the problem. You only need to solve for copper! q=m×c×∆T q = (3.50 g of copper)(.385 J/g K)(100- 22 K) q= Amount of energy (J) q = (3.50 g of copper)(.385 J/g K)(78 K) m= Mass q = 105 J c= specific heat T= change in temperature (final – initial) Answer: 105 J Problem 6 Question: A sample of metal is heated to 102.3 C and then placed into a beaker filled halfway with water at 23.4 C. Once thermal equilibrium has been reached, the temperature of the metal is 31.2 C. What is the temperature change of the water? Work: Helpful Hints: ∆ T=Final−Initial At thermal equilibrium both the metal and the water will 31.2 degrees – 23.4 degrees = 7.8 degrees be the same temperature. So 31.2 C is your final temp. Problem 7 Question: Samples of copper, aluminum, and lead of equal mass are heated to the same high temperature and then dropped into separate cups of water. Each cup holds the same volume of water initially at room temperature. Each system (metal + water + cup) is allowed to reach thermal equilibrium. Which metal will be at the lowest temperature at the thermal equilibrium? . The specific heat of water is 4.184 J/g K . A. Copper (24.5 J/mol K) B. Aluminum (.900 J/g C). C. Lead (.0305 Cal/kgK) . D. All the metals will be the same temperature when thermal equilibrium is reached E. Not enough information to determine Work: Convert all of the answer choice into the right units for specific heat (J/g C). Copper (24.5 J/mol K). 24.5 J 1 mol mol 63.55 g . =.386 J/gC Copper . . Aluminum (.900 J/g C) .900 J/gC Aluminum (already in the right units) Lead (.0305 Cal/kgK). .0306 Cal 1000 cal 4.184 J 1 kg kg 1 Cal 1 cal 1000 g . = 0.128 J/gC Helpful Hints: Answer: Lead Using your conceptual understanding you know that the element with the lowest specific heat will have the largest change in temperature *You can also plug your answers into the thermal equilibrium equation and solve for the final temperature in order to double check qmetal water q=m×c×∆T Teacher’s Comments  Grades for the test are posted and everyone was given 20 additional points because two questions were dropped  Don’t freak out if you did not do good on this test. You still have one more test to bring up your grade before the withdraw period ends  You should do the mastery and challenge problems on Mindtap for extra help! Freshman Chemistry Lecture Notes Date: Thursday, February 11, 2016 Chapter 3.5, Day 2 Problem 1 Question: When limestone, calcium carbonate, is heated it will decompose to form calcium oxide and carbon dioxide. A 2.654 g mixture composed of limestone and other inert materials is found to have a mass of 1.874 g after heating. What is the percent by mass of limestone in the mixture? Work: Step 1: Subtract the mass of the 2.654 g – 1.874 g = .78 g Co 2eft limestone mixture from the mass leftover after heating Step 2: Use the grams of CO f2und in =1.772 g CaCO 3 Step 1 to calculate grams of 100.88 CaCO 3 g CaCO 3 1 mol 1 mol 44.01g CO 2 1.772x 100=66.8 limestone∈mixture 2.654 Step 3: Divide the grams of limestone found in Step 2 by the total grams of the mixture and multiple by 100 Answer: 66.8% Problem 2 Question: A student has 480.0 mL of 0.1040 M aqueous solution of MnSO to use i4 an experiment. She accidentally leaves the container uncovered and comes back the next week to find only a solid residue. The mass of the residue is 13.83 g. Determine the chemical formula of this reaction. Work: Part 1 Step 1: 480.0 mL 1 L 0.1040 151.0 g Use molarity to convert mL to mol MnSO 4 grams of MnSO 4 1000mL 1 L 1 mol =7.538 g MnSO 4 Step 2: 13,83 g residue – 7.538 g MnSO = 4.292 g H O 2 Subtract grams of residue from the grams of MnSO fr4m Step 1 Step 3: 6.292 g 1 mol Convert grams of H O2found in H2O Step 2 to mols of H 2 18.02 g H2O = 0.3493 mol H O 2 Step 4: Helpful Hints: Divide mols of H 2 by the mols 0.3493 mol H2O of MnSO 4 =6.997H O2 .04992 mol MnS O 4 Mols of MnSO i4 found the same way you found grams in Step 1, but stop Step 5: 246.5 g/mol at the 0.1040 mols Write the hydrate formula using the ratio you calculated in Step 4: MnSO 7 4 O 2 Then calculate the molar mass Answer: 246.5 g/mol Problem 3 Question: 3.25 grams of silver nitrate, 2.67 g of sodium acetate, and 2.94 g of calcium chloride are added to 100 g of water in a beaker and mixed thoroughly. Solid silver chloride is allowed to settle to the bottom of the beaker. The water is then decanted of and the solid is thoroughly dried. If the mass of the dried solid is 2.225 g, what was the percent yield of the reaction? Step 1: Write the net ionic equation −¿ aq → AgCl(s) (equation without spectator ¿ +¿(aq)+Cl ions) Ag ¿ Step 2: 3.25g AgNO 1 mol 1 mol Determine limiting reagent 3 Ag + (reagent with smallest moles) 169.8 g 1 mol AgNO 3 =0.0191 mol Ag + 2.94gC aCl 2 mol 2 mol Cl- 110.98 1 mol g CaCl 2 =0.0530 mol Cl - +¿ Step 3: + ¿1 mol 143.35 Since Ag is your limiting 0.0191mol Ag AgCl g AgCl reagent, use its mols to calculate grams of AgCl 1 mol 1 mol Ag + AgCl =2.74 g AgCl Step 4: Determine percent yield 2.25g x100=82.2 yield experimentalx100 2.74g theoretical 82.2 yield Answer: Problem 4 Question: 123.6 mL of a 0.0500 M sulfuric acid solution are needed to neutralize 82.4 mL of a 0.0750 M strong base solution (Soln X). Only 61.8 mL of the 0.0500 M sulfuric acid solution are needed to neutralize 82.4 mL of a different 0.0750 M strong base solution (Soln Y). A) Soln X and Soln Y are the same base B) Soln X has twice the concentration of hydroxide ions as Soln Y C) Soln X and Soln Y have the same concentration of hydroxide ions D) Soln X has half the concentration of hydroxide ions as Soln Y E) Nothing can be determined from the data provided Answer: B Helpful Hints: Conceptual Problem: Both solutions have the same molarity but Soln X requires twice as much mL to neutralize the solution because it has twice as many hydroxide ions Problem 5 Question: Equal volumes of a diprotic acid solution and an alkali metal hydroxide solution are mixed. If the concentrations of the two solutions are the same, the resulting solution would be A) Acidic B) Neutral C) Basic D) Depends on the exact volumes and concentrations E) Depends of the identity of the acid and the base Answer: A Helpful Hints: Diprotic means the acid has two H ions and a base with an alkali metal (Group 1) would only have 1 OH ion. + Therefore the solution would have more H , making it slightly more acidic. Freshman Chemistry Lecture Notes Date: Thursday, February 18, 2016 Chapter 5, Day 2 Problem 1 Question: Two samples of water are mixed. After thermal equilibrium is reached, which combination results in the mixture with the lowest temperature? Work: Helpful Hints: A. 10 g of water a 20 C + 10 g of water at 30 C B. 10 g of water at 20 C + 20 g water at 30 Cnceptual Problem: You need to C. 20 g of water at 20 C + 10 g of water at 30 Cmore grams of the cooler D. 20 g of water at 20 C + 20 g of at 30 C water (20 C) and less grams of the warmer water (30 C) Answer: C To check using calculations: qCuq H2O0 q=m×c×∆T Problem 2 Question: A blacksmith typically heats and cools about a pound of iron when he or she is working. Of the iron is heated to 1300 C, what is the minimum amount of water, initially at room temperature (22 C), needed in the quenching barrel so that the temperature of the water doesn’t reach 30 C? q=m×c×∆T g J q= (lb×453.6 lb(.449g ∙°C)(30−1300℃ )=−258,656J | ironq water 258,656J=m (4.184J/g∙°C)(30−22℃ ) A. ½ gallon B. 5 gallons m= 7,727 g water C. 50 gallons D. 500 gallons  2 gallons of water Answer: B Problem 3 Question: Which requires more energy, melting 100 g of ice or converting 100 g of water to steam? Helpful Hints: Conceptual Problem: You need to have more grams of the cooler water (20 C) and less grams of the warmer A. Melting 100 g of ice Helpful Hints: B. Evaporating 100 g of water C. They’re the same It takes more energy to break apart the bonds in a liquid in order to evaporate Heat of Fusion for water= 333 J/g it into gas. It takes less energy to melt Het of Vaporization for water= 2256 J/g a solid into a liquid. Answer: B Problem 4 Question: How much heat (in kilojoules) is required to convert 15.0 g of ice at -12.0 C to steam at 120 C? Heat Capacity of ice= 2.06 J/g K . Heat capacity of water= 4.184 J/g K . Heat capacity of steam= 1.86 J/g K . Heat of Fusion for water= 333 J/g K . Heat of Vaporization for water= 2256 J/g K . Work - Not all energy is transferred as heat - Some energy is transferred as work done to or by the system - The change in total energy of a system is the sum of the heat exchanged and the work done on or by the system Problem 5 Question: In which case(s) are change in H and change in E very different? A. H 2(l)  H 2 (s) at 0 C B. H 2(s)  H O2l) at 0 C C. H 2(l) at 25 C  H 2(l) at 75 C D. H 2(l) at 75 C  H 2(l) at 25 C E. H O(l)  H O(g) at 100 C Helpful Hints: 2 2 F. H 2(g)  H O2l) at 010 C A liquid to gas phase change has the most different change in H and E Answer: E and F Problem 6 Question: Which of the following values are state functionsHelpful Hints: A. The distance from Savannah, GA to Denver, CO State functions do not depend on the B. The change in altitude from Savannah, GA to Denver, CO C. The travel time from Savannah, GA to Denver, CO path in between A and B. It only accounts for the starting and ending Answer: B point. Therefore time and distance are not state functions, because they are effected by the route you take to get from Savannah to Denver. *Energy and Temperature are State Functions *Work and Enthalpy are not Problem 7 Part 1 Question: What is the enthalpy change for the reverse reaction? H O (g)→H (g)+ O g ) ∆ H°=+241.8 Helpful Hints: 2 2 2 2 When doing the reverse reaction (switch the reactants and products), you change the sign of the enthalpy 1 H 2g → O 2(g)→H O2g) ∆ H°=¿ ? 2 Answer: - 241.8 Part 2 Question: What is the enthalpy change for the reaction balanced without fractions? H 2g → 1O 2(g)→H O2 (g) Helpful Hints: 2 In order to get rid of the fraction, multiply both sides of the equation 2H 2(g)→O 2(g)→2H O(2) by 2. Then multiply the change in ∆ H°=(−241.8)×2 Answer: -483.6 Part 3 Question: Would the enthalpy change be A. The same B. Larger (absolute value) C. Smaller (absolute value) ….if the reaction produced liquid water instead of water vapor? Answer: B Part 4 Question: If the heat of vaporization of water is 2256 J/g, what would be the enthalpy change when liquid water is produced instead of water vapor? 2 mol H 2 18 g 2256 J 1 KJ =82.2 KJ 1 mol 1 g 1000 J Answer: 82.2 KJ Freshman Chemistry Lecture Notes Date: Tuesday, February 23, 2016 Chapter 5, Day 3 Problem 1 (Review from Progress Check) Question: A 11.0% by mass solution of HCL in water had a density of 1.163 g/mL. What is the concentration of this solution in molarity? Work: 11.0 g 1 mol = .3022 mol HCl 36.4 g Helpful Hints: 100 g 1 mol 1 L = 0.08958 L H 2 11% by mass choose an amount of grams of solute (HCl) H2O and solvent (H2O) that will equal 1.1163 g 100 mL 11% .3022 = 3.37 M .08958 L For example: 11g of HCl and H 2 Answer: C Problem 2 (Review from Progress Check) Part 1 Question: Samples of three metals are prepared: cobalt, tungsten, and zinc. All of the samples have the same mass and begin the experiment at room temp, 23.3 C, Each sample is placed into a coffee-cup calorimeter that contains the same volume of water at 97.4 C. Which sample will absorb the most heat? It is probably worth noting that cobalt has the highest specific heat and tungsten has the lowest specific heat. Assume that all of the coffee- cup calorimeters have the same heat capacity. Answer: Cobalt (highest specific heat) Part 2: Tungste Question: Which sample will absorb the least heat? Answer: Tungsten (lowest specific heat) Zin c Temperature Helpful Hints: Use this graph to help understand the concept. If the element has a low specific heat, it will have a higher increase in Calorimetryre and a lower absorption of heat Heat Added/ Absorbed Coffee- Cup Calorimeter (Constant Pressure) Bomb Calorimeter (Constant Volume) Exothermic Endothermic  Temperature increase (heat  Temperature decreases (heat leaves to the surroundings is transferred into the system which increase temperature) from surroundings which  +∆T decreases temperature)  −∆T  +q surroundings  -q surroundings  -q system  −∆H  +q system +∆H  Problem 3 Question: Which of the following can be determined experimentally using a coffee cup calorimeter? Helpful Hints: A. ∆ E Coffee Cup Calorimeter B. ∆ H∨q ∆ H∨q changes . Bomb C. w ∆E Calorimeter changes . Answer: B Problem 4 Part 1 Question: 50.00 g of water at 98.2 C is added to 50.00 g of water at 23.6 C in a coffee cup calorimeter. Once thermal equilibrium has been established, the system is at 51.3 C. How much heat did the hot water lose (in J)? Work: q=m×c×∆T q= (50.00g) 4.184J∙°C 51.3−98.2℃ =−9810J ( g ) Answer: 9810 (don’t include negative) Part 2 Question: How much heat did the cold water gain (in J)? Work: Helpful Hints: J q= (50.00g)(4.184 ∙°C )51.3−23.6℃ =5790J Theoretically the q lost by the hot water g should equal the q gained by the cold water, however in an experiment, the Answer: 5790 J calorimeter absorbs some of the energy Part 3 J/°C¿ Question: What is the heat capacity of this calorimeter (in Work: 9810 – 5790 J (= J lostoto calorimeter)= 145 Helpful Hints: 51.3 (final)- 23.6 (initial) C Mass is not included in heat capacity ( Answer: 145 J/°C c= q ∆T ) Part 4 0 Question: A 21.75 g slug of an unknown metal is heated to 99.98 C in an almost boiling water bath. The slug is ohen placed into our coffee cup calorimeter, which contains 50.00 g of water at 23.4 C. Once thermaloequilibrium has been established, the temperature of the system is 27.2 C. What is the specific heat of the unknown metal? Work: q system q metal+q waterq calorimeter 0 ¿21.75)c)(27.2−99.98℃ )+(55.00g)(4.184J/°C )27.2−23.4℃ + 145J/°C )(27.2−23.4℃ ) c= .85 Answer: 0.85 J/°C Part 5 Question: What would most likely happen if you rush this experiment and heat to a lower initial temperature? A. No problem. You still get the same temperature. B. ∆ T for metal is smaller, so fewer sigfigs C. ∆ T for water is smaller, so fewer sigfigs D. ∆ T for water is larger, so more precise Answer: C Problem 5 Question: You mix 100.0 mL of 5.00 M monoprotic strong acid and 100.0 mL of 5.00 M monoprotic strong base in a coffee cup calorimeter. If both solutions were initially at 18.80 C and the temperature of the mixture after thermal equilibrium was established was 19.20 C, how much heat was released by the reaction? Assume the specific heat of the solutions is 4.20J/°C , density is 1.07 g/mL, and you’re using the same calorimeter. Work: 200 mL 1.07 g = 214 g Helpful Hints: 1 mL You mix the solutions therefore you have to add the grams of acid and base J to get 200 mL then use density to solve q=(214g (4.184 ∙°C)(19.20−18.80℃ =360 for mols g The Molarity is irrelevant because you Answer: 360 J don’t know what acid and base it is Announcements - Exam 3 registration begins tomorrow at 11 am - Exam 3 will cover Chapter 5 - Office Hours: Monday (4-5 pm) replaced by Tuesday (4-5pm) for open help (Room 570 or 573 Chemistry) Freshman Chemistry Lecture Notes Date: Thursday, February 25, 2016 Chapter 5, Day 4 Problem 1 (Continuation from last class) Part 1 Question: You mix 100.0 mL of 5.00 M monoprotic strong acid and 100.0 mL of 5.00 M monoprotic strong base in a coffee cup calorimeter. If both solutions were o initially at 18.80 C and theotemperature of the mixture after thermal equilibrium was established was 19.20 C, how much heat was released by the reaction? Assume the specific heat of the solutions is 4.20 J/°C , density is 1.07 g/mL, and you’re using the same calorimeter. Helpful Hints: You have to account for the energy that is Work: lost to the calorimeter by adding (heat 200 mL 1.07 g = 214 g capacity)(change in temperature) 1 mL q= 214g ) 4.184J ∙°C (19.20−18.80℃ =360 o o ( g ) + (145 J/ C)(19.20-18.80 C) = 417.5 J Answer: 420J Part 2 Question: What is the heat of reaction per mole of acid (in J)? Work: −420 J ∆ H= .500mol acid=−840 *Don’t forget to make q negative because it is exothermic Answer: -840 J


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