Chemistry Joke of the Day

Announcements Exam #2: Tonight!, 7:00-8:30pm, locations posted on website, bring ID Conflict Exam: 5:00-6:30pm, 335 Mechanical Engineering Building, bring ID No class tomorrow! Exam scores posted tomorrow by 5 pm Exams returned in lab on Monday after spring break Liquid nitrogen ice cream winner = highest section average!!!

Exam Format – 90 minutes total 15 Multiple Choice Questions 2 Free Response Questions 5 possible choices 50% of test grade ~3 minutes to answer each (~45 min on this half of exam) Write in answers – both qualitative and quantitative 50% of test grade ~45 minutes on this half of exam ***A periodic table & cover sheet with equations will be attached.***

Top Review Topics You Asked For Concentration of ions problems Conceptual molarity problems Stoichiometry involving gases Neutralization What to do if solubility rules aren’t covered Also… “STOICH!”

Question #1: Solubility Rules How many are soluble? Solubility Rules Most nitrate salts are soluble. Most salts of sodium, potassium, and ammonium cations are soluble. Most chloride and iodide salts are soluble. Exceptions: Ag+ and Pb2+. Most sulfate salts are soluble. Exceptions: Ca2+, Ba2+, and Pb2+. Most hydroxide salts are only slightly soluble. Soluble ones are: Na+, K+, and Ca2+. Most sulfide, carbonate, and phosphate salts are only slightly soluble. BaCO3 Na2CO3 K3PO4 Mg3(PO4)2 Na3PO4

Clicker #1 Hydrogen peroxide decomposes according to the following equation. 2H2O2(l)  O2(g) + 2H2O(l) What volume of 0.20 M hydrogen peroxide is needed to produce 5.0 L of 1.00 atm oxygen gas at 25°C? 0.102 L 0.204 L 0.510 L 1.02 L 2.04 L

Clicker #2 You have a 16.0 oz (473 mL) glass of lemonade with a concentration of 2.00 M. The lemonade sits out on your counter for a couple of days and 150. mL of water evaporates from the glass. What is the new concentration of the lemonade?   a) 1.52 M b) 2.00 M c) 2.93 M d) 6.31 M e)none of these c

Clicker #3 Consider the reaction that occurs when 4.0 L of 2.00 M silver nitrate reacts with 2.0 L of 3.00 M sodium carbonate to form a white precipitate. 2AgNO3(aq) Na2CO3(aq)  Ag2CO3(s) + 2NaNO3(aq) 2Ag+(aq) + CO32-(aq)  Ag2CO3(s) What is the concentration of sodium ions after the reaction? 0 M 1.00 M 2.00 M 3.00 M 6.00M

Clicker #4 Consider the reaction that occurs when 4.0 L of 2.00 M silver nitrate reacts with 2.0 L of 3.00 M sodium carbonate to form a white precipitate. 2AgNO3(aq) Na2CO3(aq)  Ag2CO3(s) + 2NaNO3(aq) 2Ag+(aq) + CO32-(aq)  Ag2CO3(s) What is the concentration of nitrate ions after the reaction? 0.33 M 1.00 M 1.33 M 2.00 M 4.00M

Clicker #5 Consider the reaction that occurs when 4.0 L of 2.00 M silver nitrate reacts with 2.0 L of 3.00 M sodium carbonate to form a white precipitate. 2AgNO3(aq) Na2CO3(aq)  Ag2CO3(s) + 2NaNO3(aq) 2Ag+(aq) + CO32-(aq)  Ag2CO3(s) What is the concentration of silver ions after the reaction? 0 M 1.33 M 1.67 M 2.00 M 4.00 M

Clicker #6 Consider the reaction that occurs when 4.0 L of 2.00 M silver nitrate reacts with 2.0 L of 3.00 M sodium carbonate to form a white precipitate. 2AgNO3(aq) Na2CO3(aq)  Ag2CO3(s) + 2NaNO3(aq) 2Ag+(aq) + CO32-(aq)  Ag2CO3(s) What is the concentration of carbonate ions after the reaction? 0 M 0.33 M 1.00 M 3.00 M 6.00 M

Clicker #7 100.0 mL of a 0.200 M Ba(OH)2 solution is neutralized with 0.100 M HCl. What volume of HCl is needed? Ba(OH)2(aq) + 2HCl(aq) 2H2O(l) + BaCl2(aq) 50.0 mL 100.0 mL 200.0 mL 300.0 mL 400.0 mL

Clicker #8 Consider the equation: A + 3B  4C. If 3.0 moles of A is reacted with 6.0 moles of B, which of the following is true after the reaction is complete?   A is the leftover reactant because you only need 2 moles of A and have 3. A is the leftover reactant because for every 1 mole of A, 4 moles of C are produced. B is the leftover reactant because you have more moles of B than A. B is the leftover reactant because 3 moles of B react with every 1 mole of A. Neither reactant is leftover.

Clicker #9 You pour Solution #1 and Solution #4 together in a large, empty beaker. What is the concentration of the new HCl mixture?   3.00 M 3.25 M 3.40 M 4.25 M 6.50 M

Clicker #10 Use the unbalanced equation to answer the question below. If 2.50 moles of Si were produced, how many moles of NaF were also created? a) 0.417 mol b) 2.50 mol c) 6.00 mol d) 15.0 mol e) Impossible to determine without the moles of reactants.