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Last Name Professor BEAMER First Name Which lab section are you in? (Circle One) EXAM 4 REVIEW CHEM 110 M W R Note: As Part 1 is supposed to be memorized material, you can cross-check your answers with the information in your textbook and/or from your notes. PART 2: MATH 15a) Calculate the molar mass of Na2Cr2O7. Rewrite final answer 261.98 g/mol Na2Cr2O7 (Time Range: 3 to 5 min) g/mol g/mol 2 × Na 2× 22.99 45.98 2 × Cr 2× 52.00 104.00 7×O 7× 16.00 + 112.00 261.98 g/mol Na2Cr2O7 15b) Rewrite the molar mass of the substance above in conversion factor form: 261.98 g Na2Cr2O7 = 1 mol Na2Cr2O7 (Time Range: 1 min − all levels) Page 1 of 13 EXAM 4 REVIEW CHEM 110 Identify initial value! 16) Calculate the number of molecules of NH3 in a 1.400 × 10−5 mole sample of NH3. (Time Range: 3 to 6 min) Solution/Explanation: You are converting between particles (molecules) and moles (mol). Therefore, you need to use Avogadro’s Number. You need to devise a conversion factor: 1 mol NH3 = 6.022 × 1023 NH3 molec Always start with your initial value: 1.400 × 10−5 NH3 molec 1 mol NH3 = 6.022 × 1023 NH3 molec initial value -5 1.400 × 10 ( 1 NH3 mol ) Avogadro’s Number Conversion 23 6.022 × 10 NH3 molec ( ) = 1 mol NH3 18 8.431 × 10 NH3 molec Page 2 of 13 Identify initial value! 17) A sample of CaCl2 contains 3.8 × 1025 of CaCl2 is 110.98 g/mol. EXAM 4 REVIEW CHEM 110 formula units of CaCl2. Calculate the mass (in grams) of CaCl2 in the sample. The molar mass Identify final unit: mass (grams) (Time Range: 3 to 5 min) Solution/Explanation: You are converting between particles (molecules) and mass (grams). Therefore, you need to use Avogadro’s Number. There is no direct pathway from particles to moles: mass moles particles You must go through moles first. You need two conversion factors: 110.98 g CaCl2 = 1 mol CaCl2 6.022 × 1023 CaCl2 form = 1 mol CaCl2 initial value ( 3.8 × 10 25 1 Avogadro’s Number Conversion CaCl2 form ) 1 mol CaCl2 ( ) 23 6.022 × 10 CaCl2 form = 63 mol CaCl2 molar mass 63 mol CaCl2 ( ) 1 ( 110.98 g CaCl2 ) 1 mol CaCl2 = 7003 g CaCl2 x7.0 × 103 g CaCl2x Also acceptable: initial value ( 3.8 × 10 25 1 CaCl2 form ) Avogadro’s Number Conversion molar mass 1 mol CaCl2 ( ) 23 6.022 × 10 CaCl2 form 110.98 g CaCl2 ( ) 1 mol CaCl2 = 7003 g CaCl2 x7.0 × 103 g CaCl2x Page 3 of 13 EXAM 4 REVIEW CHEM 110 Reaction for Question 18 283.88 g/mol P4O10(s) 63.01 g/mol + 12 HNO3(aq) 98.00 g/mol 4 H3PO4(aq) 108.01 g/mol + 6 N2O5(s) Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5) Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from 500.0 grams of nitric acid, HNO3(aq). Identify initial value! 18) (Time Range: 6 to 12 min) Solution/Explanation: Steps to Theoretical Yield Calculations Step 1: Convert the initial value (the KNOWN) to moles Step 2: Use mole ratios to convert moles KNOWN moles UNK Step 3: Convert moles UNK to mass UNK molar mass initial value STEP 1: * ( 500.0 g HNO3 ) 1 ( 1 mol HNO3 ) 63.01 g HNO3 = 7.935 mol HNO3 Use molar mass of the balanced equation to convert mass of HNO3 to moles of HNO3. mole ratio STEP 2: * 7.935 mol HNO3 ( ) 1 6 mol N2 O5 ( ) 12 mol HNO3 = 3.968 mol N2O5 Use the coefficients of the balanced equation to convert moles of HNO3 to moles of N2O5. molar mass STEP 3: 3.968 mol N2 O5 ( ) 1 108.05 g N2 O5 ( ) 1 mol N2 O5 = x428.7 g N2O5 OR 4.287 × 102 g N2O5x * Convert moles of N2O5 to mass of N2O5 using molar mass. * You can always put answers to calculations in standard SCINOT. (Make sure the SCINOT is correct.) See next page to see “all-in-one-step” calculation: Page 4 of 13 EXAM 4 REVIEW CHEM 110 Reaction for Question 18 283.88 g/mol P4O10(s) 63.01 g/mol + 12 HNO3(aq) 98.00 g/mol 4 H3PO4(aq) 108.01 g/mol + 6 N2O5(s) Notice that I have highlighted the coefficients of the known (HNO3) and the unknown (N2O5) 18) Calculate the theoretical yield of dinitrogen pentoxide, N2O5, that can be prepared from 500.0 grams of nitric acid, HNO3(aq). Identify initial value! (Time Range: 6 to 12 min) Solution/Explanation: Steps to Theoretical Yield Calculations Step 1: Convert the initial value (the KNOWN) to moles Step 2: Use mole ratios to convert moles KNOWN moles UNK Step 3: Convert moles UNK to mass UNK INITIAL VALUE STEP 1 STEP 2 STEP 3 initial value molar mass mole ratio molar mass 1 mol HNO3 ( ) 63.01 g HNO3 6 mol N2 O5 ( ) 12 mol HNO3 108.05 g N2 O5 ( ) 1 mol N2 O5 ( 500.0 g HNO3 ) 1 = x428.7 g N2O5 OR 4.287 × 102 g N2O5x * Again, I have no preference as to whether you break down this solution into three steps or choose the “all-in-one-step” method. Page 5 of 13 19) EXAM 4 REVIEW CHEM 110 Assume that a scientist perform this experiment and obtains a yield of 411.7 grams of N2O5. Calculate the %-yield value for this experiment. Use the value obtained in Question 18 for the theoretical yield. Note: You must rewrite the %-yield equation first with variables only as your first step. (Time Range: 2 to 4 min) Solution %-yield = experimental yield theoretical yield × 100 = = experimental yield theoretical yield 411.7 g N2 O5 428.7 g N2 O5 × 100 × 100 = 96.03 % yield N2 O5 Page 6 of 13 EXAM 4 REVIEW CHEM 110 PART 3: NON-CALCULATOR 20) Resources: You will be given a periodic table (symbols only) You will be given the memorized table of polyatomic ions chart You will be given the Activity Series Table (Top Dog) You will be given the Solubility Rules (all of them) No calculators for this section Timing: 10 min (mastery) 25 min (competence) Indicate whether the following elements are solid (s), liquid (l), or gas (g). 20a) Cu (#29) (s) 20b) Hg (#80) (l) (Time Range: 1 min maximum – all levels) 21) List the charges of the following elements when they become ions: 21a) nitrogen (#7) −3 or 3− 21c) potassium (#19) +1 or 1+ 21b) sulfur (#16) −2 or 2− 21d) iodine (#53) −1 or 1− (Time Range: 1 min 30 s maximum – all levels) IMPORTANT: For positive ions, you must write the “+” symbol! Question 21a has been corrected. It should be −3 (not +3). Sorry. Page 7 of 13 EXAM 4 REVIEW CHEM 110 22) Circle the metals that require a Roman Numeral in their names when they are present in ionic compounds. Na (#11) Fe (#26) Sn (#50) Al (#13) Zn (#30) Ba (#56) (Time Range: 2 min maximum – all levels) Na Group 1A (alkali metals) always have a charge of 1+. Therefore, Roman numerals are not required when naming salts that contain Group 1A metals. Fe Transition metal. With three exceptions, transitions metals always require a Roman numeral in their names to indicate their charge. Sn Post-Transition metal. All post-transition metals always require a Roman numeral in their names to indicate their charge. Al Group 3A. This is a little tricky. Even though there are transition metals in Group 3A, aluminum itself is not a transition metal. Therefore, it does not require a Roman numeral to indicate its charge. Zn Zinc is one of the three exceptions of the transition metals. Zinc always has a charge 2+ in ionic compounds. Therefore, a Roman numeral is not required. Here are the three exceptions again: Ba Ag (silver) is always 1+ Cd (cadmium) is always 2+ in ionic compounds Zn (zinc) is always 2+ in ionic compounds When present in ionic compounds, these three metals do not require a Roman numeral when naming them. Group 2A (alkali earth metals) always have a charge of 2+. Therefore, Roman numerals are not required when naming salts that contain Group 2A metals. Page 8 of 13 EXAM 4 REVIEW CHEM 110 23) Circle the compounds that are soluble in water. potassium sulfate V2O5 ammonium phosphate CuCl2 chromium(III) hydroxide Au2(Cr2O7)3 (Time Range: 3 min maximum – all levels) potassium sulfate K2SO4 potassium V2O5 vanadium(V) oxide vanadium(V) K+ Rule 1 = soluble Rule 6 = soluble V5+ Rule 7 = insoluble oxide O 2− Rule 5 = insoluble ammonium phosphate ammonium NH4 + Rule 1 = soluble (NH4)3PO4 phosphate PO4 3− Rule 5 = insoluble CuCl2 copper(II) chloride copper(II) Cu2+ Rule 7 = insoluble chloride Cl − Rule 4 = soluble chromium(III) Cu2+ Rule 7 = insoluble Cr(OH)3 hydroxide OH− Rule 5 = insoluble Au2(Cr2O7)3 gold(III) Au3+ Rule 7 = insoluble dichromate Cr2O7 2− Rule 7 = insoluble chromium(III) hydroxide Gold(III) dichromate sulfate SO4 2− Rule 1 beats Rule 6 Rule 5 beats Rule 7 Rule 1 beats Rule 5 Rule 4 beats Rule 7 Rule 5 beats Rule 7 Rule 7 (tie) Page 9 of 13 EXAM 4 REVIEW CHEM 110 24) Balance the following two equations: 2 24a) C4H10(l) + 9 O2(g) ∆ → 8 CO(g) + 10 H2O(g) Note: For Question 24a, I purposely did not tell you how much oxygen was present. It does not matter, since the products are already written. Sorry about the wrong answer. 3 24b) Hint: Time Limit: 25) Ca(OH)2(s) + 2 H3PO4(aq) 1 Ca3(PO4)2(s) + 6 HOH(g) Rewrite water as _____ to help you balance. 4 minutes maximum – all levels. Write the skeleton equation: (Don’t forget states: s, l, g, aq) phosphorus(s) + chlorine(g) phosphorus pentachloride(s) P(s) + Cl2(g) PCl5(s) Time Limit: 26) 2 minutes to 4 minutes Write the word equation: (Don’t forget states: s, l, g, aq) Cl2O7(s) + H2O(l) HClO4(aq) dichlorine heptoxide(s) + water(l) perchloric acid(aq) Time Limit: 2 minutes to 4 minutes Page 10 of 13 EXAM 4 REVIEW CHEM 110 27) Predict the products (chemical formulas): Do not balance equations. Include states: (s), (l), (g), or (aq) Hint: Name the products first. Write the names in the margins. Then write the chemical formulas from the names. FeBr3(aq) + Cl2(g) FeCl3(aq) + Br2(l) 27a) Time Limit: 2 minutes maximum – all levels Solution The reactants are: ionic compound + an element. Therefore, this is a Single Replacement (SR) Before anything else, name the reactants: iron(III) bromide + chlorine (g) The element (chlorine gas) is a nonmetal. Therefore, it will try to replace bromide. Check the Activity Series Table (Appendix 8): chlorine beats bromine. Therefore, chlorine and bromide switch places. Before anything else: name the products: iron(III) bromide + FeBr3(aq) chlorine iron(III) chloride + bromine Cl2(g) FeCl3(aq) + Br2(l) Note that the suffixes –ine and –ide stay put. Write the chemical formulas from the names. iron(III) chloride is a salt. Therefore, you must determine if water is hanging around, and if iron(III) chloride is soluble. The “aq” in the reactants tells you that water is present. Therefore, you must determine the solubility of iron(III) chloride. Solubility Table, Rule 4, chlorides are soluble. Therefore, iron(III) chloride is aqueous. Don’t forget states iron(III) chloride The “aq” in the reactants tells you that water is present. Therefore, you must determine the solubility of iron(III) chloride. Solubility Table, Rule 4, chlorides are soluble. Therefore, iron(III) chloride is aqueous. bromine The states of all elements should be memorized. Elemental bromine, Br2, is a liquid. Page 11 of 13 EXAM 4 REVIEW CHEM 110 XS 27b) C5H8(l) + O2(g) ∆ → CO2(g) + H2O(g) Time Limit: 1 minute maximum – all levels Solution C5H8 is a hydrocarbon (contains carbon and hydrogen). The other reactant is oxygen, and the reaction is being heated. Therefore, this is a HYDROCARBON COMBUSTION reaction. Since there is excess oxygen, the products are carbon dioxide gas and water vapor (memorized). There is no way to “figure this out.” You either have learned/memorized this, or not. K(s) + Br2(l) KBr(s) 27c) Time Limit: 1 minute maximum – all levels Solution This is a simple COMBINATION reaction. Metal + nonmetal salt Name the salt: potassium bromide (common sense) Write the chemical formula from the name. States If a product is a salt (ionic compound), the salt is a solid unless water is hanging around. There are no “aq” in the reactants, water is NOT a reactant, and water is NOT present over the reaction arrow → Water is not present, so you do not need to check for solubility. This product is a solid. H2 O Do not worry that the equation is not balanced. I asked for the unbalanced equation. Page 12 of 13 EXAM 4 REVIEW CHEM 110 28) The following reaction is NR. In three or less sentences, explain why no reaction occurs. Time Limit: 3 minutes maximum – all levels (NH4)3PO4(aq) + K2SO4(aq) (NH4)2SO4(aq) + K3PO4(aq) This is a DOUBLE REPLACEMENT (DR) reaction. In DR reactions, if all substances (reactants and products) are soluble (aqueous), then there is no reaction. All reactants and products in the reaction above are aqueous, so NR. Page 13 of 13