# HW8 - padilla(tp5647 – Homework 8 – Sutcliffe –(53770...

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Unformatted text preview: padilla (tp5647) – Homework 8 – Sutcliffe – (53770) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points How many grams of CaBr 2 are needed to prepare 4.65 L of a 7.65 M solution? 1. 0.178 g 2. 4270 g 3. 122 g 4. 7110 g correct 5. 5.26 g 6. 0.00303 g Explanation: M = 7 . 65 M V = 4 . 65 L ? g CaBr 2 = 4 . 65 L × 7 . 65 mol CaBr 2 L soln × 200 g CaBr 2 1 mol = 7110 g 002 10.0 points Which of the following is false? 1. 0.3 moles of HCl in 3.0 L of solution = 0.3 molar HCl. correct 2. 42.0 g of NaHCO 3 in 250 mL of solution = 2.0 molar NaHCO 3 . 3. 5 × 10- 2 moles of Ba(OH) 2 in 1.0 L of solution = 0.05 molar Ba(OH) 2 . 4. 11.7 g of NaCl in 0.2 L of solution = 1.0 molar NaCl. Explanation: Molarity is the moles per liter of solution. We calculate the molarity of each solution: ? M HCl = . 3 mol HCl 3 L solution = 0 . 1 M HCl (The above answer choice is the correct one. The molarity of this HCl solution is not 0.3 M as described in the answer choice and the choice is therefore false.) ? M HCl = 11 . 7 g NaCL . 2 L × 1 mol NaCl 58 . 44 g NaCl = 1 . 0 M HCl ? M Ba(OH) 2 = . 05 mol Ba(OH) 2 1 L = 0 . 05 M Ba(OH) 2 ? M NaHCO 3 = 42 g NaHCO 3 . 250 L × 1 mol NaHCO 3 84 g NaHCO 3 = 2 . 0 M NaHCO 3 003 10.0 points We wish to dilute some 18 . 4 M H 2 SO 4 solution to make 600 mL of a 0 . 1 M H 2 SO 4 solution. How much of the 18 . 4 M solution should we start with? 1. 2.70 mL 2. 1.80 mL 3. 4.60 mL 4. 4.00 mL 5. 3.26 mL correct Explanation: M 1 V 1 = M 2 V 2 V 2 = M 2 V 2 M 1 = (0 . 1 M) (600 mL) 18 . 4 M = 3 . 26087 mL 004 10.0 points Which of the following aqueous solutions padilla (tp5647) – Homework 8 – Sutcliffe – (53770) 2 should NOT form a precipitate with aque- ous Ba(NO 3 ) 2 ? 1. K 2 CO 3 2. KOH correct 3. K 2 SO 4 4. K 3 PO 4 Explanation: All of the answer choices are soluble com- pounds and would dissociate to form ions in solution. These ions could react with the Ba 2+ and NO- 3 ions from Ba(NO 3 ) 2 . All an- swer choices contain the cation K + that could combine with NO- 3 to form KNO 3 , a soluble compound. Next we look at the possible prod- ucts of the combination of Ba 2+ with anions from the answer choices. BaCO 3 , Ba 3 (PO 4 ) 2 , and BaSO 4 are all insoluble compounds and would form precipitates. Ba(OH) 2 is soluble. A mixture of Ba(NO 3 ) 2 and KOH would not form a precipitate. 005 10.0 points What volume of 0.30 M NaOH is required to react completely with 70.0 mL of 0.24 M H 2 SO 4 ? 1. 124.0 mL 2. 49.6 mL 3. 99.1 mL 4. 56.0 mL 5. 112.0 mL correct Explanation: [NaOH] = 0.30 M V H 2 SO 4 = 70 mL [H 2 SO 4 ] = 0.24 M The balanced equation for this neutraliza- tion reaction is H 2 SO 4 + 2 NaOH → Na 2 SO 4 + 2 H 2 O We determine the moles of H 2 SO 4 present: ? mol H 2 SO 4 = 0 . 07 L soln × . 24 mol H 2 SO 4 1 L soln = 0 . 0168 mol H 2 SO 4 Using the mole ratio from the balanced chemical equation we calculate the moles of...
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HW8 - padilla(tp5647 – Homework 8 – Sutcliffe –(53770...

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