You push on a crate, and it starts to move but you don't. .... If there is more than one object causing energy transformations, use a separate bar for each. (b) Repeat using a system of just the block and incline. (c) Why can't you use the ... extern
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CHEM 139: Acids, Bases, & Stoichiometry Practice Problems 1. Phosphoric acid reacts with strontium hydroxide to produce a precipitate. a. Write the balanced chemical equation for the reaction. b. What mass of precipitate is produced when 25.0 mL of 0.750M strontium hydroxide react with 45.0 mL of 0.500M phosphoric acid? c. What volume of a 1.50M strontium hydroxide solution is required to completely neutralize 50.0 mL of a 0.675M phosphoric acid solution? 2. Nitric acid can be prepared as follows: 3 NO2(g) + H2O(l) → 2 HNO3(aq) + NO(g) a. Calculate the mass of NO2 gas required to produce 250.0 mL of 1.50M nitric acid. b. Calculate the molarity of a nitric acid solution prepared by bubbling 10.0 g of nitrogen dioxide gas in 500.0 mL of deionized water. (Assume the total volume of the resulting solution is also 500.0 mL.) c. Calculate the pH of a solution prepared by diluting 1.5 mL of a 0.100M nitric acid solution with 500.0 mL of deionized water. 3. A student prepares a calcium hydroxide solution by mixing calcium oxide with water, CaO(g) + H2O(l) → Ca(OH)2(aq) a. Calculate the molarity of a calcium hydroxide solution prepared by completely reacting 1.50 g of calcium oxide with 2.5 L of deionized water. b. Calculate the pH of the solution described in part a. 4. Tartaric acid, H2C4H4O6(aq), is a diprotic acid often present in wines. It precipitates from solution when the wine ages. a. Write the balanced chemical equation for the reaction between tartaric acid and sodium hydroxide. b. Calculate the molarity of tartaric acid in a sample of white wine if 25.0 mL of wine required 49.51 mL of 0.05026M sodium hydroxide for complete neutralization. c. Calculate the mass percent concentration of tartaric acid in the wine described in part b given that the density of the wine is 1.03 g/mL.