Solution Stoichiometry Chem Worksheet 15 6 – 19 Answers and solutions to textual problems 9.1 a. δ O δ + δ + H H In the water molecule, oxygen has a partial negative charge and hydrogen has a partial positive charge. b δ δ + O H δ + δ + δ H H O δ + H 9.2 a. The surface tension of tightly packed water molecules on the surface is sufficient to support the weight of small insects. B. Surfactant is a compound that breaks hydrogen bonds that contribute to surface tension. 9.3 The component in the lowest amount is the soluble substance. The larger amount is halal. A. sodium chloride solution; water, solvent b. water, solution; Ethanol, solvent c. oxygen, dissolved substance; Nitrogen, solvent 9.4, the component in the lowest amount is the soluble substance. The larger amount is halal. A. silver, solvent; living matter, solution b. water, solvent; Sugar, solution c. iodine, solution; Alcohol, solvent 9.5 a. Potassium chloride is an ionic solution soluble in water (polar solvent). Iodine, a non-polar solute, is soluble in carbon tetrachloride (a non-polar solvent). C. Sugar, a polar solute, is soluble in water, which is a polar solvent. D-benzene, a non-polar solute, is soluble in carbon tetrachloride, which is a non-polar solvent. 9.6a. Vegetable oil, non-polar substance must be soluble in hexane (non-polar substance). Benzene, a non-polar substance must be soluble in hexane c. Lithium nitrate, an ionic substance soluble in water, (polar solvent) d. Sodium sulfate, a water-soluble ionic substance (polar solvent) 9.7 K+ and I ions are adsorbed on the surface of a solid by polar water molecules, where the hydration process surrounds the dissociated ions with water molecules. 9.8 The Li + and Br ions on the crystal surface separate as polar water molecules begin to surround each ion in the hydration process. Ions in solution are hydrated. 9.9 The molar mass of Na 2 SO 4 10 H 2 O is 322. The percentage of water is calculated as = 55.9% 322.
2 Chapter 9 Answers and Solutions 9.10 MgSO 4 7H 2 O % H 2 O = mass of 7 H 2 O x 100 = 126 g x 100 = 51.1 % H 2 O molar mass of MgSO 4 7H 2 O g 9.11 KF when Salt dissolves in water. The weak acid HF exists mostly as a single molecule along with some ions when dissolved in water. NaOH completely dissociates into ions when dissolved in water. As CH 3 OH dissolves in water, it remains molecular. Strong electrolytes dissociate into ions. LiBr Li + + Br b. NaNO 3 Na + + NO 3 c. FeCl 3 FeCl d. Mg(NO 3 ) 2 Mg NO a. In solution, a weak electrolyte exists mostly as molecules with several ions. B. Sodium bromide is a strong electrolyte and forms ions in solution. C. A non-electrolyte does not dissociate and only forms molecules in solution a. Na 2 SO 4 (s) 2Na + (aq) + SO 2 4 (aq) Sodium sulfate is a strong electrolyte in solution and exists only as ions in solution. b C 2 H 5 OH(l) C 2 H 5 OH(ak) Only molecules A non-electrolyte exists only as one molecule in solution. C. HCN(ak) H + (ak) + CN (ak) Most molecules, some ions A weak electrolyte exists as molecules and some ions in solution a. A strong electrolyte because only ions are present in K 2 SO 4 solution b. The electrolyte is weak because both ions and molecules are present in the NH 4 OH solution. C. There is no electrolyte because there are only C 6 H 12 O 6 molecules in the solution 9.18 a. A non-electrolyte b. strong electrolyte c. weak electrolyte 9.19 a. 1 mol K + 1 Ek = 1 K equivalent + 1 mol K + b. 2 moles of OH 1 Ek = 2 equivalent to 1 mole of OH c. 1 mol Ca 2 + 2 Ek = 2 is equivalent to 1 mol Ca 2 + 2 d. 3 mol CO 3 2 Ek = 6 equivalent to 1-2 mol CO a. 1 mole of Mg 2+ = 2 equivalent b. 0.5 mol H + = 0.5 equivalent c. 4 moles of chlorine = 4 equivalent d. 2 mol Fe 3+ = 6 Equiv 9.21 a g Cl 1 mol Cl 1 Ek = Equiv. 35.5 g Cl 1 mol Cl b g Fe 3+ 1 mol Fe 3+ 3 Ek = equivalent to 55.9 g Fe 3+ 1 mol Faith 3+ c. 4.0 g Ca 2 + 1 mol Ca 2 + 2 Eq = 0.20 Equiv. 40.1 g Ca 2 + 1 mol Ca 2 + d. 1.0 g H + 1 mol H + 1 Eq = 1.0 Equiv. 1.0 g H + 1 mol H +
Solution Stoichiometry Chem Worksheet 15 6
3 9.22 a g Na + 1 mol Na + 1 equivalent = equivalent to 23.0 g Na + 1 mol Na + b. 8.0 g OH 1 mol OH 1 equivalent = 0.47 equivalent 17.0 g OH 1 mol OH c g Mg 2 + 1 mol Mg 2 + 2 equivalent = 1.65 equivalent 24.3 g Mg 2 + 1 mol Mg 2 + d. 4.0 g Al 3 + 1 mol Al 3 + 3 equivalent = 0.44 equivalent 27.0 g Al 3 + 1 mol Al L 3.0 mek 1 Ek 1 mol Mg g Mg 2 + = 0.18 g Mg 2 + 1 L 1000 mek Mg 2 Ek 1 Ek 1 mol Cl 35.5 g Cl = 20, g Cl 1 L 1000 mec 1 Ek 1 mol Cl L 154 mek 1 Ek 1 mol Na g Na + = 3.5 g Na + 1 L 1000 mec 1 Ex 1 mol Na L 154 mec 1 Ex 1 mol Cl 35.5 g Cl = 5.5 g Cl 1 L 1000 mec 1 Ex 1 mol Cl L 40. mec 1 Ex 1 mol K g K + = 2.3 g K + 1 L 1000 mek 1 Ek 1 mol K L 40 .mek 1 Ek 1 mol Cl 35.5 g Cl = 2.1 g Cl 1 L 1000 mek 1 Ek 1 mol Cl 9.27 The total anion equivalents must equal the cation equivalents in any solution. mek of anion = 40. mek Cl / L + 15 mek HPO 4 2 / L = 55 mek/L mek Na + = mek anion = 55 mek Na + / L 9.28 Total equivalents of cations must be equivalent of anions in each way solution 147 mek Na + / L + 4 mek K + / L + 4 mek Ca 2+ /L = 155 mek/l cation = 155 mek Cl / L 9.29 a. The solution must be saturated because the excess solution has not dissolved. b The solution is not saturated because it dissolves the sugar cube a. The solution is not saturated because the salt dissolves. B. The solution must be saturated because undissolved solutes remain at the bottom of the beaker. It is unsaturated because 34.0 g of KCl is the maximum that will dissolve in 100 g of H 2 O at 20 °C. b Adding 11.0 g of NaNO 3 to 25 g of H 2 O is the same as adding 44.0 g of NaNO 3 to 100 g of H 2 O. At 20°C, 88.0 g of NaNO 3 can be dissolved, so the solution is not saturated. C. Adding grams of sugar to 125 grams of H 2 O is 320 grams per 100 grams of H 2 O. At 20 degrees Celsius, only grams of sugar can be dissolved, which is less than 320 grams. The sugar solution is saturated and there is excess sugar. At 50°C, 42.6 g of KCl is soluble in 100 g of H 2 O, which means that 21.3 g of KCl is soluble in 50 g of H 2 O. Therefore, adding 25.0 g of KCl to 50 g exceeds the solubility and a saturated solution is obtained. Adding g of NaNO 3 to 75 g of H 2 O is equal to adding g of NaNO 3 to 100 g of H 2 O. This solution is saturated because g of NaNO 3 is greater than the solubility of g that can be dissolved in 100 g of solution. Water at 50°C At 50°C, grams of sugar can be dissolved in 100 grams of H 2 O. Adding 80.0 g of sugar to 25 g of H 2 O equals 320 g of sugar to 100 g of H 2 O. This is greater than the solubility, meaning the solution is saturated.
Worksheet Balancing Equations.
4 Chapter 9 Answers and Solutions 9.33 a g KCl 200 g H 2 O = 68.0 g KCl (this dissolves at 20 C) 100 g H 2 O at 20 C 68.0 g KCl can be dissolved in 200 g H 2 O did b Since 80.0 g of KCl dissolves at 50 °C
Solution stoichiometry worksheet, chem solution, ap chemistry stoichiometry worksheet, chem worksheet, stoichiometry problems worksheet answers, worksheet for basic stoichiometry, chemistry stoichiometry worksheet, stoichiometry worksheet and answers, stoichiometry calculations worksheet, chemistry stoichiometry worksheet answers, solution stoichiometry calculator, stoichiometry worksheet