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Hydrogen bonding… (a) occurs only between water molecules

Hydrogen bonding… (a) occurs only between water molecules (b) is stronger than covalent bonding (c) can occur between NH 3 and H 2 O ( d) results from strong attractive forces in ionic compounds. Which of the following is not a general property of solutions ?

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Hydrogen bonding… (a) occurs only between water molecules

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  1. Hydrogen bonding… (a) occurs only between water molecules (b) is stronger than covalent bonding (c) can occur between NH3 and H2O (d) results from strong attractive forces in ioniccompounds

  2. Which of the following is not a general property of solutions? (a) a homogeneous mixture of two or more substances (b) variable composition (c) dissolved solute breaks down to individual molecules (d) the same chemical composition, the same chemical properties, and the same physical properties in every part

  3. Which procedure is most likely to increase the solubility of most solids in liquids? (a) stirring (b) pulverizing the solid (c) heating the solution (d) increasing the pressure

  4. The addition of a crystal of NaClO3 to a solution of NaClO3 causes additional crystals to precipitate. The original solution was (a) unsaturated (b) dilute (c) saturated (d) supersaturated

  5. If NaCl is soluble in water to the extent of 36.0 g NaCl / 100 g H2Oat 20 oC, then a solution at 20 oC containing 45.0 g NaCl/ 150 g H2O would be… 45.0 g NaCl = X g NaCl 150 g H2O 100 g H2O X = 30.0 g NaCl is unsaturated

  6. If 5.00 g NaCl is dissolved in 25.0 g H2O, the percent NaCl by mass is 5.00 g NaCl =5.00 5.00 g NaCl + 25.0 g H2O 30.0 = 16.7 %

  7. How many grams of 9.0% AgNO3 solution will contain 5.3 g AgNO3? 9.0 = 5.3 100 X X = 59

  8. What mass of BaCl2 will be required to prepare 200. mL of 0.150 M solution? 0.150 mol x 0.200 L = 0.0300 mol L x 208.3g/mol = 6.25 g BaCl2

  9. How many grams of a solution that is 12.5 % by mass AgNO3 would contain 0.400 mol of AgNO3? 0.400 mol x 169.9 g = 68.0 g AgNO3 mol 12.5 = 68.0 g AgNO3 100 X X = 544 g of solution

  10. How much solute is present in 250. g of 5.0% K2CrO4 solution? 5.0 = X 100 250. g solution X = 12.5 gK2CrO4

  11. First find moles of each reactant molarity = moles liters molarity x liters= moles 0.642 mol x 0.0805 L = 0.0517 mol Ba(NO3)2 L 0.743 mol x 0.0445 L = 0.0331 mol KOH L 0.0517 mol 0.0331 mol 1 Ba(NO3)2 + 2 KOH → 1 Ba(OH)2 + 2 KNO3

  12. 0.0517 mol 0.0331 mol 1 Ba(NO3)2 + 2 KOH → 1 Ba(OH)2 + 2 KNO3 1 Ba(NO3)2 = 0.0517 1 Ba(OH)2 x = 0.0517 mol Ba(OH)2 2 KOH = 0.0331 mol 1 Ba(OH)2 x = 0.0166 molBa(OH)2 --limiting reactant = KOH 0.0166 mol Ba(OH)2 x 171.3g = 2.84g Ba(OH)2 1 mol

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