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Hydrogen bonding… (a) occurs only between water molecules

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Presentation on theme: "Hydrogen bonding… (a) occurs only between water molecules"— Presentation transcript:

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 ionic compounds

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 H2O at 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 = 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 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 X X = g solution

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

9 How many grams of a solution that is 12
How many grams of a solution that is 12.5 % by mass AgNO3 would contain mol of AgNO3? 0.400 mol x g = g AgNO3 mol = g AgNO3 X X = 544 g of solution

10 How much solute is present in 250. g of
5.0% K2CrO4 solution? = X g solution X = 13 g K2CrO4

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

12 mol mol 1 Ba(NO3) KOH → 1 Ba(OH) KNO3 1 Ba(NO3)2 = 1 Ba(OH) x = mol Ba(OH)2 2 KOH = mol 1 Ba(OH) x = mol Ba(OH)2 --limiting reactant = KOH mol Ba(OH)2 x 171.3g = 2.84g Ba(OH)2 1 mol


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