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.

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Christopher G. Hamaker, Illinois State University, Normal IL
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Presentation transcript:

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? (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

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

The addition of a crystal of NaClO 3 to a solution of NaClO 3 causes additional crystals to precipitate. The original solution was (a) unsaturated (b) dilute (c) saturated (d) supersaturated

If NaCl is soluble in water to the extent of 36.0 g NaCl / 100 g H 2 O at 20 o C, then a solution at 20 o C containing 45.0 g NaCl / 150 g H 2 O would be… 45.0 g NaCl = X g NaCl 150 g H 2 O100 g H 2 O X = 30.0 g NaCl is unsaturated

If 5.00 g NaCl is dissolved in 25.0 g H 2 O, the percent NaCl by mass is 5.00 g NaCl= g NaCl g H 2 O30.0 = 16.7 %

How many grams of 9.0% AgNO 3 solution will contain 5.3 g AgNO 3 ? 9.0 = X X = 59

What mass of BaCl 2 will be required to prepare 200. mL of M solution? mol x L = mol L x 208.3g/mol = 6.25 g BaCl 2

How many grams of a solution that is 12.5 % by mass AgNO 3 would contain mol of AgNO 3 ? mol x g = 68.0 g AgNO 3 mol 12.5 = 68.0 g AgNO X X = 544 g of solution

How much solute is present in 250. g of 5.0% K 2 CrO 4 solution? 5.0 = X g solution X = 12.5 g K 2 CrO 4

First find moles of each reactant molarity = moles liters molarity x liters = moles mol x L = mol Ba(NO 3 ) 2 L mol x L = mol KOH L mol mol 1 Ba(NO 3 ) KOH → 1 Ba(OH) KNO 3

mol mol 1 Ba(NO 3 ) KOH → 1 Ba(OH) KNO 3 1 Ba(NO 3 ) 2 = Ba(OH) 2 x = mol Ba(OH) 2 2 KOH = mol 1 Ba(OH) 2 x = mol Ba(OH) 2 --limiting reactant = KOH mol Ba(OH) 2 x 171.3g = 2.84g Ba(OH) 2 1 mol