1441answertest2

1441answertest2 - Answer Key 1441 Test Two Chapters 4 &...

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Unformatted text preview: Answer Key 1441 Test Two Chapters 4 & 5 ( Zumdahl) 1. (a) H + + OH- H 2 O (b) Ag + + Cl- AgCl (c) HC 2 H 3 O 2 + OH- H 2 O + C 2 H 3 O 2- 2. M = moles/L 3M = moles / 0.2 L moles = 0.6 moles NaOH (0.6 mole NaOH/1)(40g NaOH/1 mole NaOH) = 24g 3. M = moles/L M = (30g NaOH/1)( 1 mole NaOH / 40 g NaOH) / 0.3 L = 2.5 M 4. M 1 V 1 = M 2 V 2 (3M)(300 ml) = (1.5 M)(V 2 ) V 2 = 600 ml Since you start with 300 ml of solution and want to end with 600ml, the water necessary is the difference 600 ml 300 ml = 300 ml water 5. Ca(OH) 2 + 2HCl CaCl 2 + 2 H 2 O moles = (V) (M) = 0.4 L X 1.5 M HCl = 0.6 moles of HCl (0.6 mole HCl/1)(1 mole Ca(OH) 2 / 2moles HCl) * = 0.3 mole Ca(OH) 2 M = moles/ L 2M = 0.3 mole Ca(OH) 2 / L L = 0.15 L or 150 ml * The numbers in this step come from the balanced equation. 6. (4.75 g NaCl/1)(1 mole NaCl/58.44 g NaCl) = 0.0813 mole NaCl (0.575 g CaCl 2 /1)(1 mole CaCl 2 /110.98 g CaCl 2 ) = 0.00518 mole CaCl 2 NaCl Na + + Cl- 0.0813 mole 0.0813 mole 0.0813 mole CaCl...
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This note was uploaded on 10/13/2009 for the course CHEM 1441 taught by Professor Tanizaki during the Spring '08 term at UT Arlington.

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1441answertest2 - Answer Key 1441 Test Two Chapters 4 &...

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