H 4 4 d C 2 H 2 1 e C 2 H 4 2 f K 2 Cr O 4 6 g K 2 Cr 2 O 7 6 h KMn O 4 7 i

H 4 4 d c 2 h 2 1 e c 2 h 4 2 f k 2 cr o 4 6 g k 2 cr

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H 4 =-4 d. C 2 H 2 =-1 e. C 2 H 4 =-2 f. K 2 Cr O 4 =+6 g. K 2 Cr 2 O 7 =+6 h. KMn O 4 =+7 i. NaHC O 3 =+4 j. O 2 =0 k. NaI O 3 =+5 l. K 2 O =-1 m. P F 6 =+6 n. Au Cl 4 =+3 (Reference: Chang 4.47) 5. For the following reactions, identify the type (combination, combustion, decomposition, displacement, methathesis [double-displacement)] neutralization), balance the reaction, and for the redox reactions identify the oxidizing agent and reducing agent. (12 points) (a) CH 4 + O 2 CO 2 + H 2 O Combustion, CH4 + 2 O2 = CO2 + 2H2O (b) HCl + NaOH H 2 O + NaCl Neutralization HCl + NaOH = H2O + NaCl (c) H 2 + O 2 H 2 O Redox reactions 2H2 + O2 = 2 H2O oxidizing agent=02 reducing agent=H2 (d) CaCO 3 + HCl CaCl 2 + CO 2 + H 2 O Decomposition CaCO3 +2HCl = CaCl2 + CO2 + H2O 2 Copyright © 2014 by Thomas Edison State College. All rights reserved.
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(e) Zn + HCl ZnCl 2 + H 2 Displacement Zn + 2 HCl = ZnCl2 + H2 (f) NaCl Na + Cl 2 Decomposition 2NaCl = 2Na+ + Cl2 6. Calculate the molarity of each of the following solutions: (4 points) a. 29.0 g of ethanol (C 2 H 5 OH) in 545 mL of solution Number of mole of C2H5OH = 29.0 / (12*2+1.0*6+16.0) = 0.630434782mole Molarity = number of mole/volume = 0.630434782 / (545/1000) = 1.26moldme- 3 = 1.26M b. 9.00 g of sodium chloride (NaCl) in 86.4 mL of solution Number of mole of NaCl = 9.00 / (58.44) = 0.154 mol Molarity = number of mole/volume = 0.154 / (86.4/1000) = 1.78 M 7. Calculate the mass of KI in grams required to prepare 5.00 × 10 2 mL of a 2.80 M solution. (3 points) 5x10^2 ml = 500 ml = .5L 2.8 = moles/.5L moles=1.4 (1.4)(166) = 232.4 232.4 grams KI
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  • Mole, Sodium, Chemical reaction, Sodium chloride, Sodium hydroxide

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