All cations are assumed to be in aqueous solution and the common anion is the

All cations are assumed to be in aqueous solution and

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All cations are assumed to be in aqueous solution, and the common anion is the nitrate ion. (8 points) A- Separation of K + from Ag + ions when we added chloride ions Cl - then AgCl is insoluble and will precipitate out while KCl being soluble will remain in the solution. 2 Copyright © 2014 by Thomas Edison State College. All rights reserved.
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LOVEJOY B- Add carbonate ions OH-, Pb(OH)2 being insoluble will from precipitate. The Ba(OH)2 is soluble and remains in the solution. C- (NH4)2 CO3 being soluble will remain in the solution whereas CaCO3 is an insoluble and will precipitate out. D- Add sulfate ions. The 2+ ions will react to from different compounds with the 2- sulfate ion. (Reference: Chang 4.24) Excellent work here! Each of your choices shows the selective precipitation result. 4. Give the oxidation number of the underlined atoms in the following molecules and ions: (7 points ) a. Cl F = x + (-1)=0 = x = +1 = Cl = +1 b. I F 7 x + 7(-1) =0 x+7 I=+7 c. C H 4 x + 4(+1)=0 x = -4 C = -4 d. C 2 H 2 2(x) + 2 (+1)=0 x= -1 C= -1 e. C 2 H 4 2 (x) + 4 (+1) = 0 x = -2 C = -2 f. K 2 Cr O 4 2 (+1) + 2x + 7(-7) = 0 x = + 6 Cr = + 6 g. K 2 Cr 2 O 7 2(+1) + 2x + 7(-2) =0 3 Copyright © 2014 by Thomas Edison State College. All rights reserved.
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LOVEJOY 2x=12 X=+6 Cr = + 6 h. KMn O 4 X+ 1 + (-2*4)=0 X= +7 Mn = +7 i. NaHC O 3 (+1)+(+1)+X+(3x-2)=0 +2 -6 + X = 0 -4+X=0 X= +4 C= + 4 j. O 2 O= 0 k. NaI O 3 3(-2) + 1+X =0 X=+5 I= +5 l. K 2 O 2(+1) + x = 0 X=-2 O=-2 m. P F 6 X + 6(-1)= 0 P = +6 n. Au Cl 4 X+4(-1)=0 X= 4 Au= +4 (Reference: Chang 4.47) Excellent! Each of these are correctly identified 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) 4 Copyright © 2014 by Thomas Edison State College. All rights reserved.
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LOVEJOY (a) CH 4 + O 2 = CO 2 + H 2 O combustion / CH 4 + 2O 2 CO 2 + 2 H 2 O Reaction and Redox Reaction Oxidizing agent = O 2 Reducing agent = CH 4 (b) HCl + NaOH = H 2 O + NaCl Neutralization / HCl + NaOH → H2O + NaCl correct! (c) H 2 + O 2 = H 2 O Redox reaction / 2H2 + O2 2H2O Oxidizing agent=O2 and reducing agent = H2 yes, and combination (d) CaCO 3 + HCl = CaCl 2 + CO 2 + H 2 O Methathesis (double-displacement) / CaCO3 + 2HCl → CaCl2 + CO2 + H2O correct (e) Zn + HCl = ZnCl 2 + H 2 Displacement + redox / Zn + 2HCl ZnCl 2 + H 2 Oxidizing agent=Zn and reducing agent = H Oxidizing agent = HCl Reducing agent = Zn (f) NaCl = Na + Cl 2 Decomposition / 2NaCl =2Na + Cl 2 and Redox Reaction Oxidizing agent = Na + Reducing agent = Cl - 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 Moles of C2H5OH=29.0g x 1mol C2H5OH/ 46.024g C2H5OH =.630 moles of C2H5OH =.630 mol/.545L = 1.15M correct!
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