PhET Simulation - Sugar and Salt Solutions - Assignment Example

Chemistry Assignment Lecturer: Affiliation: Due PhET Simulation - Sugar and Salt Solutions Part A Partial charges 2. Solute 3. Ions 4. Positively, Negatively 5. CaCl2, NaCl 6. Sugar 7. Partial Charges Sample Exercise 4.1 Part A 2.0 Problem 4.19 Part A HCOOH, H+, HCOO- Part B HCOOH (aq) H+ (aq) + HCOO- (aq) Net Ionic Equations FeCl2 (aq) + 2NaOH (aq) --> Fe (OH) 2(s) + 2NaCl (aq) Fe2+ + 2OH- --> Fe (OH) 2(s) Part B MgSO4 (aq) + Pb(NO3)2(aq) --> PbSO4(s) + Mg(NO3)2(aq) Pb2+ + SO4^2- --> PbSO4(s) Sample Exercise 4.

3 Part A Yes Sample Exercise 4.4 Part A Only barium nitrate precipitates. Problem 4.36 Part A KOH, NH3, HNO3, KClO2, H3PO3, CH3COCH3 Oxidation-Reduction Reactions Part A N2 (g) +3H2 (g) -----> 2NH3 (g) N2 is a redactor and its oxidized in the reaction N2 (0) - 2x3e- ---------> 2N (3+) Part B 3Fe (NO3)2 (aq) +2Al (s) --------> 3 Fe(s) +2 Al (NO3) 3 (aq) Al is a redactor and its oxidized in the reaction Al (0) - 3e- -------->Al (+3) Part C Cl2 (aq) + 2 NaI (aq) ------> I2 (aq) + 2NaCl (aq) I is a redactor and its oxidized in the reaction 2xI (-1) - 2x1e- ---------> I2(0) Part D PbS (s) + 4H2O2 (aq) ------> PbSO4(s) + 4 H2O (l) S is a redactor and its oxidized in the reaction. S (-2) - 8e- ----------> S(+6) Activity Series A: no reaction, B: no reaction, C: reaction, D: reaction Part B Y Q W Z X Sample Exercise 4.

10 Part A Sodium Problem 4.52 ZnCl2 (aq) + 2NaOH (aq) → Zn (OH) 2(s) + 2NaCl (aq) Br2 (l) + 2K(s) → 2KBr(s) CH3CH2OH (l) + 3O2 (g) → 3H2O (l) + 2CO2 (g) Part B P4(s) + 10HClO (aq) + 6H2O (l) → 4H3PO4 (aq) + 10HCl (aq) --- Cl is reduced from +1 to -1 Br2 (l) + 2K(s) → 2KBr(s) --- Br is reduced from zero to -1 CH3CH2OH (l) + 3O2 (g) → 3H2O (l) + 2CO2 (g) --- oxygen is reduced from zero to -2 Part D ZnCl2 (aq) 2NaOH (aq) -> Zn (OH)2(s) +2NaCl(aq)- precipitation Ion Concentration Part A K2S → 2 K {+} + S {2-} (0.15 M K2S) x (2 mol K {+} / 1 mol K2S) = 0.

3 M K {+} Part B It has the same concentration as the ion. Part C 24.59g Sample Exercise 4.13 1.84×10^−3 M Problem 4.68 Mol = mass/molar mass = 0.136/176.12 = 7.72 * 10^-4 Molarity = mol/Volume (L) = 7.72 * 10^-4/0.2588 = 29.82 * 10^-4 = 0.002982 M (mol/L)