Create a 1 page page paper that discusses phet simulation. Chemistry Assignment Lecturer: Affiliation: Due PhET Simulation – Sugar and Salt Solutions Part A Partial charges 2. Solute 3. Ions4. Positiv

  

Create a 1 page page paper that discusses phet simulation. Chemistry Assignment Lecturer: Affiliation: Due PhET Simulation – Sugar and Salt Solutions Part A Partial charges 2. Solute 3. Ions4. Positively, Negatively5. CaCl2, NaCl6. Sugar7. Partial Charges Sample Exercise 4.1Part A2.0 Problem 4.19Part AHCOOH, H+, HCOO-Part BHCOOH (aq) H+ (aq) + HCOO- (aq)Net Ionic EquationsFeCl2 (aq) + 2NaOH (aq) –&gt. Fe (OH) 2(s) + 2NaCl (aq)Fe2+ + 2OH- –&gt. Fe (OH) 2(s)Part BMgSO4 (aq) + Pb(NO3)2(aq) –&gt. PbSO4(s) + Mg(NO3)2(aq)Pb2+ + SO4^2- –&gt. PbSO4(s)Sample Exercise 4.3Part AYes Sample Exercise 4.4Part AOnly barium nitrate precipitates.Problem 4.36Part AKOH, NH3, HNO3, KClO2, H3PO3, CH3COCH3Oxidation-Reduction ReactionsPart AN2 (g) +3H2 (g) —–&gt. 2NH3 (g)N2 is a redactor and its oxidized in the reactionN2 (0) – 2x3e- ———&gt. 2N (3+)Part B3Fe (NO3)2 (aq) +2Al (s) ——–&gt. 3 Fe(s) +2 Al (NO3) 3 (aq)Al is a redactor and its oxidized in the reactionAl (0) – 3e- ——–&gt.Al (+3)Part CCl2 (aq) + 2 NaI (aq) ——&gt. I2 (aq) + 2NaCl (aq)I is a redactor and its oxidized in the reaction2xI (-1) – 2x1e- ———&gt. I2(0)Part DPbS (s) + 4H2O2 (aq) ——&gt. PbSO4(s) + 4 H2O (l)S is a redactor and its oxidized in the reaction.S (-2) – 8e- ———-&gt. S(+6)Activity SeriesA: no reaction, B: no reaction, C: reaction, D: reactionPart BY Q W Z XSample Exercise 4.10Part ASodiumProblem 4.52ZnCl2 (aq) + 2NaOH (aq) → Zn (OH) 2(s) + 2NaCl (aq)Br2 (l) + 2K(s) → 2KBr(s)CH3CH2OH (l) + 3O2 (g) → 3H2O (l) + 2CO2 (g)Part BP4(s) + 10HClO (aq) + 6H2O (l) → 4H3PO4 (aq) + 10HCl (aq) — Cl is reduced from +1 to -1Br2 (l) + 2K(s) → 2KBr(s) — Br is reduced from zero to -1CH3CH2OH (l) + 3O2 (g) → 3H2O (l) + 2CO2 (g) — oxygen is reduced from zero to -2Part DZnCl2 (aq) 2NaOH (aq) -&gt. Zn (OH)2(s) +2NaCl(aq)- precipitationIon ConcentrationPart AK2S → 2 K {+} + S {2- }(0.15 M K2S) x (2 mol K {+} / 1 mol K2S) = 0.3 M K {+}Part BIt has the same concentration as the ion.Part C24.59gSample Exercise 4.131.84×10^−3 MProblem 4.68Mol = mass/molar mass = 0.136/176.12 = 7.72 * 10^-4Molarity = mol/Volume (L) = 7.72 * 10^-4/0.2588 = 29.82 * 10^-4 = 0.