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. 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) --&gt. Fe (OH) 2(s) + 2NaCl (aq)

Fe2+ + 2OH- --&gt. Fe (OH) 2(s)

Part B

MgSO4 (aq) + Pb(NO3)2(aq) --&gt. PbSO4(s) + Mg(NO3)2(aq)

Pb2+ + SO4^2- --&gt. 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) -----&gt. 2NH3 (g)

N2 is a redactor and its oxidized in the reaction

N2 (0) - 2x3e- ---------&gt. 2N (3+)

Part B

3Fe (NO3)2 (aq) +2Al (s) --------&gt. 3 Fe(s) +2 Al (NO3) 3 (aq)

Al is a redactor and its oxidized in the reaction

Al (0) - 3e- --------&gt.Al (+3)

Part C

Cl2 (aq) + 2 NaI (aq) ------&gt. I2 (aq) + 2NaCl (aq)

I is a redactor and its oxidized in the reaction

2xI (-1) - 2x1e- ---------&gt. I2(0)

Part D

PbS (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 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) -&gt. 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.