# Thall

1.   Run       [A]        [B]             [C]            Rate
1       1.00 M     1.00 M      1.00 M       0.150  M/s
2       2.00         2.00           2.00           4.80
3       1.50         1.00           1.00           0.338
4       1.50         3.00           1.00           9.13

(a) Write rate equation and describe how orders determined.        (6)
(b) Calculate rate constant.                                                           (3)
(c) Determine rate when [A] = [B] = [C] = 3.00M                       (3)

2.  Reaction first-order in A.                                       (15)
(a) Calculate k if 30.0% A remains after 10.0 min.
(b) Calculate time needed for 95.0% of A to react.
(c) Calculate % A remaining after 17.5 min.

3.  At 300K the rate constant is 0.100 s-1 and at 350K the rate constant is 0.800 s-1.     (10)
(a) Determine activation energy.
(b) Determine rate constant at 400 K.

4.  A2(g) +  B2(g)  2 AB(g)        Kc = 16
3.00 mol A2, 3.00 mol B2, and 3.00 mol AB added to a  1.00 L container.  (8)
Calculate [A2], [B2], and [AB] at equilibrium.

5.   A(g)  +  B(g)  AB(g)      T = 300K
(a) 2.00 mol each A, B, and AB added to a 1.00 L container. Calculate [A], [B], and [AB]
when equilibrium reached (Kc = 1.0).         (8)
(b) Calculate Kp.                                             (4)

6.  2 A(g)  +  2 B(g) A2B2(g) + heat     (Ea = 10 kJ DH = -90 kJ)
(a) State and account for shift when temperature decreased.                (3)
(b) State and account for shift when everything doubled.                       (3)
(c) State and account for shift when volume increased.                        (3)
(d) State and account for shift when [B] doubled and [A2B2] tripled.     (3)
(e) Draw energy profile to scale labeling values for Ea and DH.           (3)
(f) Explain why transition state resembles reactant.                              (3)
(g) Explain why K>1.                                                                        (3)

7.  Provide brief explanation:     (16)
(a) A small increase in temperature results in large increase in rate of reaction.
(b) The concentration of solids & liquids are omitted from equilibrium expression.
(c) For a two step reaction, the rate depends on the slow step.
(d) The initial concentration has no influence on the half-life of a first order reaction.

1. (a)  rate = k[A]2[B]3
runs 1&2 establish 5th order overala
runs 3&4 establish 3rd order in B (if triple [B], rate 27x)
runs 1&3 establish 2nd order in A (increase [A] by 1.5x, rate 2.25x)

(b)  rate = k[A]2[B]3
0.150 M/s = k [M5k = 0.150 M-4s-1

(c)  rate = [0.150][3]5 = 36.5 M/s

2.  (a) ln (A0/At) = kt
ln (100/30) = k(10 min)      k = 0.12 min-1
(b) ln (100/5) = (0.12)t              t = 25 min
(c) ln (100/At) = (.12)((17.5) At = 12.2  or 12.2%

3.  (a) Ea = (R)[(T1T2)/(T2-T1)]ln (k2/k1) = (0.00831)[(350)(300)/(50)]ln 8 = 36.3 kJ/mol

(b) ln (k2/k1) =  (Ea/R)[(T2- T1)/(T1T2)]  = (36.3/.00831)[(400-300)/(300)(400)]  = 3.64
(k2/0.10 s-1) = 38       k2 = 3.8 s-1

4.   A2(g) +  B2(g)  2 AB(g)
Eq:  3-x         3-x           3+2x

(3+2x)/(3-x) = 4       x = 1.50     [A2] = [B2] = 1.50M         [AB] = 6.00M

5.    A  +  B   AB      T = 300K
2-x    2-x      2+x

(a)  (2+x)/(2-x)2 = 1
x2 - 5x +2 = 0          x = 0.44        [A] = [B] = 1.56M      [AB] = 2.44M

(b) Kp = Kc(RT)Dn
= (1)[(0.0821)(300)]-1 = 0.041

6.   (a) Reaction exothermic, SHIFTS RIGHT when Temperature decreased
(b) K =[A2B2]/[A]2[B]2 = 2/(22)(22) = 1/8   SHIFTS RIGHT
(c) If pressure decreased, favors greater # gas molecules, SHIFTS LEFT
(d) Q = 3/(12)(22) = 3/4     SHIFTS RIGHT
(e)

(f) Reactant closer to transition state in terms of energy.
(g) When equilibrium reached, there will be much more product than reactant  (see energy profile).

7. (a) Increasing the temperature dramatically increases the number of effective collisions.
(b) These concentrations are constant and can be represented in K.
(c) The slow step has the higher energy barrier and it is assumed the slow step takes about the same
time as the slow and fast steps combined.
(d) Half-life defined as the time required for half the initial concentration to react. Therefore, it is
independent of initial concentration.

1. Construct table and provide CA/CB (if any): CrO4-2   HC2O4-1   NH4+1   HClO4    PCl3     (10)
2. List theories (Arrenhius, Bronsted-Lowry, or Lewis) that are applicable: (10)
(a) Br + Br Br2              (b) HClO3 + H2 H3O+1 + ClO3-1
(c) Cl+1 + Cl-1 Cl2          (d) H + Cl  HCl           (e) HNO3 + H2 H3S+1 + NO3-1

3. (a) For HAc/Ac-1       Prove Kw = Ka. Kb      (12)
(b) Explain how  HAc/NaAc works as buffer.
(c) Using Ka/Kb, explain why NH4CN is basic in water.

4. Calculate pH resulting from mixing:     (16)
(a) 40 mL of 1.00 M HCl and 60 mL of 1.00 M NaOH.
(b) 50 mL of 1.00 M HCN and 30 mL of 1.00 M NaOH.

5. Calculate pH:    (20)
(a) 3.00 M H3PO4    (b) 10.0 M Na3PO4    (c) 0.100 M NaClO  (d) 0.50M NaHCO3

6. When 2.95 g of C3H9N diluted to exactly 1 liter, pH of 11.40 results.    Calculate Kb     (8)
7.  Calculate pH, [HF] & [F-1] for 0.200 M HF.      (8)

8. Calculate Ka if 1.0M weak acid ionizes 3.0%. (8)

9.  1.00 L of buffer contains 1.00 mol HAc and 0.50 mol NaAc:     (12)
(a) Calculate pH after adding 0.25 mol OH- (assume volume remains 1.00L)
(b) Calculate pH after adding 0.25 mol H+ (assume volume remains 1.00L)

1.           CrO4-2          HC2O4-1      NH4+1       HClO4      PCl3
CA  HCrO4-1      H2C2O4     none             none         HPCl3+1
CB      none              C2O4-2       NH3             ClO4-1       none

2.  (a) none    (b) Arrenhius/BL/Lewis    (c) Lewis     (d) none     (e) BL/Lewis

3. (a)     HAc  H+1  +  Ac-1       Ac-  HAc  +  OH-1
Ka = [H+1][Ac-1]/[HAc]Kb = [HAc][OH-1]/[Ac-1]Ka. K = [H+1][OH-1]  =  Kw
(b)  If additional base added, H+1 combines with OH-1 and shift takes place to the RIGHT
If additional acid added, Ac-1 combines with H+1  and shift takes place to the LEFT

(c)  NH4+1 H+1 + NH3     Ka = 5.6x10-10
CN-1  HCN + OH-1    Kb = 1.6x10-5      Solution basic because  Kb > Ka

4.   (a) In:  0.040 mol   0.060 mol
HCl     +    NaOH   ®NaCl   +   H2O
Eq:     0             0.020 mol
NaOH in excess by 0.020 mol     [OH-1] = 0.020mol/0.10L = 0.20M   pOH = 0.70   pH = 13.30

(b) In:  0.050 mol   0.030 mol
HCN     +    NaOH NaCN   +   H2O
Eq:   0.020 mol        0            0.030 mol

HCN/NaCN buffer now exists
Ka = [H+][CN-]/[HCN] = [H+][0.030mol/0.080L]/[0.020mol/0.80L] = 6.2x10-10
[H+] = 4.1x10-10 M          pH = 9.38

5.  (a)   H3POH+1  +  H2PO4-1      Ka = 7.1x10-3 = x2/3    x = [H+1] = 0.15 M    pH = 0.84
3-x             x              x
(b)   PO4-3HPO4-2   +   OH-1     Kb = 2.22x10-2 = x2/10    x = [OH-1] = 0.47 M     pOH = 0.33
10-x            x                 x                                                                                  pH = 13.67

(c)   ClO-1 HClO   +   OH-1         Kb = 3.3x10-7 = x2/0.1     x = [OH-1] = 1.8x10-4    pOH = 3.74
0.1-x         x               x                                                                                       pH = 10.26

(d)   HCO3-1 H+1  +  CO3-2              Ka = 4.7x10-11
HCO3-1   H2CO +  OH-1          Kb = 2.2x10-8     Since Kb > Ka, solution BASIC
0.5-x ~ 0.5       x               x
x2/0.5 = 2.2x10-8       x = [OH-1] = 1.1x10-4
pOH = 3.96             pH = 10.04

6.   mol B = 2.95/59.0 = 0.0500M    [B] = 0.0500 M     when pH = 11.40,  [OH-1] = 0.0025 M
B   =    BH+1   +   OH-1
0.0475M    0.0025M      0.0025M
Kb = [BH+1][OH-1]/[B] = (0.0025)2/0.0475 = 1.3x10-4

7.     HF   H+1  +  F-1       Ka = 6.8x10-4
0.2-x          x          x

x2/(0.2-x) = 6.8x10-4 = x2/0.2
x = 0.012      pH = 1.92    [F-1] = 0.012M     [HF] = 0.19M

8.   HA   H+1  +  A-1
0.97         0.03    0.03      Ka = [H+1][A-1]/[HA] = (0.03)2/0.97 = 9.3x10-4

9.   (a) Addition of 0.25M OH-1 causes 0.25M HAc to shift RIGHT
at equilibrium:  [HAc] = 0.75M      [Ac-1] = 0.75M
Ka = [H+1][Ac-1]/[HAc] = [H+1][0.75]/[0.75] = 1.8x10-5
[H+1] = 1.8x10-5     pH = 4.74
(b) Addition of 0.25M H+1 causes 0.25M Ac-1 to shift LEFT
at equilibrium:  [HAc] = 1.25M      [Ac-1] = 0.25M
Ka = [H+1][Ac-1]/[HAc] = [H+1][0.25]/[1.25] = 1.8x10-5
[H+1] = 9.0x10-5     pH = 4.05

1.   Calculate solubility for PbI2 in g/L (Ksp = 8.7x10-9)                      (10)
2.   Calculate molar solubility for PbSO4 in 0.50 M Na2SO4           (10)
3.   Calculate Ksp if 0.062 g Ca3(PO4)2soluble in 1.00 L of H2O        (10)
4.   Calculate Ksp for M(OH)3 if saturated solution gives pH of 9.00   (10)
5.   Calculate pH range for separating 1.0M Mg+2 from 1.0M Zn+2. (10)
6.   If 50.0mL 0.020M Ca+2 mixed with 50.0mL 0.030M CrO4-2,   (10)
will precipitate form?   Explain with calculation!
7.  4.00 M Na3PO4 solution has density of 1.20 g/mL.                     (12 )
Calculate:  (a) weight % Na3PO4      (b) molality     (c) mole fraction Na3PO4
8.   The vapor pressure of pure water is exactly 600 torr.                  (10)
Solution consisting of 108 g H2O and 54.0 g of nonvolatile solute displays
vapor pressure of 500 torr. Calculate FW of solute.
9.  Solution prepared using 33.3 g CaCl2 and exactly 400 g H2O.       (10)
Determine boiling & freezing points of this solution.
10. Determine F.P. if 3.00m CaBr2 solution ionizes 80.0%.               (10)

1.  PbI Pb+2  +  2 I-1         K = 8.7x10-9 = [Pb+2][I-1]2 =(x)(2x)2 = 4x3
x            x          2x            x = 0.0013
(0.0013 mol/L)(461 g/mol) = 0.60 g PbI2

2.   PbSOPb+2  +  SO4-2          Ksp = 6.3x10-7 = [Pb+2][SO4-2] = (x)(x+0.50) @ 0.50x
x             x         x+0.50          x = 1.3x10-6 M

3.   Solubilty Ca3(PO4)2 = (0.062g/L)/(310 g/mol) = 2.0x10-4 M
Ca3(PO4)2 3 Ca+2     +    2 PO4-3
2.0x10-4 M         6.0x10-4 M     4.0x10-4 M         Ksp = [Ca+2]3[PO4-3]2 = [6.0x10-4]3[4.0x10-4]2 =3.5x10-17

4.  When pH=9.00,  [OH-1] = 1.0x10-5
M(OH)3 M+3    +    3 OH-1
3.3x10-6 M    1.0x10-5 M                      Ksp = [M+3][OH-1]3 =  [3.3x10-6][1.0x10-5]3 = 3.3x10-21

5.   Mg(OH)2(s)  Mg+2    +   2  OH-1
1.0M                      x M
Mg(OH)2   Ksp = 7.1x10-12 = [Mg+2][OH-1]2
7.1x10-12 = [1.0][OH-1]2
[OH-1] = 2.7x10-6 M     pOH = 5.58    pH =8.42   pH range equal or less than 8.42

6.  After mixing:  [Ca+2] = 0.010M        [CrO4-2] = 0.015M         Ksp for CaCrO4 = 7.1x10-4
Ion Product = [Ca+2] [CrO4-2] = [0.010] [0.015] = 1.5x10-4
Since IP < Ksp, NO PPT forms!

7.   Assume 1.00 L or 1200 g total mass
contains 4.00 mol Na3PO4 or 656 g Na3PO4   and  544g H2O (1200 - 656)
(a)  % Na3PO4 = (656/1200)x100 = 54.7
(b)  m = 4.00mol/0.544kg = 7.35
(c)  mol H2O = 544/18 = 30.2       mol fraction Na3PO4 = 4.00/34.2 = 0.117

8.      Psol  =  Pw Xw
500 torr = (600 torr)Xw     Xw = 0.833         nw = 108/18 = 6.00      ns = 54.0/FW

Xw = nw/(nw + ns) = 6.00/(6.00 + 54.0/FW) = 0.833        FW =  45.0 g/mol

9.   mol CaCl2 = 33.3/111 = 0.300          m = 0.300/0.400kg = 0.750
DTb = imKb = 3(0.750)(0.512) = 1.15     BP = 101.2°C
DTf = imK = 3(0.750)(-1.86) = -4.19    FP = -4.2° C

10.  Assume 100 particle CaBr2       CaBr2 Ca+2   +   2Br-1
20              80                160   =  260       i = 260/100 = 2.60

DTf= imKf = (2.60)(3.00)(1.86) = 14.5    F.P. = normal F.P. -  DTf = -14.5°C

1. Use DG/DH/DS to explain:                                                              (12)
(a) Water does not solidify at 10°C.
(b) Salts dissolve more readily in hot water than in cold water.
(c) Endothermic crystal formation.

2. Determine temperatures range needed to become spontaneous.       (9)
Provide brief explanation!
(a) DH = 0 kJ            (b) DH = -0.500 kJ          (c) DH = +300 kJ
DS = -20 J/K          DS = -500 J/K              DS = +300 J/K

3. CaS(s) + 2 O2(g) ® CaSO4(s)                                                     (16)
CaS(s):    DHf = -482 kJ/mol     DGf = -477 kJ/mol
CaSO4(s) DHf = -1434 kJ/mol     DGf = -1321 kJ/mol
Determine: (a) DHrxn  (b) DGrxn  (c) DSrxn (d) Is reaction favored at high/low temp? Explain!

4.     NH4Cl(s) ®   NH3(g)   +   HCl(g)                                              (16)
NH3:       S° = 192 J/K/mol       DHf° = -46 kJ/mol
HCl:        S° = 187 J/K/mol        DHf° = -92 kJ/mol
NH4Cl:    S° = 95 J/K/mol        DHf° = -314 kJ/mol
Determine:     (a) DS°rxn      (b) DH°rxn     (c) DG°rxn     (d) K

5.  Using Tables 20.3/20.4, estimate enthalpy of formation for N2H4.   (10)

6. Write balanced ox/red equations and determine E° net:     (10)
(a) AuBr3   ®   Au   +   Br2             (b) SnCl2   +   Zn   ®   ZnCl2   +   Sn

7. Write balanced anode/cathode/overall equations for electrolysis:  (10)
(a) LiNO3(aq)  (produce H2/O2)            (b) KBr(aq)   (produce H2)

8. (a) Determine mass Ni obtained if Ni+2 subjected to 3.50 A for 35.0 minutes.  (10)
(b) Determine minutes required for 15.0 g Cu+2 to deposit if current 4.00 A.

9. Design galvanic cell using Al/Ag electrodes. Indicate anode, cathode, direction of     (8)
current, and cell solutions. Provide balanced anode/cathode equations and calculate potential.

1. (a) when water solidifies:  DH  negative (favorable)      DS  negative  (not favorable)
water does not solidify at 10°C because TDS>DH
(b)  DH ? DS positive  (favorable)
because DS favorable, want TDS as large as possible
(c)  DH positive DS negative  for endothermic crystal formation
endothermic crystal formation not possible!

2.  (a) Never (b) T<1K   (c)T>1000K

3. (a)DH = -DHCaS  +  DHCaSO4  = -(-482kJ)  +  (-1434kJ)  =  -952 kJ
(b)  DG = -(-477kJ)  +  (-1321kJ) =  -844kJ
(c) DS = [DH - DG]/T  =  [-952kJ - (-844kJ)]/298K = -0.362kJ/K = -362J/K
(d) Since enthalpy driven, favored at low temperature

4. (a)  DS = -95J/K + 192J/K + 187J/K = +284J/K
(b)  DH = +314kJ + (-464kJ) + (-92kJ) = +176kJ
(c)  DG = DH - TDS = 176kJ - (298K)(0.284kJ/K) = +91.4kJ
(d)  ln K = -DG/RT = (-91.4)/[(0.00831)(298)] = -36.9      K = 9.4x10-17

5.  DHf(N)  =   2(473 kJ)              form 4 N-H bonds   4(-388 kJ)
DHf(H)  =   4(218 kJ)              form 1 N-N bond        -240kJ
+1818 kJ                                                -1792kJ
DH = +1818kJ - 1792kJ = 26kJ

6.  (a) Ox: 6 Br-1® 3Br2 + 6e-1   -1.08 V     (b) Ox: Zn°® Zn+2 + 2e-1      +0.76 V
Red: 2 Au+3 + 6e-1 ® 2Au° +1.50 V         Red: Sn+2 + 2e-1 ® Sn°     -0.14 V
+0.42 V                                                      +0.62 V

7. (a) anode: 2H2O ® 4H+1 + O2 + 4e-1                (b) anode: 2Br-1 ® Br2 + 2e-1
cathode: 4H2O + 4e-1 ® 4OH-1 + 2H2              cathode: 2H2O + 2e-1® 2OH-1 +  H2
overall: 2H2O ® 2H2 + O2                           overall: 2H2O + 2Br-1 ®  H2 + Br2+ 2OH-1

8.   (a)  Ni+2  +  2e-1   ®  Ni°                                Q = I x t = (3.50)(2100) = 7350 coul = 0.0762F
0.0762F     0.0381 mol (2.24g Ni)

(b)  Cu+2  +  2e-1   ®  Cu°                  t = Q/I = 45500/4.00 = 11380 s = 190 min
0.472F      0.236 mol
or 45500 coul

9.

1. Reaction first-order in A.       (12)
Calculate:    (a) Rate constant k if 65.0% A remains after 6.00 min.
(b) Time needed for 75.0% of A to react.  (c) % A remaining after 25.0 min.  (d) t½

2.   A2(g) + B2(g) 2 AB(g)
(a) 10.0 mol A2 and 10.0 mol B2, added to a 1.00 L container. At equilibrium 5.00 mol A   (4)
and 5.00 mol B2 remain. Calculate equilibrium constant K.
(b) If 5.00 mole AB added to the system already at equilibrium in part (a),      (8)
determine [A2], [B2], and [AB] when new equilibrium reached.

3.   Calculate pH:   (a) 0.100 M NaClO    (b) 1.00 M NH4CN      (12)

4. If 1.00 L of a buffer contains 1.00 mol HAc and 0.500 mol NaAc, calculate:    (15)
(a) pH of buffer.
(b) pH after adding 0.30 mol NaOH to the buffer (assume volume remains 1.00L).
(c) pH after adding 0.30 mol HCl to the buffer (assume volume remains 1.00L).

5. Calculate Ksp for Pb(OH)2 if 0.00100 g soluble in 1.00 L of water.      (10)

6.  Solution prepared from 95.3g MgCl2 and 450g H2O.  Calculate:          (12)
(a) MgCl2 molality  (b) solution freezing point   (c) solution vapor pressure (pure water 600 torr)

7.           SnCl4(g) + 2 H2O(g) ® SnO2(s) + 4 HCl(g)         (15)

DHf         (-472)       (-286)          (-581)          (-92)     ( kJ/mol)
DGf         (-432)       (-237)          (-520)          (-95)      ( kJ/mol)

Determine: (a) DH°rxn     (b) DG°rxn    (c) DS°rxn     (d) K     (e) account for DS°

8. Design galvanic cell using Zn/Ni electrodes. Indicate anode, cathode, direction of current, and cell   (12)
solutions. Provide balanced anode/cathode equations and calculate potential.

9. (a) Write balanced ox/red equations and determine overall E° for CuCl2 ® Cu + Cl2
(b) If CuCl2 subjected to 8.0 amp for 5 hours, calculate grams of Cu obtained.

1. (a)  ln(A0/A) = kt
ln 1.54 = 0.43 = k(6.00 min)         k = 0.072 min-1
(b) ln(A0/A) = kt
ln (100/25) = ln 4 = 1.39 = (0.072 min-1)t     t = 19.3 min
(c) ln(A0/A) = kt
ln(100/A) = (0.072 min-1)(25.0 min) = 1.80
100/A = 6.05    A = 16.5 or 16.5%
(d) t½ = 0.693/.072min-1 = 9.63 min

2.  (a)      A2 + B22AB
Eq:  5M     5M        10M       K = 102/52 = 4.0

(b)      A2 + B22AB
Eq:  5+x    5+x      15-2x      K = (15-2x)2/(5+x)2 = 4   or  (15-2x)/(5+x) = 2
15-2x = 10+ 2x      x = 1.25
[A2] = [B2] = 6.15M     [AB] = 12.5M

3.    (a) ClO-1HClO   +  OH-1
Eq:  0.1-x           x              x           Kb = 4.0x10-8 = x2/0.1    x = [OH-1] = 2.0x10-4
pOH = 3.40     pH = 10.60

(b)  CN-1HCN   +  OH-1       Kb = 2.0x10-5      soluton basic (Kb > Ka)
NH4+1NH3  +  H+1                  Ka = 5.6x10-10
CN-1HCN   +  OH-1
Eq:   1-x           x              x           Kb = 2.0x10-5 = x2/1    x = [OH-1] = 4.5x10-3
pOH = 11.650   pH = 2.35

4.  (a)  [H+1] = (Ka)([HAc])/[AC-1] = (1.8x10-5)(1.00M)/(0.500M) = 3.6x10-5  pH =4.44

In:  1.00M                  0.50M
(b)     HAc H+1   +  Ac-1           [H+1] = (Ka)([HAc])/[AC-1] = (1.8x10-5)(0.70M)/(0.80M) = 1.6x10-5
Eq:  0.70M                  0.80M                                                                                               pH =4.80

(c)   [H+1] = (Ka)([HAc])/[AC-1] = (1.8x10-5)(1.30M)/(0.20M) = 1.2x10-4     pH = 3.93

5.     [Pb(OH)2] = 0.00100/241 = 4.1x10-6 M                       Pb(OH)2  Pb+2    +   2OH-1
4.1x10-6 M      4.1x10-6 M    8.2x10-6 M
Ksp = [Pb+2][OH-1]2 = (4.1x10-6 M)(8.2x10-6 M)2 = 2.8x10-16

6.  (a) mol MgCl2 = 1.00    m = 1.00mol/0.450kg = 2.22m
(b) DTf = imkf = (3)(2.22)(-1.86) = -12.4°C
(c) mol ions = 3.00    mol H2O = 25.0     XH2O = 25/28 = 0.89
P =(XH2O)(PH2O)= (0.89)(600 torr) =  534 torr

7.  (a) DH° = +85 kJ   (b) DG° = +6 kJ   (c) DS° = +265 J/K   (d) K = 0.089   (e) 3 gas particles ® 4 gas particles

8.

9.  (a)  RED:  Cu+2  +  2e-1 ® Cu°    +0.34 V
OX: 2Cl-1® Cl2  + 2e-1        -1.36 V        E° = -1.02 V

(b)  Q = I x t = (8.0A)(5)(3600s) = 144,000 C = 1.5 F         Cu+2 + 2e-1® Cu°
1.5F    0.75 mol    48g Cu°