Solutions

Solutions

CHAPTER 1 (a) 10.73 ft3 psi lb-mol−1 ◦ R−1 , (b) 0.082056 atm L mol−1 K−1 0.67 Pa 2.8 atm (a) 71 µm, (b) 246 µm (a) Your mass on the Moon is the same as on Earth—72.6 kg. The scale would read a “weight” of one-sixth that: 12.1 kg. Force is the equivalent of weight in SI with the unit of N. Therefore, your “weight” on the Moon would be 119 N, (b) 72.6 kg 1.11 (a) a mass fraction of 23.6 %, (b) 273 K, (c) 12.7 kJ mol−1 , (d) 320 ng g−1 , (e) 0.001 23 mg or 1.23 µg, (f) 9500 kt or 9.5 Gg, (g) 4.2 MPa, (h) P V = nRT , (i) 20.1 V ± 0.3 V or (20.1 ± 0.3) V 1.1 1.3 1.5 1.7 1.9

CHAPTER 2 2.1 (a) 3.2 × 102 (320), (b) 15.3, (c) 1530, (d) 0.26 2.3 We reject the last data point 10. μ = 28.5, σ = 9.9, z = −1.87, P (z < −1.87) = 0.5 + 0.469 = 0.969, α = 0.03 2.5 Assuming a normal distribution, P = 0.034 2.7 V = 0.04 mV for α = 0.05 and 0.015 mV for α = 0.01. 2.9 (a) μ = 110, σ 2 = 1200., σ = 40, (b) no 2.11 PO2 = 0.150 atm, O2 = 0.001 atm 2.13 t = 23 min, t = 2 min 2.15 (a) VC = 50.9 mL, (b) E = 0.05, (c) I = 0.01, (d) Reading errors and not respecting significant figures 2.17 (a) μ = 160, σ = 30, (b) 17 %, (c) Nube can reject 90 min 2.19 η = 25 mPa s, η = 1 mPa s 2.21 57 min 385

386 Solutions

2.23 (a) μ = 20, σ = 3, (b) Yes, we can reject 12 because:     xm − μ 12 − 20.28 P =P = P (−2.46) = 2(0.493) = 0.986 σ 3.36 and 1 − 0.986 <

1 . 2(10)

(c) μ= ρ−ρo =

(7850 − 850) 13 · 4π  



1.5 100

3

· 9.81 · 20.28

1.5 6π 100 · 0.4

= 170 Pa s

2ρ + 2ρo = 4 kg

 2  ρ−ρo 2 t + ρ − ρo t  2    3.36 2 4 = + = 0.16 7000 20.28

μ = μ



μ = 30 Pa s

CHAPTER 3 3.1 (a) Five, one and one, (b) Although the model has different values for the exponents, these are not strictly fitted coefficients since they are not allowed to vary. The equation is nonlinear with four fitted parameters. 3.3 (a) H0 : μ1 − μ2 = 0 H1 : μ1 − μ2 = 0. (b) Assuming that the standard deviation of the rice is the same, then df = 2n − 2 = 18 and the pooled sample standard deviation is σx¯ 2 −x¯ 2 = 1 2   2 2 2 2 s1 /n + s2 /n = 0.17 /10 + 0.25 /10 = 0.31 X¯ 1 − X¯ 2 − t (α, df )σx¯ 2 −x¯ 2 < μ1 − μ2 < X¯ 1 − X¯ 2 + t (α, df )σx¯ 2 −x¯ 2 1

2

1

2

6.90 − 6.44 − 2.1 · 0.31 < μ1 − μ2 < 6.90 − 6.44 + 2.1 · .31 − 0.19 < μ1 − μ2 < 0.19 Thus we cannot reject the null hypothesis and so they are equally as long.

Solutions 387

TABLE QS3.5A Eight Experiments Using Two-Level Full Factorial Design −1 −1 −1 −1 1 1 1 1

−1 −1 1 1 −1 −1 1 1

−1 1 −1 1 −1 1 −1 1

TABLE QS3.5B Eight Experiments Using Three-Level Full Factorial Design −1 −1 −1 0 0 1 1 1

−1 0 1 −1 1 −1 0 1

1 0 −1 0 0 −1 0 1

3.5 (a) See Table QS3.5A. (b) See Table QS3.5B. (c) There is a confounding factor for (a) with the second and third models. The columns corresponding to β0 and β11 X12 are parallel. There is also a confounding factor for (b) with the first and third models. The columns corresponding to β3 X3 and β12 X1 X2 are parallel. (d) Part (a) with the third model and the base case presents no confounding. Part (b) with the third model and the base case does present confounding. 3.7 (a) There will be four parameters and the model will be of the following form: E = a1 + a2 T + a3 F + b. (b) A minimum of four experiments are required to obtain the four parameters. Eight experiments are needed for a full factorial design. It is outlined in Table QS3.7. (c) The new model will have six parameters and be of the following form: E = a1 + a2 T + a3 F + a12 T + a13 F + b.

388 Solutions

TABLE QS3.7 Full Factorial Design for the Four Parameters Exp

E

T

F

1

1

1

1

2

1

1

−1

3

1

−1

1

4

1

−1

−1

5

−1

1

1

6

−1

1

−1

7

−1

−1

1

8

−1

−1

−1

TABLE QS3.13 Factorial Design for the Relationship between Temperature, Pressure, and Quantity of Solvent to Produce Pesticide Exp

X1

X2

1

1

1

2

1

−1

3

−1

1

4

−1

−1

(d) A minimum of six experiments must be performed to obtain the six parameters. 25 experiments are required for a full factorial design, which is 32. 3.9 P = 140 + 0.905Lt − 45.1Ct − 0.187Lt Ct R 2 = 0.9889 3.11 (a) The resistance depends on the concentration. (b) 20 %. (c) No 3.13 (a) See Table QS3.13. (b) ⎛ ⎞ 1 1 1 1 ⎜ ⎟ 1 −1 −1 ⎟ ⎜ 1

=⎜ ⎟. ⎝ 1 −1 1 −1 ⎠ 1 −1 −1 1 (c) Y = θ

Solutions 389

⎛ ⎜ ⎜ ⎜ ⎝

Y (1) Y (2) Y (3) Y (4)

⎞

⎛

⎟ ⎜ ⎟ ⎜ ⎟=⎜ ⎠ ⎝

1 1 1 1

1 1 −1 −1

1 −1 1 −1

1 −1 −1 1

⎞⎛ ⎟⎜ ⎟⎜ ⎟⎜ ⎠⎝

β0 β1 β2 β12

⎞ ⎟ ⎟ ⎟ ⎠

3.15 (a) Null hypothesis: Ho : μ1 = μ2 = μ3 = μ4 , and the alternative hypothesis is that the means differ. (b) Yes. (c) Yes 3.17 (a) α = 2.8, (b) R 2 = 0.993 3.19 Emf = 1.53 + 0.0436T R 2 = 0.972 Emf = 0.045T R 2 = 0.94 Emf = 0.0365T + 0.0000416 R 2 = 0.9993 3.21 (a) (b) (c) (d) (e)

β0 439 495 423 271 191

β1 Ug −40 −43 −35 −38 −26

β2 Gs

β2 Ug Gs

0.28 0.4

−0.02

R2 0.98 0.90 0.57 0.992 0.9927

CHAPTER 4 4.1 270 kg = 0.100 m3 2700 kg m−3   0.010 2 XA,wire = π = 0.00785 m2 2 V=

V 0.100 m3 = = 12.7 m XA 0.00785 m2  1/3 0.1 m3 Lcube = = 0.464 m 0.00785 m2 Lwire =

XA,cube = (0.464 m)2

390 Solutions

From Pouillet’s law: ρAl L L = σAl · XA XA Rwire = 46.2 µ R=

Rcube = 0.0616 µ 4.2 R= ρ=

50. mV V = = 51 m I 0.97 A

RXA = 0.0240 µ m or L

σ = 41. × 106 S m−1 .

The metal is most likely lead. 4.3 EY = σ/ F 400 N σ= = π = 5.09 MPa 2 2 XA 4 (0.01) m =

L 0.00001 m = L0 0.2 m EY = 10.2 GPa

4.4 Assuming the thermal expansion is linear over the temperature range and for a block ρρo = VVo , L = [1 + kT (T − T0 )] L0   1/3  ρ ρVo 1/3 = Lo L= ρ ρ   7700 1/3 = 1 m = 1.00239 m 7765 1.00239/1.00 − 1 = 2.99 × 10−5 ◦ C−1 kT = 100 − 20 4.5 Converting the given resistance to resistivity:   RXA 5 π4 (0.001 m)2 = = 7.854 × 10−6 m ρ= L 0.50 m At 75 ◦ C:

Solutions 391

ρ = ρ0 [1 + α(T − T0 )]   = 7.854 × 10−6 1 + 2.52 × 10−3 (75 − 25) = 8.84 × 10−6 m L 0.50 m R=ρ = 8.84 × 10−6 m π = 5.61 . 2 XA 4 (0.001 m) Correcting for XA and L as the wire elongates with the increased T :   L = 0.50 1 + 8.42 × 10−5 (75 − 25) = 0.502 m   D = 0.001 1 + 8.42 × 10−5 (75 − 25) = 0.001004 m π XA = (0.001004 m)2 = 7.92 × 10−7 m2 . 4 From Pouillet’s law, the resistance of the expanded wire is: R=

8.84 × 10−6 · 0.502 = 5.61 m. 7.92 × 10−7

Since temperature dilates the gauge so little, its effect on the resistance is negligible. 4.6 kT = 8.00 × 10−6 ◦ C−1 , T0 = 0 ◦ C ρ = ρ0 [1 + kT (T − T0 )] ρ −6 ◦ C−1 (300 − 0) = 0.003 ρ0 = 1 + 8.00 × 10

−4 = FG1R0 R = 0.003 4 = 4.12 × 10 4.7 RT = 200 + 400 = 600 Vi = 0.020 A · 600 = 12 V. 4.8 F = 0.1 kg 9.81 m s2 = 0.981 N

σ = EY =

F XA XA = EY F = 5 × 108 Pa · 0.0001 · 0.981 N = 1.96 × 10−5 m2   XA 1/2 D=2 = 5 mm π

CHAPTER 5 5.1 1n 1 2 2 ˜ Mvrms P = nMv rms = 3 3V

392 Solutions 2 Mvrms = 3kN T 1n · 3kN T P= 3V P V = nkN T = nRT

R = kN 5.3 5.5 5.7 5.9 5.11 5.13 5.15 5.17 5.19 5.21 5.23 5.24

P = 0.32 atm, PO2 = 0.067 atm 126.6 kPa (a) 771 mbar, (b) 20.2 m, (c) 0.16 m, (d) 5.6 mm P = 1.1 MPa, P = 1800 Pa (a) PA = 0.637 atm, PB = 2.81 atm, PC = 0.877 atm, PD = 16.0 atm, (b) P = 1.08 atm 18.74 kPa (a) 2 atm, (b) The Nonette, (c) 85 m, (d) Same height t = 0.41 mm (a) H = 1.96 kPa, H = 6.13 kPa, H = 10.3 kPa, (b) Glycerol 0.0326 m (a) 0.007 mm, (b) 20. kPa, (c) 2.0 m, (d) 0.030 mm, (e) 0.06 − 6 %@ FS (a) z=

1 − (P /Po )1/5.26 = 1045 m (1050 m). 0.0000226

(1)

(b) P = Po = 0.5 mmHg  1/2 2  2 ∂z ∂z P + z = P = 0.5 mmHg ∂P ∂Po o 1/2  z = 6.92 + 6.92 = ±10 m. (c) z= (d)

1 − (711/760)1/5.26 = 557 m (560 m). 0.0000226

√ P = t (α, n − 1) s/ n = 2.26 · 2.8 = 6.3 mmHg ∂z 0.19 P =  1.19 · 6.3 = 86.5 m ∂P 0.0000226 · 711 627 711 1/2  z = 86.52 + 86.52 = ±120 m.

(e) Yes, we can.

(2)

Solutions 393

TABLE QS6.17 Absolute and Relative Errors for Each Device (T and wT in ◦ C, relative error in %) Exp. therm.

Type T

Type K

T

wT

Rel. err.

T

wT

Rel. err.

T

wT

Rel. err.

21

0.5

2

21.4

0.1

0.6

21.4

0.1

0.6

25

0.5

2

26.0

0.1

0.5

25.7

0.1

0.5

29

0.5

2

29.2

0.1

0.4

29.7

0.1

0.4

33

0.5

2

33.4

0.1

0.4

35.0

0.1

0.3

39

0.5

1

38.3

0.1

0.3

38.5

0.1

0.3

42

0.5

1

42.4

0.1

0.3

42.5

0.2

0.6

47

0.5

1

46.9

0.1

0.2

46.5

0.1

0.3

51

0.5

1

50.4

0.1

0.2

49.9

0.1

0.2

CHAPTER 6 6.1 6.3 6.5 6.7 6.9

(a) T = 45.4 ◦ C, (b) P2 = 173.79 kPa (a) Toluene, (b) Galinstan, (c) Toluene, 10X (a) No, (b) 2.822 mV t (a) r = 23 (α2 −α1 )(T −T0 ) , (b) 2.43 

6.11 6.13 6.15 6.17

R R



2 =

1 R 0 R0

2



a + 2bT + T 1 + aT + bT 2

2

−0.1275 + 0.0415T + 1.226 × 10−7 T 2 , R 2 = 0.99994 J: 0.663 mV, T: 0.513 mV, K: 0.519 mV, and J: 8 ◦ C, T: 8.25 ◦ C, K: 8.5 ◦ C (a) 12.183 mV, (b) 205.7 ◦ F (96.5 ◦ C), (c) 13.145 mV (a) T: T = 24.01E + 25.03, K: T = 24.41E + 24.99, (b) See Table QS6.17, (c) Assuming the experiment was well conducted, the thermocouples should have a greater error (wT ) than calculated in (b).

CHAPTER 7 7.1 (a) P = 21 Pa, (b) Q = 5.40 L3 s−1 , (c) Q = 1.3 % 7.3 Pipe 1: 106 L min−1 at STP, pipe 2: 139 L min−1 at STP. The flow rate is higher for the second pipe. 7.5 (a) 657 Pa, (b) Turbulent, (c) 1 % 1/2   ρb b 7.7 Q = XA · um = XA C2gV − 1 ρf D Ab 1/2   2·9.81·4.2×10−9 3970 −6 Q = 25.1 × 10 − 1 0.1·3.14159×10−6 1.17

394 Solutions

7.9 7.11 7.13 7.15 7.17 7.19 7.21

Q = 45 L min−1 m ˙ = 53 g min−1 u = 6 m s−1 , m ˙ = 88 kg s−1 0.051 m3 s−1 (a) 27.4 % of the height, (b) Steel: 43 % of the height, tantalum: 30 % of the height 0.55 % ∼ 6 % Pdyn = 0.45 mbar, u1 = 0.0166 m s−1 14 m s−1 (a) Above NRe = 20 000 the orifice coefficient Co is 0.61. NRe =

ρud m ˙ =4 μ πdo μ

m ˙ min = 0.00300 kg s−1 . (b) m ˙ max = 6m ˙ min = 0.0180 kg s−1 1 P = 2ρ



 1/2 2 m ˙ 1 − β4 = 71 500 Pa. Co X˙ A2

(3)

(c) The sensitivity increases by a factor of 17 but the energy loss increases and the range is much smaller. (d) P =

 2 1 0.018(1 − 0.54 )1/2 = 29 500 Pa. 2 · 4 0.95 · 0.0000378

(4)

(e) (Q /Qmin ) = 0.07 (Q /Q)max = 0.002. (f) P1 α −4 − β 4 = P2 1 − β4

(5)

CHAPTER 8 8.1 (a) 0.010 m s−1 , (b) u = 0.0006 m s−1 , (c) 17 800 kg m−3 , (d) 17.1 karats, (e) Yes, (f) Take more data points, prolong the time the falling ball falls, use a more viscous oil. 8.3 (a) q T =k A x kT x = q/A

Solutions 395



x



   q/A 2 T 2 + T q/A  2  0.2 = 0.052 + + 0.012 · 10 cm = 0.53 cm. 15

x = x

k k

2



+

◦ (b) T2 = 2 Aq x k = 2T1 = 30 C. (c) The uncertainty of the temperature measurement is incorrect, or the heat flux is 2.2 or 10 % instead of 1 %. −1 K−1 and 1 8.5 k2 = − TTmm −T −T2 k1 , so k2 = 3.5 W m

   2  2  2   ∂k2 ∂k2 ∂k2  T 1 + T 2 + T m + k2 = ∂T1 ∂T2 ∂Tm

(6)

∂k2 k1 = ∂T1 Tm − T2 ∂k2 T1 − Tm = k1 ∂T2 (Tm − T2 )2   Tm − T1 ∂k2 1 = −k1 − ∂Tm (Tm − T2 )2 Tm − T2 k2 = 0.09 W m−1 K−1 8.7 (a) Ti − T o Q=  Ri 

Ri =

  1 ln (ro /ri ) ln ro,ss /ri,ss 1 + + + 2πkr L 2πkss L 2πro,ss Lho 2πri Lhi

The heat transfer resistance across the metal wall is negligible and since we have the exterior wall temperature, we calculate the overall heat loss considering the inside convection term and the term across the insulation. Q=

960 − 125 = 912 W m−1 . 0.716 + 0.199

(b) Q = ho SA T 912 = 7.6 W m−2 K−1 . ho = π(0.36 + 0.019)(125 − 25)

396 Solutions

(c) 1 = 0.115 W m−1 K−1 2πro ho Q = ho SA T 912 = 7.6 W m−2 K−1 . ho = π(0.36 + 0.019)(125 − 25) Rho =

(d) 508 ◦ C (e) kl = 0.005 % 8.9 Sapphire. 8.11 (a) Air, (b) T, F, F, (c) T, T, T, (d) 1C, 2A, 3B, 4D, (e) F, T, F, F, (f) gases, liquids, plastics, alloys, metals, (g) F, T, F (4540−970)·2·0.012 ·9.81 0 )V g 8.13 (a) 13.3, 0.54, (b) μ = (ρ−ρ ·13.3 = 25.9 Pa s. 6πrL t = 9·0.4  (c)  = k(α)σ = 2.57 · 0.54 = 1.4 s, (d) Wμ /μ = (Wρ /ρ)2 + (Wt /t)2  = (20/3570)2 + (1.4/13.4)2 = 0.10 = 10 %, Wμ = 0.1·25.9 = 2.6 Pa s (e) X − μ 14.1 − 13.3 = = 1.48 σ 0.54 P (Z < 1.48) = 0.431

a=

5(1 − 2 · 0.431) = 0.691 > 0.5, therefore accept. t is 13.4 s, 12.9 s, 14.1 s, 12.7 s, and 13.2 s.

CHAPTER 9 9.1 (a) 836 plates, (b) 0.066 cm, (c) 1.57, (d) 1502 plates, (e) 84.61 cm 9.3 (a), (c), (e) (but separation will be poorer because the contact time will be shorter 9.5 (a) 11.3 − 3.45 = 2.27 3.45 11.63 − 3.45 ‘ knh = = 2.37 3.45 12.1 − 3.45 ‘ kmcp = = 2.51. 3.45

‘ k3mp =

(b) u = L/to = 4.35 cm min−1 The equivalent diameter of the particle is Vcone = 3(b · h) = (πr 2 h)/3 = (πdp2 h)/12  dp = 12Vπhcone = 1.63 mm

Solutions 397

φ = SAsphere /SAcone = 0.90 We calculate the HETP from van Deemter’s equation HETP = 2φdp +

8k ‘ df2 2φDgas u = 3.8 mm. ++ 2 u π (1 + k ‘ )2 Dliq

(7)

(c) Nth = L/HETP = 15/0.38 = 40 plates. (d) Nth = 5.54(tr /w1/2 )2 w1/2 = 

w1/2

tr Nth 5.54

= 0.612 min

tr w1/2 = 2.35 √ Nth    2  2  2.35 2.35t R =  + −  1/2 tR 3/2 Nt h (Nth 2Nth

w1/2 = 0.612 ± 0.003 min. 9.7 (a) t0 = 0.8 min, t1 = 3.2 min, t2 = 6 min, t3 = 18 min, k1 = 3, k2 = 6.5, k3 = 21.5, (b) Nth = 5.54(tr /w1/2 )2 , so Nth,1 = 5700, Nth,2 = 3200, Nth,3 = 55 400, (c) R = 9.4, the minimum value is 1.25 although 1.5 is accepted, shorten the column, increase the temperature, increase the linear velocity of the mobile phase, and decrease the amount of stationary phase. (d) SS = 5 % 9.9 (a) Reduce the analysis time. (b) Solute diffusion is faster than in N2 , so the balance between the mobile phase and the stationary phase is reached more quickly. For DCT it is easier to detect higher resolution. Conductivity. (c) Detectable: alcane, organic; undetectable: oxygen, nitrogen, carbon dioxide, etc. (d) 8 or 9, k  is 13.5, 18, and 21, α1,2 = 1.33, α1,3 = 1.56, α2,3 = 1.17, R2,1 = 1.89 > 1.25 so it is well resolved, R3,2 = 1.13 < 1.25 so it is not well resolved N2

9.11 (a) Nth = 2544, 4th = 12.61, L = 13 cm, (b) k  = 1.5, (c) R = 7.41 > 1.5, there is separation.

CHAPTER 10 10.1 0.64 (S = π4 dp2 ) 10.3 H = 1.08, H = 0.02, φ = 1 10.5 Vs = π6 dp3 Vpy = 13 h3 SA,sp = πdp2    √ Spy = h2 + 4 12 h h2 + (h/2)2 = h2 (1 + 5)

k k  +1

= 0.6,

398 Solutions

Vs = Vpy 1 3 π 3 h 6 dp = 3 dp = 3 π2 h πd 2 |v

=v

2/3 2

√ h = 0.72 φ = sp Spsp p = π(2/π) (1+ 5)h2 10.7 One—sphericity refers to the outer surface only 10.9 1.69 mm2 , 0.847 mm 9 i=1 dp,i dN 10.11 (a) dp = 13.62 µm, (b) d¯N L =  = 18.12 µm, (c) d¯SV = 9 9

3 i=1 dp,i dN 9 2 d i=1 p,i dN

10.13 10.15 10.17 10.19 10.21

i=1

dN

= 22.83 µm, (d) No

(a) d50 = 250 µm, (b) DN [3, 2] = 210 µm, DN [1, 0] = 161 µm dnl = 0.136, dns = 0.137, dnv = 0.139, dsv = 0.143 ut = 0.242 m s−1 , ut = 0.099 m s−1 , ut = 0.506 m s−1 (a) 105 µm, (b) iii, (c) i, (d) ii, (e) ii, (f) dp,eq = 0.968dp,s , φ = 0.794 (a) D[2, 0] = 2.24 mm, (b) P = 122 kPa, (c) The definition of the mean diameter should be based on the drag. There are many small particles. The gas velocity is 2.0 m s−1 at the inlet, but the average speed is higher. The temperature gradient is important, therefore the gas velocity is higher. The uncertainty of the variables is greater than estimated. (d) P = 0.24

CHAPTER 11 11.1 (a) 2858 mL, 221.3 mL min−1 , 52.5 min, 0.982, (b) 18.23 ppm. The concentration is too high. The desulfurization unit must be revised to improve process efficiency. 11.2 (a) C = 0.0016 mol L−1 , (b) C  = 0.0008 mol L−1 11.3 See Fig. QS11.3 11.4 The Structure FWHM is: π = 2.18 × 10−3 β = 0.225◦ − 0.1◦ = 0.125◦ · 180

FIGURE QS11.3 IR spectra solution.

Solutions 399

According to Scherrer’s equation, D=

Kλ 0.91 · 1.5406 = 65.8 nm = β cos θ 2.18 · 10−3 · cos 24.562 2

So the crystal size of the sample is 65.8 nm. 11.5 See Fig. QS11.5 11.6 We have identified each of the peaks together with position in the chemical structure (Fig. QS11.6). For example, beginning from left to right, the broad peak at 7.6 ppm (labeled 10) belongs to hydroxyl group on the aromatic ring. The singlet at 7.5 ppm (labeled 3) belongs to the shielded hydrogen bonding to the amide. The doublet at 7.3 ppm (labeled 5 and 7) are two equivalent protons in the aromatic ring with a lower electronic cloud. The doublet at 6.7 ppm (labeled 6 and 8) are the two remaining protons on the aromatic ring. They are further to the right (compared to protons 5 and 7) because the hydroxyl group (OH) pulls less electronic charge from than the amide group (NHCO). Finally, the singlet at 2.1 ppm (labeled 11) integrates three protons and belongs to the methyl group.

FIGURE QS11.5 Theoretical paracetamol NMR spectrum.

FIGURE QS11.6 Theoretical paracetamol NMR spectrum.

400 Solutions

FIGURE QS11.7 Theoretical d-limonene NMR spectrum.

11.7 Each of the peaks we have identified according to their chemical shift in the d-limonene molecule (Fig. QS11.7). The triplet peak at 5.4 ppm (labeled 3) represents one proton with two vicinal protons on the ring. The quintuplet at 2.35 ppm (labeled 7) represents one proton with 4 vicinal protons on the ring. The hydrogens on the methyl groups are in the beta position so they do not contribute to the multiplicity of the signal. The singlets at 1.75 ppm and 1.65 ppm each belong to the methyl groups labeled 10 and 1. The shielding increases the chemical shift of 10 with respect to 1. The peaks labeled 6 and 6 correspond to the diasterotopic protons in the aliphatic ring. The multiplicity reflects three vicinal protons plus two that are in the beta position (long range). The signal split in two sextets because each proton is non-equivalent and sees a different chemical environment. The broad signals from 2.2 ppm to 1.9 ppm (labeled 5 , 4 , 4 and 5 ) represent many overlapping protons. By exclusion, we assign them to the remaining diasterotopic protons in positions 4 and 5. 11.8 The fluid is water with n3 = 1.33, γ = 28◦ n1 sin α = n2 sin β c c → = 1.33 v3 = n3 v3 n2 sin β = n3 sin γ n2 sin β γ = sin−1 n3 ◦ = 28