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Remaining Time: 58 minutes, 15 seconds. Question Completion Status: A 100.0 ml solution of 0.0600 M...
A 45.0 mL sample of 0.0100 M Cu2+ buffered at pH 8.00 is titrated with 0.0250...
A 45.0 mL sample of 0.0100 M Cu2+ buffered at pH 8.00 is titrated with 0.0250 M EDTA. Calculate pCu2+ at each of the added volumes of EDTA. For the Cu2+−EDTA complex, logKf is 18.78. The fraction of free EDTA in the Y4− form, αY4−, can be found in the table below. 0 mLpCu2+= 6.00 mLpCu2+= 18.0 mLpCu2+= 23.0 mLpCu2+= pH αY4- 0 1.3 x 10-23 1 1.4 x 10-18 2 2.6 x 10-14 3 2.1 x 10-11 4 3.0...
A 30.0 mL sample of 0.0200M Cu2t buffered at pH 9.00 is titrated with 0.0300 M...
A 30.0 mL sample of 0.0200M Cu2t buffered at pH 9.00 is titrated with 0.0300 M EDTA. Calculate pCut at each of the added volumes of EDTA. For the Cu2+-EDTA complex, log Kr is 18.78. The fraction of free EDTA in the Y form, ay can be found in this table. O mL pCu2+ 6.67 mL pCu2+ 20.0 mL pCu2 Values of ay for EDTA a PH ay4- 0 1.3 x 10-23 1 1.4 x 10-18 2 2.6 x 10-14...
Page 243, Practical Question TABLE 12-1 Values of ye for EDTA at 25°C and -0.10 M...
Page 243, Practical Question TABLE 12-1 Values of ye for EDTA at 25°C and -0.10 M pH • Calculate (Ca2+] in 0.10 M CaY2- at pH 8.00, logKf(Ca2+ - EDTA) = 10.65. (answer: 2.3x10-5 M). 1.3 X 10-23 1.4 X 10-18 2.6 X 10-14 2.1 X 10-11 3.0 x 10 2.9 x 10-7 1.8 X 10- 3.8 X 10+ 4.2 x 10 Ca2+ + Y4- Cay2- 0.041 Conditional formation constant [MY-4) K = ayun Ky = m**][**] 0.30 0.81 0.98...
Calculate pFe2+ at each of the points in the titration of 25.00 mL of 0.02219 M...
Calculate pFe2+ at each of the points in the titration of 25.00 mL of 0.02219 M Fe2+ by 0.03689 M EDTA at a pH of 7.00. The values for log K and Qyt can be found in the chempendix. 11.50 mL pFe2+ = the equivalence point, V. pFe2+ = 19.50 mL pFe2+ = Values of dy'. IUI EDI. UL dy4- 1.3 x 10-23 1.4 x 10-18 2.6 x 10-14 2.1 x 10-11 SD 0 Vau WNO 3.0 x 10-9 2.9...
The metal ion Mn+ was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL...
The metal ion Mn+ was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL of 0.0500 M metal ion (Mn+) buffered at a pH of 9.00. a) Calculate the equivalence volume (Veq). b) What is the concentration of the free metal ion at volume V = Veq/2. c) If the conditional formation constant, K’f = 5.4 X 1010, calculate the concentration of free metal ion at volume V = Veq.
Calculate pFe2+ at each of the points in the titration of 25.00 mL of 0.02166 M...
Calculate pFe2+ at each of the points in the titration of 25.00 mL of 0.02166 M Fe2+ by 0.03644 M EDTA at a pH of 5.00. The values for log K and (y can be found in the chempendix. 13.00 mL pFe2+ = the equivalence point, V. pFe? 17.50 mL pFe2+ = Formation Constants > Formation Constants for metal-EDTA Complexes Ion Li+ Ion T13+ Bi3+ log Kf 35.3 27.8a Na+ K+ Be2+ Mg2+ Ca2+ Sr2+ Ba2+ Ce3+ Pr3+ Nd3+ Pm3+...
(10) (3) 4. The metal ion Mnt was titrated with 0.0500 M EDTA. The initial solution...
(10) (3) 4. The metal ion Mnt was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL of 0.0500 M metal ion (M9+) buffered at a pH of 9.00. a) Calculate the equivalence volume (Veg). b) What is the concentration of the free metal ion at volume V = Veg/2. c) If the conditional formation constant, K'r = 5.4 X 1010, calculate the concentration of free metal ion at volume V = Veq. (4)
W ackboard Remaining Time: 23 minutes, 58 seconds. Question Completion Status: QUESTION 15 Which of the...
W ackboard Remaining Time: 23 minutes, 58 seconds. Question Completion Status: QUESTION 15 Which of the following is FALSE for a 0.150 M solution of a weak base with a kb - 1.40x10? The equilibrium concentration of hydroxide ions is 4.58x10-3M The solution would have a pH = 11.66. The equilibrium concentration of the conjugate acid would be 9.16x103 M. The conjugate acid would have a Ka - 7.14x10-11 The equilibrium concentration of the base would be 0.145 M. QUESTION...
X Blackboard → XC O A Remaining Time: 58 minutes, 46 seconds. Question Completion Status: Question...
X Blackboard → XC O A Remaining Time: 58 minutes, 46 seconds. Question Completion Status: Question 3 What is the (min, max) relationship between "Loans Payable" and "Pay Interest", respectively? a. (1, M):(0.1) b. (1, 1); (1,M) (1, 1), (0, M) d. (OM): (1,M) e. None of the above > A Click Submit to complete this assessment
хс Remaining Time: 09 minutes. 53 seconds Question completion Status What is the resistance of an...
хс Remaining Time: 09 minutes. 53 seconds Question completion Status What is the resistance of an Aluminum transmission line of length 3 km that has a cross sectional area of 3.14x10 *m? at temperature of 30°C Table 18.1 Resistivities and Temperature Coefficients at 20°C ρ (Ω-m) Conductors Silver Copper Gold wse the data in the table below Aluminum Tungsten Iron Lead Platinum Manganin Constantan Mercury Nichrome 1.59 x 10-* 1.67 x 10 2.35 x 10- 2.65 x 10- 5.40 x...
A 30.0 mL sample of 0.0200M Cu2t buffered at pH 9.00 is titrated with 0.0300 M EDTA. Calculate pCut at each of the added volumes of EDTA. For the Cu2+-EDTA complex, log Kr is 18.78. The fraction of free EDTA in the Y form, ay can be found in this table. O mL pCu2+ 6.67 mL pCu2+ 20.0 mL pCu2 Values of ay for EDTA a PH ay4- 0 1.3 x 10-23 1 1.4 x 10-18 2 2.6 x 10-14...
Page 243, Practical Question TABLE 12-1 Values of ye for EDTA at 25°C and -0.10 M pH • Calculate (Ca2+] in 0.10 M CaY2- at pH 8.00, logKf(Ca2+ - EDTA) = 10.65. (answer: 2.3x10-5 M). 1.3 X 10-23 1.4 X 10-18 2.6 X 10-14 2.1 X 10-11 3.0 x 10 2.9 x 10-7 1.8 X 10- 3.8 X 10+ 4.2 x 10 Ca2+ + Y4- Cay2- 0.041 Conditional formation constant [MY-4) K = ayun Ky = m**][**] 0.30 0.81 0.98...
Calculate pFe2+ at each of the points in the titration of 25.00 mL of 0.02219 M Fe2+ by 0.03689 M EDTA at a pH of 7.00. The values for log K and Qyt can be found in the chempendix. 11.50 mL pFe2+ = the equivalence point, V. pFe2+ = 19.50 mL pFe2+ = Values of dy'. IUI EDI. UL dy4- 1.3 x 10-23 1.4 x 10-18 2.6 x 10-14 2.1 x 10-11 SD 0 Vau WNO 3.0 x 10-9 2.9...
Calculate pFe2+ at each of the points in the titration of 25.00 mL of 0.02166 M Fe2+ by 0.03644 M EDTA at a pH of 5.00. The values for log K and (y can be found in the chempendix. 13.00 mL pFe2+ = the equivalence point, V. pFe? 17.50 mL pFe2+ = Formation Constants > Formation Constants for metal-EDTA Complexes Ion Li+ Ion T13+ Bi3+ log Kf 35.3 27.8a Na+ K+ Be2+ Mg2+ Ca2+ Sr2+ Ba2+ Ce3+ Pr3+ Nd3+ Pm3+...
(10) (3) 4. The metal ion Mnt was titrated with 0.0500 M EDTA. The initial solution contained 100.0 mL of 0.0500 M metal ion (M9+) buffered at a pH of 9.00. a) Calculate the equivalence volume (Veg). b) What is the concentration of the free metal ion at volume V = Veg/2. c) If the conditional formation constant, K'r = 5.4 X 1010, calculate the concentration of free metal ion at volume V = Veq. (4)
W ackboard Remaining Time: 23 minutes, 58 seconds. Question Completion Status: QUESTION 15 Which of the following is FALSE for a 0.150 M solution of a weak base with a kb - 1.40x10? The equilibrium concentration of hydroxide ions is 4.58x10-3M The solution would have a pH = 11.66. The equilibrium concentration of the conjugate acid would be 9.16x103 M. The conjugate acid would have a Ka - 7.14x10-11 The equilibrium concentration of the base would be 0.145 M. QUESTION...
X Blackboard → XC O A Remaining Time: 58 minutes, 46 seconds. Question Completion Status: Question 3 What is the (min, max) relationship between "Loans Payable" and "Pay Interest", respectively? a. (1, M):(0.1) b. (1, 1); (1,M) (1, 1), (0, M) d. (OM): (1,M) e. None of the above > A Click Submit to complete this assessment
хс Remaining Time: 09 minutes. 53 seconds Question completion Status What is the resistance of an Aluminum transmission line of length 3 km that has a cross sectional area of 3.14x10 *m? at temperature of 30°C Table 18.1 Resistivities and Temperature Coefficients at 20°C ρ (Ω-m) Conductors Silver Copper Gold wse the data in the table below Aluminum Tungsten Iron Lead Platinum Manganin Constantan Mercury Nichrome 1.59 x 10-* 1.67 x 10 2.35 x 10- 2.65 x 10- 5.40 x...
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