TestBankChapter17
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TestBankChapter17

Course Number: CHEM 105BLG, Spring 2008

College/University: USC

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General Chemistry, 9e (Petrucci) Chapter 17 Additional Aspects of Acid-Base Equilibria 1) The common ion in a mixture of a weak base and a strong base is the hydronium ion. Answer: True F Diff: Answer: 2 Refere nce: Sectio n 17 T True F Diff: 2hsoluti Reference: eon Section 17 chan 1 Tr pges H slight ly owith f addit ion of aa small bamou unt of f acid f or ebase. Answer: True F Diff: 2 and...

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Chemistry, General 9e (Petrucci) Chapter 17 Additional Aspects of Acid-Base Equilibria 1) The common ion in a mixture of a weak base and a strong base is the hydronium ion. Register to View Answer F Diff: Answer: 2 Refere nce: Sectio n 17 T True F Diff: 2hsoluti Reference: eon Section 17 chan 1 Tr pges H slight ly owith f addit ion of aa small bamou unt of f acid f or ebase. Register to View Answer F Diff: 2 and Reference: sits Section 17 t conju 2 A id r gate obase nwill gform a abuffe cr. Register to View Answer F Diff: Answer: 2 Refere nce: Sectio n 17 T True F Diff: 2hrange Reference: eof Section most 17.2 Tange cacidobase l indic oators r is 1 pH cunit. h Register to View Answer F Diff: Answer: 1 Refere nce: Sectio n 17 A True F Diff: 1nsoluti Reference:of on Section 17 t HCN 3 Ia h(aq) ewith NaO t H(aq) i , the t equiv r alenc ae t point i occur os at a npH great oer f than 7. Register to View Answer F Diff: 1oion, Reference: r the Section end 17.4 F titrat apoint nneed s to amatc ch the cequiv ualenc re apoint t. e Register to View Answer F Diff: 1 Reference: Section 17 4 A acid-strong base will produce a longer vertical section of a weak titration curve than will a strong acid-strong base. A n s w e r : True F Diff: Answer: 1 Refere nce: Sectio n 17 A True F Diff: 1 proti Reference: sc acid Section 17 asuch 5 A poly l as t NaH CO3 ocann f ot acid aas an acid. Register to View Answer F Diff: 1 Refere nce: Sectio n 17 H It will lower the pH. B) The pH will C) not change. The solution D) becomes hotter. The pH cannot be measured. E) It will raise the pH. Register to View AnswerDiff: 1 Reference: Section 17-1 14) How will addition of sodium chloride affect the pH of a HCl solution? A) It will lower B) the pH. The pH will C) not change. The solution D) becomes hotter. The pH cannot be measured. E) It will raise the pH. Register to View AnswerDiff: 1 Reference: Section 17-1 15) What is the [H3O+] of a solution measured to be 0.20 M in sodium acetate and 0.40 M in acetic acid? [Ka = 1.8 10-5] A) 1.8 10-5 B) 9.0 10-6 C) 3.6 10-5 D) 7.2 10-5 E) 4.7 Register to View AnswerDiff: 1 Reference: Section 17-1 16) What is the concentration of the acetate ion of a solution measured to be 0.20 M acetic acid and 0.20 M in hydrochloric acid? [Ka for acetic acid = 1.8 10-5] A) 3.6 10-5 B) 9.0 10-6 C) 1.8 10-5 D) 7.2 10-5 E) 0.20 Register to View AnswerDiff: 1 Reference: Section 17-1 17) Ten of 0.10 M NH3(aq) (K = 1.8 10-5) is mixed with 10 mL of 0.10 M NH4Cl. Neglecting the milliliters differences between activities and concentrations, the resulting solution: A) has a pH = B) 4.74 has a [H+] of C) approximately 10-3 M has a [NH4+] greater than that of the NH4Cl(aq) D) has an [OH-] E) of 1.8 10-5 M is acidic Register to View AnswerDiff: 1 Reference: Section 17-1 18) If some NH4Cl is added to an aqueous solution of NH3: A) the pH of the solution will increase B) the pH of the solution will decrease C) the solution D) will not have pH the pH of the solution will not change E) NH4Cl cannot be added to NH3 Register to View AnswerDiff: 1 Reference: Section 17-1 19) In 0.100 M HC2H3O2(aq), [H3O+(aq)] = [C2H3O2-(aq)] = 1.3 x 10-3 M. If a few drops of concentrated HCl(aq) are added to this solution, the C2H3O2-(aq) concentration is A) < 1.3 x 10-3 M. B) > 1.3 x 10-3 M. C) = 1.3 x 10-3 M. D) 0.100 M. Register to View AnswerDiff: 2 Reference: Section 17.1 20) Which of the following would NOT be a good buffer pair? A) ammonium B) chloride and ammonium hydroxide sodium C) chloride and sodium hydroxide boric acid and sodium borate D) potassium E) carbonate and potassium bicarbonate All of these are good buffer pairs. Register to View AnswerDiff: 1 Reference: Section 17-2 21) A weak acid (Ka = 10-5) and the sodium salt of its anion in solution in equimolar amounts has: A) pH > 7 B) pH < 7 C) pH = 7 D) depends only on concentration E) none of these Register to View AnswerDiff: 1 Reference: Section 17-2 22) Which of the following would NOT be considered a buffer solution? A) 0.1 M B) HC2H3O2/0.1 M NaC2H3O2 0.1 M NH3/0.1 M NH4Cl C) 0.1 M D) H2SO3/0.1 M NaHSO3 0.1 M E) HNO3/0.1 M NaNO3 0.1 M H3PO4/0.1 M NaH2PO4 Register to View AnswerDiff: 1 Reference: Section 17-2 23) Which of following dilutions will cause the pH of a 0.20 M NH3/0.02 M NH4+ buffer to drop to below the pH 7.00? A) 1 : 10 B) 1 : 100 C) 1 : 1,000 D) 1 : 10,000 E) none of these Register to View AnswerDiff: 2 Reference: Section 17-2 24) The Hasselbach equation, used to calculate the pH of simple conjugate-pair buffer systems, would Henders be expressed for an ammonia/ammonium chloride buffer, for which K (NH ) is 1.8 10-5, as: b 3 onA) pH = 4.74 + B) log([NH3]/[NH4+]) pH = 4.74 + C) log([NH4+]/[NH3]) pH = 9.25 + D) log([NH3]/[NH4+]) pH = 9.25 + E) log([NH4+]/[NH3]) pH = 14.0 - log(1.8 10-5) Register to View AnswerDiff: 2 Reference: Section 17-2 25) What is the buffer range (for an effective 2.0 pH unit) for a benzoic acid/sodium benzoate buffer? [Ka for benzoic acid is 6.3 10-5] A) 8.8 - 10.8 B) 7.4 - 9.4 C) 5.3 - 7.3 D) 4.7 - 6.7 E) 3.2 - 5.2 Register to View AnswerDiff: 3 Reference: Section 17-2 26) Phenol red indicator changes from yellow to red in the pH range from 6.6 to 8.0. State what color the indicator will assume in the following solution: 0.20 M KOH A) red B) yellow C) red-yellow D) mixture The indicator is its original color. E) There is not enough information to answer this question. Register to View AnswerDiff: 2 Reference: Section 17-3 27) Phenol red indicator changes from yellow to red in the pH range from 6.6 to 8.0. State what color the indicator will assume in the following solution: 0.1 M NaCl A) red B) yellow C) red-yellow D) mixture The indicator is its original color. E) There is not enough information to answer this question. Register to View AnswerDiff: 2 Reference: Section 17-3 28) Phenol red indicator changes from yellow to red in the pH range from 6.6 to 8.0. State what color the indicator will assume in the following solution: 0.10 M NH4NO3 A) red B) yellow C) red-yellow D) mixture The indicator is its original color. E) There is not enough information to answer this question. Register to View AnswerDiff: 2 Reference: Section 17-3 29) Phenol red indicator changes from yellow to red in the pH range from 6.6 to 8.0. State what color the indicator will assume in the following solution: 0.10 M HC2H3O2 A) red B) yellow C) red-yellow D) mixture The indicator is its original color. E) There is not enough information to answer this question. Register to View AnswerDiff: 2 Reference: Section 17-3 30) What factor governs the selection of an indicator for a neutralization titration? A) the final B) volume of the solution the volume of titrant C) the molarity of the standard solution D) the pH at the stoichiometric (equivalence) point E) the solubility of the indicator Register to View AnswerDiff: 2 Reference: Section 17-3 31) Phenolph thalein may be used as an indicator for the titration of: A) a weak base B) with a strong acid a weak acid C) with a weak base any acid and base D) a weak acid E) with a strong base phenolphthalei n cannot be used as an indicator Register to View AnswerDiff: 2 Reference: Section 17-3 32) Which of following is NOT typical of a titration curve for the titration of a strong acid by a strong base? the A) The beginning pH is low. B) The pH change is slow until near the equivalence point. C) At the D) equivalence point, pH changes by a large value. Beyond the E) equivalence point, pH again rises slowly. The equivalence point would be at a pH less than 3.5. Register to View AnswerDiff: 2 Reference: Section 17-3 33) Phenol red indicator changes from yellow to red in the pH range from 6.6 to 8.0. State what color the indicator will assume in the following solution: 0.10 M NaCN A) red B) yellow C) red-yellow D) mixture The indicator is its original color. E) There is not enough information to answer this question. Register to View AnswerDiff: 2 Reference: Section 17-3 Use the information below to answer the following question(s). methyl orange: red at pH < 3.1: orange at pH 3.1-4.4: yellow-orange above pH 4.4 litmus: red at pH < 4.5: purple at pH 4.5-8.3: blue above pH 8.3 thymol blue: yellow at pH < 8.0: green at pH 8.0-9.6: blue above pH 9.6 trinitrobenzene: colorless at pH < 12: yellow at pH 12.0-1: orange above pH 14.0 34) Which of indicators from the following table would be most appropriate for the titration of 0.30 M acetic the pH acid (Ka = 1.8 10-5) with 0.15 M sodium hydroxide? A) methyl orange B) litmus C) thymol blue D) trinitrobenzen e E) none of these Register to View AnswerDiff: 3 Reference: Section 17-3 35) Which of indicators from the following table would be most appropriate for the titration of 0.30 M the pH hydrochloric acid with 0.15 M sodium hydroxide? A) methyl orange B) litmus C) thymol blue D) trinitrobenzen e E) none Register to View AnswerDiff: 3 Reference: Section 17-3 36) Which of indicators from the following table would be most appropriate for the titration of 0.30 M the pH hydroxlamine (Kb = 9.1 10-9) with 0.15 M hydrochloric acid? A) methyl orange B) litmus C) thymol blue D) trinitrobenzen e E) none Register to View AnswerDiff: 3 Reference: Section 17-3 37) Which of following is NOT typical of a titration curve for the titration of a weak acid by a strong base? the A) There is an B) initial increase in the pH as the titration begins. Over a long C) section of the curve before the equivalence point, pH changes slightly. At the halfD) neutralization point, pH = pKa for the acid used. The steep part of the curve, near the equivalence point, is over a large pH range. E) The equivalence point would be at a pH greater than 7.0. Register to View AnswerDiff: 3 Reference: Section 17-3 38) Which condition characterizes the stoichiometric point of a neutralization titration? A) Equivalent B) amounts of acid and base have reacted. The pH is C) exactly 7.0 The indicator changes color. D) A slight excess of titrant is present. E) A slight excess of indicator is present. Register to View AnswerDiff: 1 Reference: Section 17-4 39) Choose the correct statement. A) 30 mL of 2 B) molar H3PO4 will exactly react with 15 mL of a 2 molar NaOH solution. One liter of 1 molar HCl will exactly neutralize 2 liters of 0.5 molar NaOH. C) A 1 molal D) solution always contains exactly 1 mole in a liter of solution. One liter of a 1 of an acid always exactly neutralizes one liter of a one molar solution of a base. molar solution E) A 1 molar solution requires 2 moles of H2SO4 per liter of solution. Register to View AnswerDiff: 1 Reference: Section 17-4 40) The solution that is added from the burret during a titration is the: A) buffer B) titrant C) indicator D) base E) none of these Register to View AnswerDiff: 1 Reference: Section 17-4 41) In the of 50.0 mL of 0.0200 M C6H5COOH(aq) with 0.100 M KOH(aq), what is/are the major species in titration the solution after the addition of 5.0 mL of NaOH(aq)? A) C6H5COOH, B) C6H5COO-, and Na+ C6H5COOH C) C6H5COOD) and Na+ C6H5COOH, OH-, and Na+ Register to View AnswerDiff: 2 Reference: Section 17.4 42) For the determine whether the solution at the equivalence point is acidic, basic or neutral and why: following HCl is titrated with NH3(aq) titration, A) acidic because of hydrolysis of NH4+ B) basic because C) of hydrolysis of NH3 acidic because of hydrolysis of ClD) acidic because of hydrolysis of HCl E) neutral salt of strong acid and strong base Register to View AnswerDiff: 2 Reference: Section 17-4 43) For the titration, determine whether the solution at the equivalence point is acidic, basic or neutral and following why: KOH is titrated with HI(aq) A) basic because of hydrolysis of K+ B) basic because of hydrolysis of KOH C) acidic because of hydrolysis of OHD) acidic because of hydrolysis of HI E) neutral salt of strong acid and strong base Register to View AnswerDiff: 2 Reference: Section 17-4 44) The titration curve for 10.0 mL of 0.100 M H3PO4(a q) with 0.100 M NaOH(a q) is given below. Estimate the pKa2 of H3PO4. A) 7.2 B) 4.8 C) 9.8 D) 2.2 Register to View AnswerDiff: 2 Reference: Section 17.4 45) In a solution labeled "0.010 M H2SO4(aq)," the pH is A) ~1.9. B) = 2.0. C) ~2.2. D) = 1.0. Register to View AnswerDiff: 2 Reference: Section 17.4 46) In the ation of 50.0 mL of 0.1 M BOH (a weak base with Kb = 1.6 10-7) with 0.10 M H2SO4, the neutraliz most correct description of the solution at the mid-point of the titration, i.e., half neutralized, is: A) a solution in B) which [B+] essentially equals [BOH] a solution in C) which [B+] essentially equals 4 10-4 M a solution is 75 mL and contains some undissociated BOH molecules and some B+, OH-, and whose volume HSO4- ions D) a solution E) containing SO42- and B2+ ions a solution in contact with BSO4 solid Register to View AnswerDiff: 2 Reference: Section 17-4 47) Choose the expression that gives the molar concentration of a H2SO4 solution if 24.3 mL of a 0.105 M NaOH solution is required to titrate 60 mL of the acid: A) (60 B) 24.3)/0.105 (60 2)/(24.3 0.105) C) (24.3 D) 0.105)/(60) (24.3 E) 0.105)/(60 2) (60 0.105)/(2 24.3) Register to View AnswerDiff: 2 Reference: Section 17-4 48) Why is it NOT practical to titrate ammonia with acetic acid? A) The change in pH near the end point is not large enough to be accurately detected. B) There is no C) known indicator which changes color at the right pH. The reaction is too slow. D) There is not E) enough difference between the ionization constants of ammonia and acetic acid. Ammonia does not react with acetic acid. Register to View AnswerDiff: 2 Reference: Section 17-4 49) A unknown acid had a pH of 3.70. Titration of a 25.0 ml aliquot of the acid solution required 21.7 solution ml of 0.104 M sodium hydroxide for complete reaction. Assuming that the acid is monoprotic, of an what is its ionization constant? A) 9.0 10-2 B) 2.0 10-4 C) 4.4 10-7 D) 3.6 10-9 E) 2.7 10-11 Register to View AnswerDiff: Reference: 2 Section 17-4 50) For the titration, determine whether the solution at the equivalence point is acidic, basic or neutral and following why: nahco3(aq) titrated with NaOH(aq) A) basic because of excess OHB) acidic because of hydrolysis of HCO3C) acidic because of hydrolysis of Na+ D) neutral salt of strong acid and strong base E) basic because of hydrolysis of CO32- Register to View AnswerDiff: 3 Reference: Section 17-4 51) A 0.10 M solution of Na2HPO4(aq) has a pH of A) 9.79. B) 4.68. C) 7.20. D) 12.38. Register to View AnswerDiff: 2 Reference: Section 17.5 52) In a prepared by mixing equal volumes of 0.20 M acetic acid and 0.20 M hydrobromic acid, the solution common ion is: A) BrB) C2H3O22C) H2C2H3O2 D) H2Br+ E) H3O+ Register to View AnswerDiff: 1 Reference: Section 17-1 53) Determin of the following solution. Initial concentrations are given. e the pH [HC2H3O2] = 0.250 M, [HCl] = 0.120 M [Ka = 1.8 10-5] A) 0.60 B) 0.92 C) 0.43 D) 4.74 E) 13.08 Register to View AnswerDiff: 1 Reference: Section 17-1 54) Determin [C2H3O2-] of the following solution. Initial concentrations are given. e the [HC2H3O2] = 0.250 M, [HI] = 0.120 M [Ka = 1.8 10-5] A) 8.6 10-6 B) 3.8 10-5 C) 1.8 10-5 D) 0.25 E) 0.37 Register to View AnswerDiff: 1 Reference: Section 17-1 55) Determin of the following solution. Initial concentrations are given. e the pH [HF] = 1.296 M, [HCl] = 1.045 M, Ka for HF is 6.6 10-4 A) 14 B) 3.1 C) 3.2 D) -0.019 E) 0.60 Register to View AnswerDiff: 1 Reference: Section 17-1 56) Calculate a 1.00 L solution of 0.100 M NH3(aq) after the addition of 0.010 mol HCl(g). For NH3, pKb = the pH of 4.74. A) 10.21 B) 9.26 C) 8.31 D) 11.46 Register to View AnswerDiff: 2 Reference: Section 17.1 57) Determin of the following solution. Initial concentrations are given. e the [F-] [HF] = 1.296 M, [NaF] = 1.045 M, Ka for HF is 6.6 10-4 A) 1.046 B) 2.344 C) 5.3 10-4 D) 8.2 10-4 E) 0.251 Register to View AnswerDiff: 2 Reference: Section 17-1 58) Calculate a 1.00 L solution of 0.100 M NH3(aq) after the addition of 0.010 mol NH4Cl(s). For NH3, pKb = the pH of 4.74. A) 10.26 B) 9.26 C) 8.26 D) 11.56 Register to View AnswerDiff: 2 Reference: Section 17.1 59) What is a solution prepared by mixing equal volumes of 0.10 M hydrochloric and 0.1 M hydrofluoric the pH of acid? [K for HF is 6.6 10-4] a A) 1.0 B) 1.3 C) 1.6 D) 2.2 E) 5.0 Register to View AnswerDiff: 2 Reference: Section 17-1 60) Assumin volume change on mixing, what mass of ammonium chloride should be added to 250.0 mL of g no 0.25 M ammonia to produce a solution of pH 10.70? [Kb for ammonia is 1.8 10-5] A) 3.7 mg B) 120 mg C) 40 mg D) 30 mg E) 80 mg Register to View AnswerDiff: 3 Reference: Section 17-1 61) Determin of the following solution. Initial concentrations are given. e the pH [HC2H3O2] = 0.250 M, [NaC2H3O2] = 0.120 M [Ka = 1.8 10-5] A) 5.1 B) 4.4 C) 8.9 D) 9.6 E) 7.0 Register to View AnswerDiff: 1 Reference: Section 17-2 62) Determin of the following solution. Initial concentrations are given. e the pH [HF] = 1.296 M, [NaF] = 1.045 M, Ka for HF is 6.6 10-4 A) 10.73 B) 3.27 C) 3.18 D) 3.09 E) 11.91 Register to View AnswerDiff: 1 Reference: Section 17-2 63) A weak acid has Ka = 4.2 10-3. If [A-] = 2.0 M, what must [HA] be so that [H+] = 2.1 10-3 M? A) 1.5 M B) 2.0 M C) 1.0 M D) 0.5 M E) none of these Register to View AnswerDiff: 2 Reference: Section 17-2 64) A weak acid has Ka = 1.00 10-3. If [HA] = 1.00 M what must be [A-] for the pH to be 2.7? A) 0.50 M B) 2.0 M C) 2.7 M D) 0.37 M E) none of these Register to View AnswerDiff: 2 Reference: Section 17-2 65) A buffer by adding 2.4 g of ammonium nitrate to 100.0 mL of 0.30 M ammonia (K = 1.8 10-5). To this b was solution was then added 10.0 mL of 0.30 M sodium hydroxide, which caused a pH change of: prepared A) 3.0 pH units B) 0.30 pH units C) 0.90 pH units D) 0.09 pH units E) zero Register to View AnswerDiff: 3 Reference: Section 17-2 66) A pH 4.88 buffer was prepared by dissolving 0.10 mol of benzoic acid (Ka = 6.3 10-5) and 0.50 mol of sodium benzoate in sufficient pure water to form a 1.00 L solution. To a 70.0 mL aliquot of this solution was added 2.00 mL of 2.00 M HI. What was the pH of the new 72.0 mL solution? A) 2.84 B) 3.16 C) 3.36 D) 4.64 E) 4.88 Register to View AnswerDiff: 3 Reference: Section 17-2 67) The of a solution may be defined as the number of moles of H+ that will change the pH of 1.00 L of buffer the buffer by 1.00 pH units. What is the buffer capacity of a solution which is 0.10 M in acetic capacity acid (K = 1.8 10-5) and 0.30 M in sodium acetate, in units of mol (H+) per liter? a A) 0.20 B) 0.10 C) 0.087 D) 0.009 E) zero Register to View AnswerDiff: 3 Reference: Section 17-2 68) For HClO2, Ka = 1.2 10-2. What is the pH of a solution in which [NaClO2] = 0.193 M and [HClO2] = 0.203 M? A) 3.32 B) 0.11 C) 1.92 D) 1.94 E) 1.90 Register to View AnswerDiff: 1 Reference: Section 17-2 69) An acid = 1 10-6. At what pH would this acid and its corresponding salt make a good buffer? has a Ka A) 6 B) 7 C) 8 D) 5 E) 4 Register to View AnswerDiff: 1 Reference: Section 17-2 70) A has [HC7H5O2] = 0.100 M and [Ca(C7H5O2)2] = 0.200 M. Ka = 6.3 10-5 for HC7H5O2. The solution solution volume is 5.00 L. What is the pH of this solution after 5.00 ml 10.0 M HCl is added? A) 4.80 B) 4.86 C) 4.65 D) 4.70 E) 4.75 Register to View AnswerDiff: 2 Reference: Section 17-2 71) A [HC7H5O2] = 0.100 M and [Ca(C7H5O2)2] = 0.200 M. Ka = 6.3 10-5 for HC7H5O2. The solution solution volume is 5.00 L. What is the pH of the solution after 10.00 ml of 5.00 M NaOH is has added? A) 4.80 B) 4.86 C) 4.65 D) 4.70 E) 4.75 Register to View AnswerDiff: 2 Reference: Section 17-2 72) What is a buffer solution prepared by dissolving 25.5 g NaC2H3O2 in a sufficient volume of 0.550 M the pH of HC H O to make 500.0 mL of buffer? 232 A) 4.74 B) 4.68 C) 4.91 D) 4.57 E) 480 Register to View AnswerDiff: 2 Reference: Section 17-2 73) A k states that to prepare a particular buffer solution mix 63.0 mL of 0.200 M HC2H3O2 with 37.0 handboo mL of 0.200 M NaC2H3O2. What is the pH of this buffer? (Ka = 1.8 10-5) A) 4.74 B) 4.51 C) 4.98 D) 5.33 E) 7.00 Register to View AnswerDiff: 2 Reference: Section 17-2 74) What is solution made by dissolving 2.16 g of sodium benzoate (NaC6H5CO2) in a sufficient volume of the pH of 0.033 M benzoic acid solution to prepare 500.0 mL of buffer? [Ka for benzoic acid is 6.3 10-5] a A) 4.23 B) 4.37 C) 4.64 D) 5.77 E) 6.30 Register to View AnswerDiff: 2 Reference: Section 17-2 75) If 30.0 mmol HCl(g) is added to 1.00 L of a buffer that is 0.340 M NH3(aq) and 0.290 M NH4Cl(aq), what are the final concentrations of NH3(aq) and NH4Cl(aq), respectively? Assume no volume change. A) 0.310 M and B) 0.320 M 0.310M and C) 0.290 M 0.370 M and D) 0.290 M 0.370 M and 0.320 M Register to View AnswerDiff: 2 Reference: Section 17.2 76) What is the pH of a buffer that is 0.88 M HCN(aq) and 0.53 M NaCN(aq)? The Ka of HCN is 6.2 x 10-10. A) 8.99 B) 9.21 C) 9.43 D) 4.79 Register to View AnswerDiff: 2 Reference: Section 17.2 77) A buffer M C6H5COOH(aq) and 0.282 M Na(C6H5COO)(aq). Calculate the pH after the addition of is 0.282 0.150 moles of nitric acid to 1.0 L of the buffer. For C H COOH, pK = 4.20. 65 a A) 3.69 B) 4.20 C) 4.71 D) 3.87 Register to View AnswerDiff: 2 Reference: Section 17.2 78) All of the following will result in the production of a buffer except A) mixing equal B) volumes of 0.10 M NaC2H3O2(aq) and 0.10 M HCl(aq). mixing equal C) volumes of 0.10 M NaC2H3O2(aq) and 0.050 M HCl(aq). mixing equal D) volumes of 0.10 M NaC2H3O2(aq) and 0.10 M HC2H3O2(aq). mixing equal volumes of 0.10 M HC2H3O2(aq) and 0.050 M NaOH(aq). Register to View AnswerDiff: 2 Reference: Section 17.2 79) How of 0.200 M acetic acid are mixed with 13.2 mL of 0.200 M sodium acetate to give a buffer with many mL pH = 4.2? A) 37 mL B) 18 mL C) 3.8 mL D) 13 mL E) 46 mL Register to View AnswerDiff: 2 Reference: Section 17-2 80) What mass of sodium acetate should be dissolved in 250.0 mL of 0.30 M acetic acid to form a buffer of pH 5.0? [Ka for acetic acid is 1.8 10-5] A) 11 g B) 8.0 g C) 7.5 g D) 5.0 g E) 1.4 g Register to View AnswerDiff: 2 Reference: Section 17-2 81) What is the change in pH after addition of 10.0 mL of 1.0 M sodium hydroxide to 90.0 mL of a 1.0 M NH3/1.0 M NH4+ buffer? [Kb for ammonia is 1.8 10-5] A) 1.0 pH unit B) 0.1 pH unit C) 0.01 pH unit D) 0.001 pH unit E) zero Register to View AnswerDiff: 3 Reference: Section 17-2 82) A pH 9.56 buffer was prepared by mixing 2.00 moles of ammonia (Kb for ammonia is 1.8 10-5) and 1.00 mol of ammonium chloride in water to form a solution with a volume of 1.00 L. To a 200.0 mL aliquot of this solution was added 10.0 mL of 10.0 M sodium hydroxide. What was the resulting pH? A) 9.28 B) 9.56 C) 9.95 D) 10.50 E) 13.7 Register to View AnswerDiff: 3 Reference: Section 17-2 83) A k states that to prepare a particular buffer solution mix 39.0 mL of 0.20 M NaH2PO4 with 61.0 handboo mL of Na HPO . What will be the pH of this buffer? 2 4 A) 7.2 B) 7.4 C) 7.0 D) 6.8 E) 6.6 Register to View AnswerDiff: 3 Reference: Section 17-2 84) The ds are available as 0.10 M aqueous solutions: pyridine (pKb = 8.82), triethylamine (pKb = following 3.25), HClO , NaOH, phenol (pK = 9.96), HClO (pK = 7.54), and NH (pK = 4.74). Identify 4 a a 3 b compoun two solutions that could be used to prepare a buffer with a pH of approximately 5. A) pyridine and B) HClO4 triethyamine C) and HClO4 phenol and D) NaOH HClO and NaOH Register to View AnswerDiff: 3 Reference: Section 17.2 85) Twenty- five milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH at equivalence? A) 7.0 B) 6.2 C) 7.5 D) 8.6 E) It cannot be determined Register to View AnswerDiff: 1 Reference: Section 17-4 86) What will the pH at the neutralization point of 0.00812 M Ba(OH)2 be when titrated with HCl? A) 7.0 B) 12.2 C) 8.0 D) 9.0 E) 6.0 Register to View AnswerDiff: 1 Reference: Section 17-4 87) What volume in mL of 2.0 M H2SO4 is needed to neutralize 7.8 g of Al(OH)3(78.0 g/mol)? Al2(SO4)3 and water are the titration products. A) 150 B) 75 C) 300 D) 500 E) none of these Register to View AnswerDiff: 1 Reference: Section 17-4 88) Twenty- milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH when 30 ml of NaOH five have been added? A) 2.0 B) 12.0 C) 12.2 D) 1.8 E) 12.3 Register to View AnswerDiff: 2 Reference: Section 17-4 89) 25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH before any NaOH is added? Ka for acetic acid = 1.8 10-5. A) 1.0 B) 5.7 C) 2.9 D) 4.3 E) 2.2 Register to View AnswerDiff: 2 Reference: Section 17-4 90) What is the equivalence point for the titration of 0.20 M nitrous acid by 0.20 M sodium hydroxide? [Ka the pH at for nitrous acid is 4.5 10-4] A) 5.83 B) 7.00 C) 8.17 D) 9.00 E) 10.67 Register to View AnswerDiff: 2 Reference: Section 17-4 91) Determin of the following solution. Initial concentrations are given. e the pH [NaCHO2] = 0.815 M, [HBr] = 0.105 M, Ka (HCHO2) = 1.8 10-4 A) 2.91 B) 4.63 C) 4.57 D) 9.43 E) 11.09 Register to View AnswerDiff: 2 Reference: Section 17-4 92) Twenty- milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH before any NaOH is five added? A) 0.40 B) 2.5 C) 0.1 D) 1.0 E) 25 Register to View AnswerDiff: 2 Reference: Section 17-4 93) Twenty- milliliters of 0.10 M HCl is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH has five been added? A) 1.4 B) 1.2 C) 1.0 D) 2.0 E) 1.6 Register to View AnswerDiff: 2 Reference: Section 17-4 94) If 25 mL of 0.20 M NaOH is added to 50 mL of 0.10 M HC H O (K = 1.8 10-5), what is the pH? 232 a A) 5.33 B) 10.21 C) 8.78 D) 13.56 E) 1.34 Register to View AnswerDiff: 2 Reference: Section 17-4 95) 25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 15 ml of NaOH have been added? Ka for acetic acid = 1.8 10-5. A) 7.0 B) 4.3 C) 4.7 D) 4.9 E) 4.6 Register to View AnswerDiff: 2 Reference: Section 17-4 96) 25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH at the equivalence point? Ka for acetic acid = 1.8 10-5. A) 7.0 B) 10.6 C) 5.3 D) 8.7 E) 9.4 Register to View AnswerDiff: 2 Reference: Section 17-4 97) 25 ml of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 30 ml of NaOH have been added? Ka for acetic acid = 1.8 10-5. A) 12.0 B) 2.0 C) 12.2 D) 8.7 E) 12.3 Register to View AnswerDiff: 2 Reference: Section 17-4 98) What volume in ml of 0.05 M NaOH would have to be added to 50.0 ml of 0.10 M H2SO4 in order to affect complete neutralization of the acid? A) 50 ml B) 20 ml C) 100 ml D) 150 ml E) 10 ml Register to View AnswerDiff: 2 Reference: Section 17-4 99) 1.80 an impure mixture containing sodium carbonate required 84.0 ml of 0.125 M H2SO4 for grams of complete neutralization. What percent of the mixture is sodium carbonate? A) 62% B) 31% C) 120% D) 57% E) 1.1% Register to View AnswerDiff: 2 Reference: Section 17-4 100) Determin of the following solution. Initial concentrations are given. e the pH [NH3] = 1.20 M, [KOH] = 0.320 M, Kb (NH3) = 1.8 10-5 A) 13.5 B) 8.7 C) 9.8 D) 11.7 E) 12.2 Register to View AnswerDiff: 3 Reference: Section 17-4 101) What is a 0.30 M trisodium phosphate solution? [K for monohydrogen phosphate ion is 4.2 10-13] a the pH of A) 13.3 B) 12.7 C) 10.5 D) 9.8 E) 8.6 Register to View AnswerDiff: 1 Reference: Section 17-5 102) What is the pH of a 1.0 M solution of Na3AsO4. Ka1 = 6 10-3, Ka2 = 1 10-7, Ka3 = 3 10-12 ? A) 2.5 B) 7.0 C) 8.2 D) 11.5 E) 5.8 Register to View AnswerDiff: 1 Reference: Section 17-5 103) What is the pH of a 1.0 M solution of Na2SO3? Ka1 = 1.3 10-2, Ka2 = 6.2 10-8 ? A) 6.8 B) 7.2 C) 7.0 D) 10.4 E) 3.6 Register to View AnswerDiff: 1 Reference: Section 17-5 104) What is a 1.0 M solution of trisodium phosphate? K = 7.1 10-3, K = 6.3 10-8, K = 4.2 10a1 a2 a3 the pH of 13 A) 1.6 B) 7.0 C) 12.4 D) 6.2 E) 7.8 Register to View AnswerDiff: 1 Reference: Section 17-5 105) Calculate the pH of an aqueous solution that is 1.0 M Na CO . K for H CO is 4.7 10-11. 2 3 a2 2 3 A) 8.8 B) 10.3 C) 3.7 D) 12.2 E) 1.8 Register to View AnswerDiff: 2 Reference: Section 17-5

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USC - BISC - 120LG
2) Inheritance of chromosomes and DNA 15.1 - 15.3 and 16.11We now know that Mendels hereditary factors are located on chromosomesFig. 15.12Early studies of meiosis showed that the behavior of chromosomes parallels the behavior of Mendels
USC - BISC - 120LG
Quiz 1 starts today! Available on Blackboard 11 AM today (Wednesday) to 11 AM Thursday (24h) No time limit (except the 24h time limit) Can log out and log back in Be sure to Submit answers (not just Save them) and wait until Blackboard acknowledg
USC - BISC - 120LG
The Case For and Against EvolutionCh. 22.3 + R1 (on Blackboard)Teach both. You know, dont be afraid of information.- Sarah Palin, 2006You are here2The movie Flock of Dodos by USC Prof Randy Olson tries to explain why many Americans dont beli
USC - BISC - 120LG
Quiz 2 Available on Blackboard Wednesday Sept 10 (11AM) to Thursday Sept 11 (11AM) Covers lectures 4, 5 and 61Evolutionary Mechanisms I23.1-23.2Populations evolve, individuals do notXhttp:/www.youtube.com/watch?v=faRlFsYmkeY3The Modern
USC - BISC - 120LG
Reminders Quiz 2 available on Blackboard 11AM today to 11AM tomorrow (24h) Sept 12 (Friday) is the last day to register, add classes and drop without a W Last years Midterm 1 is now posted under Course Information. Note that format will be slightl
USC - BISC - 120LG
Speciation (24.1 - 24.2) Two basic patterns of speciation Anagenesis (phyletic or vertical evolution) transforms one species into another Cladogenesis (branching evolution) is the splitting of a gene pool, giving rise to one or more new species2
USC - BISC - 120LG
Phylogenetic Systematics26.1-26.3Molecular systematics is revealing surprises!Fig. 26.22More surprises!EukaryaArchaeaEubacteria3Some definitions Phylogeny = the evolutionary history of a species or group of related species Systema
USC - BISC - 120LG
Insights from Molecular Systematics26.4 - 26.62Molecular trees span both short and long periods of time because molecules evolve at different rates3Conserved sequences like rRNA resolve deep divergences4Rapidly evolving sequences like mt
USC - BISC - 120LG
Announcements SI review session tonight 7-9 PM in THH 201 and 301 Extra Hardy Weinberg questions (and answers) posted under Course Information / Exam Preparation Dr. Edmands extended office hours this week Monday 11 AM - 12:30 PM Tuesday 9 - 1
USC - BISC - 120LG
Evolution of Social Behavior51.4, 51.5 and R31Evolution of Social Behavior Sexual Selection Kin Selection &amp; Social Interactions Human Sociobiology2Sexual Selection Selection for mating success May oppose natural selection May lead to
USC - BISC - 120LG
Exam 1 on Monday! (Sept 29) During lecture period, room assignments are posted on Blackboard 9 AM students must take the 9 AM exam, 10 AM students must take the 10 AM exam Covers lectures 2-13 (14 will be on the Final) Multiple choice, matching,
UCSD - ECON - 1
January 07, 2009 We learn economic due to scarcity (limited resources) need to make choices and have trade-offs Scarcity principle: theres boundless needs and wants, resources are limited having move of one thing means having less of another t
UCSD - ECON - 1
January 9, 2009 Economist agents are rational: their objective is to make benefits through certain goals Example: cost-benefit principle One listens to a CD, but there are 2 songs that one hates. However, one is comfortably seated and does not wa
UC Irvine - BIOCHEM - 98
Bio 98A Midterm Exam 2/11/08ID # _TA Use OnlyA term used to describe a molecule that contains both a hydrophobic and a hydrophilic region . a b c d e clathrate amphipathic micelle bipolar amphoteric(1) _The predominate form of an important
UC Irvine - BIOCHEM - 98
AB
UC Irvine - BIOCHEM - 98
Bio. 98 Midterm ReviewInstructors: Poulos/Cumsky Peer Tutor: Michael ChauFormat of this sessionThis review session:Focuses on: A few important concepts per lecture Common questions associated with topics Does NOT focus on: Memorization-type
UC Irvine - BIOCHEM - 98
2/9/07Bio 98A MidtermName _score)If one combines 0.1 moles of NaH2PO4 and 0.2 moles of Na2HPO4 in 900 ml of water, the final pH will be _7.16 or 7.2_. -log R = D, so log (0.2/0.1) = 0.301, The basic form prevails around the pKa of 6.86 So
UC Irvine - BIOCHEM - 98
2/1/06 )Bio 98A Midterm ExamName _ (4) _For the following two ligands, what type of forces could be involved in their non-covalent binding to a protein's active site? (do not include van der Waals forces) leucine ionic or charge-charge or elect
Arizona - CHEM - CHEM 151
General Chemistry CHEM 151Week 2UA GenChemWeek 2 Reading AssignmentWeek 2 Section 5.8 Kinetic Molecular theory : a Model for Gases 5.9 Mean Free Path Tools of Chemistry equations and graphs for gases Section 5.1 pressure and Units 5.2 Pressure
Arizona - CHEM - CHEM 151
WELCOME to the UNIVERSITY OF ARIZONA FUNDAMENTALS OF CHEMISTRYCHEM 151 and 103AUA GenChemDr. Wayne WesolowskiDoc Weso Dr. W ProfessorUA GenChemMeet Someone !Share with someone you DONT know. Your name; Your major; Why you are taking
Arizona - CHEM - CHEM 151
The Following statement is from Dr. Anne Padias, Director of Academic Services for the Chemistry Department. The exams for all sections of Chem 151, including the students registered for Chem 103a, are scheduled for the following Mondays at 5 5:50 p
Arizona - CHEM - CHEM 151
General Chemistry CHEM 151Week 4UA GenChemWeek 4 Reading Assignment See D2LUA GenChemCounting ParticlesIs it practical, in a lab situation, to physically COUNT the particles? NO! The particles are too small! So how do we typically measure
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 5UA GenChemWeek 5 Reading Assignment See D2L CONTENTUA GenChemUnit 2: Zooming InWhat are they made of? Why are they joined in this way?They could be atoms, molecules, ionsWhat holds them together? Why are
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 6UA GenChemPhotoelectron SpectroscopyAnother spectroscopic technique can be used to determine number of electrons with a determined energy in an atom. KEhIonization Energy = h - KE The nth Ionization Energy is
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 7UA GenChemElectrons in AtomsElectrons in an atom can be classified into two main groups:CORE electrons: Strong attraction to protons in the nucleus (closer) Unavailable for exchange or bonds. VALENCE electrons:
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 8UA GenChemIonic Compounds: PropertiesHard High Melting Very High Boiling Do not Conduct Electricity as solids, but conduct when melted ( ions then can move) Shatter when struck. UA GenChemA Powerful Model
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 9UA GenChemREVIEWING CHEMICAL BONDINGIn a chemical reaction between two atoms, their valence electrons are reorganized. Atoms lose or share electrons because that leads to a more stable state (full electron shells
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 10UA GenChemMolecular GeometryValence Shell Electron Pair Repulsion Theory The most important factor in determining geometry is the relative repulsion between electron groups around the central atom Molecules adop
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 11UA GenChemForces and Particles INTRAMOLECULAR forces WITHIN the molecule (particle) which hold atoms together, determine the electron and molecular geometries. NN has strong intramolecular forces INTERMOLECULA
Arizona - CHEM - Chem 151
General Chemistry CHEM 151Week 12UA GenChemAqueous SolutionsA variety of molecular and ionic compounds are soluble in water. Their solubility depends on the strength of the intermolecular forces between particles in the system. solventA soluti
Cornell - NTRES - 2010
Feeding rate attack rate handling time prey encounter encounters per min0.2 0.5 1 0.08 Change this one, this equals to 5 sec handling time in minutes. 5 seconds is 5/60 minut FR Functional response 1 0.2 0.48 2 0.39 0.92 3 0.57 1.33 4 0.75 1.71 5 0
N.C. State - ECE - 220
iECE220 Homework Assignment #5 SolutionsProblem 7.5 Consider the differential equation dv(t) + 2.0 106 v(t) = 10u(t) dt (0.1)with the initial condition v(0) = 0. (a) Find vc (t). (b) Find vp (t). (c) Find the total solution v(t). (d) Plot all
N.C. State - ECE - 220
iECE220 Homework Assignment #1 SolutionsProblem 2.15. Sketch the following signals v1 (t) = (t)[et + cos(2106 t)] v2 (t) = (t 5) + (t + 5) v3 (t) = (t)e3t [u(t) u(t + 2)] sin(21000t) Solution See gures 0.1 through 0.3. Note the value of the sig
N.C. State - ECE - 220
iECE220 Homework Assignment #2 SolutionsProblem 3.18. For safety reasons, electrical devices limit the output voltage vout (t) to a maximum allowed value A, regardless of the value of the input voltage vin (t). Mathematically, we can represent su
N.C. State - ECE - 220
iECE220 Homework Assignment #3 SolutionsProblem 6.21. Let a and b be real-valued constants. Let 2 x1 2 3a 1 2 2 x2 = b 1 x3 1 34 Determine, if possible, values for a and b that will make the system have no solution (i.e., inconsistent),
N.C. State - ECE - 220
iECE220 Homework Assignment #4 SolutionsProblem 4.33 Write the following signals in phasor notation: v1 (t) = 4 cos(23000t + 4/5) v2 (t) = 60 cos(2400t ) v3 (t) = Re[10ej(21000t/3) ] v4 (t) = 3.4 cos(500t) v5 (t) = 3.5 sin(400t) v6 (t) = 100 v7
N.C. State - ECE - 220
iECE220 Homework Assignment #6 SolutionsProblem 7.10 Consider the differential equation obtained from an RLC circuit d2 v(t) R dv(t) 1 1 + + v(t) = vs (t) (0.1) 2 dt L dt LC LC where R = 10, L = 1, C = 2 and the initial conditions are v(0) = 0, v
N.C. State - ECE - 220
iECE220 Homework Assignment #7 SolutionsProblem 8.1 Find the Laplace transform of v(t) = 4e6t cos(7t)u(t) by direct integration. (Hint: using Eulers identity may be helpful.) Solution We haveV (s) =04e6t cos(7t)u(t)est dt e(s+6)t cos(7t)dt
N.C. State - ECE - 220
iECE220 Homework Assignment #8 SolutionsProblem 9.11 Without using any integrals, determine the Fourier series coefcients, ak , bk and k , of the following functions. Note the frequency that corresponds to each index k. The time variable is measu
N.C. State - ECE - 220
iECE220 Homework Assignment #9 SolutionsProblem 10.1 Using direct integration, nd the Fourier transform of the signal s(t) = e25t cos(2100t + /4)u(t) You can check your result by using the Laplace transform table in Chapter 8. Solution s(t) = e25
N.C. State - ECE - 220
ECE 220 Spring 2008 Problem Lab #08 Solutions* Note, all gures are at the end of the document. I. Laplace transforms, use of MATLAB laplace() functionRead the help le for the MATLAB function laplace(). Question 1. Use laplace() to evaluate the La
N.C. State - ECE - 220
ECE220 Problem Lab #09 Review for Final Exam Date: Week of April 21, 2008The goal of this lab is to review material in preparation for the nal exam. Make sure you also review the problems we covered in the preparation labs for tests 1, 2, and 3. Wo
N.C. State - ECE - 220
ECE220 Problem Lab #3 Getting Practice complex numbers and functions Date: Week of February 5, 2007I. Basic concepts in complex numbers 1. 2+j2Magnitude: Phase:2 + j 2 = 2 2 + 2 2 = 2 2 = 2.82842 + j 2 = / 4 = 0.785Q1 - Show Coordinates2
N.C. State - ECE - 220
Solutions, ECE220 Problem Lab #1 Getting practice with MATLAB, periodic signals and mathematical background skills Date: Week of January 21, 2008The goals of this lab are to: (a) make sure you are up to speed with the required/assumed math backgrou
N.C. State - ECE - 220
Solutions, ECE220 Problem Lab #2 Signal Plots, Vectors, Matrices, and Linear Equations Date: Week of January 28, 2008The main goal of this lab is to give you practice on: (a) signal calculation, processing and plotting, and, (b) vectors, matrices a
N.C. State - ECE - 220
ECE220 Problem Lab #5 Solutions Dierential equations and transfer functions Date: Week of February 25, 2008The goal of this lab is to (a) summarize and reinforce all the methods we presented in chapter 7, about solving rst order dierential equation
N.C. State - ECE - 220
ECE220 Solution Lab #6 Review for Test 2 Date: Week of March 10, 2008The goal of this lab is to review material in preparation for Test 2. Work out as many problems as you can. If you have worked on a problem before, you may skip it. Do not worry i
N.C. State - ECE - 220
N.C. State - ECE - 220
N.C. State - ECE - 220
N.C. State - ECE - 220
N.C. State - ECE - 220
N.C. State - ECE - 220
ECE 220 Test 3 Solutions14 April 2008Problem 1. (20 points) A system is described by the dierential equation given below: dv(t) + 4v(t) = vs (t) dt where vs (t) is the input. Let vs (t) = 10 cos(5t) and v(0) = 0. Determine the solution in the Lapla
N.C. State - ECE - 220
ECE 220 Lab9 SolutionMade by Rong GuoProblems 7.20 myfunction.m -%Refer to matlab script7.3 on Page381. function yd=myfunction(t,v) A=[3 2;6 0]; b=[0;cos(2*pi*60*t)]; yd=A*v+b; return --pro7x20.m -clear all [t,v]=ode23('myfunction',[0,1],[-1/6;0
N.C. State - ECE - 220
N.C. State - ECE - 220
N.C. State - ECE - 220
4.384.394.404.42The phase and magnitude plots obtained are as shown:5.105.11C1 = [1 5 6] C2 = Not possible. Matrix dimensions do not agree. Atranspose= [1 02 10 C3= [-2 6 1] C4 = Matrix dimensions do not agree. -1 3 1 1]5.125.205.
N.C. State - ECE - 220
Plot for Problem 7.56 x 10-6Vc(t) Vp(t) V(t) 420-2-4-600.20.40.60.811.2 x 10-5Plots for Problem 7.63 Driving force Particular solution 210-1-2-3012345 Time (t)6789103 yc (t) 2.5 yp(t)
N.C. State - ECE - 220
N.C. State - ECE - 220
Name:ID#4Signature2 2_This Exam is &quot;Closed-Book.&quot; &quot;Notes,&quot; in any form, are NOT allowed. Calculators cannot be used in &quot;complex number mode.&quot; Calculators can ONLY be used to perform &quot;real number arithmetic and trig&quot; operations. At the end o
N.C. State - ECE - 220