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The net force on the 1.0 nC charge in FIGURE P22.48 is

Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36) | 4th Edition | ISBN: 9780134081496 | Authors: Randall D. Knight (Professor Emeritus) ISBN: 9780134081496 191

Solution for problem 22.84 Chapter 22

Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36) | 4th Edition

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Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36) | 4th Edition | ISBN: 9780134081496 | Authors: Randall D. Knight (Professor Emeritus)

Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36) | 4th Edition

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Problem 22.84

The net force on the 1.0 nC charge in FIGURE P22.48 is zero. What is q?

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Exam 3 Study Guide 1. CH4(g) + 2O2(g)→CO2(g) + 2H2O(g). This reaction is exothermic. What would happen to the equilibrium if the temperature in the system was raised 2. State the definitions of Bronsted-Lowry acids and bases and Lewis acids and bases. 3. You have two solutions: one solution is 0.25 M HCl, the other solution is 1.0 M acetic acid. Which solution is stronger 4. True or false: water can function as an acid and a base. 5. How does a catalyst affect equilibrium in a chemical reaction 6. Given the reaction N2H4(g)⁶N2(g) + 2H2(g) and the following ∆G˚ values, calculate the value of K at 298 K. ∆G˚ N2H4(g) = 159.4 kJ/mol and ∆G˚ N2(g) = ∆G˚H2(g) = 0.0 kJ/mol. 7. Using molecular structure, explain the trend in strength of oxoacids (with the same central atom). 8. If the pH of blood is 7.4 and the pH of milk is 6.5, how much more acidic is milk than blood 9. Rank the following acids from strongest to weakest based on their Ka values. HNO2: Ka = 4.0 x 10^-4, H2CO3: Ka = 4.3 x 10^-7, HF: Ka = 7.2 x 10^-4. 10. A solution has a pOH of 2.7. Is this solution acidic, basic, or neutral 11. 0.213 kg of NaOH is added to 6500 mL of water. Calculate the pH of the solution. 12. You have a solution of HCl with a pH of 3.67. Calculate the H3O+ concentration. 13. Out of the following list of acid and bases, name which are strong and which are weak: HBr, CH3COOH, NaOH, NH3, HNO3, Ba(OH)2. 14. Identify the Bronsted-Lowry acid and base in the following equation and explain why it is the acid/base: CO3 2-(aq) + H2O(l)→HCO3-(aq) + OH-(aq). 15. Identify the conjugate acid and conjugate base in the reaction in the previous question. 16. Rainwater has a pH of 5.6. Is rainwater an acid, a base, acidic, or basic 17. Write the equation for the equilibrium constant for the autoionization of water and state its value under standard conditions. 18. You are given this reaction: CO(g) + H2O(g)⁶CO2(g) + H2(g) Kc=0.64 You start with 1.0 mol CO and 1.0 mol CO2 in a 1 L flask. You then add 2.0 mol H2O. What is the concentration of H2 at equilibrium 19. A 7.2 L vessel contains the following composition at equilibrium: 48.2 g CaCO3, 24.3 g CaO, and 26.5 g CO2. Given the equation CaCO3(s)→CaO(s) + CO2(g), calculate Kc for this reaction. 20. The following reaction is being studied: N2(g) + 3H2(g)→2NH3(g). At a certain temperature, Kp = 0.0011. A vessel is filled with N2 at 15.07 atm, H2 at 33.46 atm, and NH3 at 22.30 atm. Predict the expected changes in the compositions the next time they are measured. Answers 1. In exothermic reactions, heat is treated like a product. If we increase the temperature, this shifts the equilibrium to the left, favoring the reverse reaction. 2. Bronsted-Lowry definition: acids donate protons and bases accept protons. Lewis definition: acids accept electrons and bases donate electrons. 3. The solution of 0.25 M HCl is stronger because HCl is a strong acid whereas acetic acid is a weak acid. 4. True. An example is in the equation for the autoionization of water, H2O(l)(acid) + H2O(l)(base)⁶H3O+(aq) + OH-(aq), where H2O functions as both an acid and a base. 5. A catalyst lowers the activation energy of a reaction to increase the rate at which it proceeds, but it does not affect equilibrium or cause any shifts between reactants and products. 6. Use ∆Grxn = -RT(lnK) and rearrange to K = e^(-∆Grxn/RT) to solve for K. ∆G˚rxn = [0(1) + 0(2)] - [159.4(1)] = -159.4 kJ/mol *Calculate ∆G˚rxn K = e^[159.4 x 10^3/(8.314 x 298)] *Convert kJ/mol to J/mol K = 8.76 x 10^27 7. Oxoacids increase strength with each addition of an O atom because the electron density is distributed more evenly across the molecule, resulting in a more stable acid. 8. 10^-6.5 M = 10^0.9 = 10 (rounded) 10^-7.4 M Milk is about 10 times more acidic than blood. It can also be said that blood is about 10 times more basic than milk. 9. The higher the Ka value of an acid, the stronger that acid is, so rank them in order of largest to smallest Ka value: HF, HNO2, H2CO3. 10. A solution with a low pOH is the same as a solution with a high pH, so a solution with a pOH of 2.7 is basic. 11. 213 g NaOH = 5.325 mol *Convert kg to g and g to mol 39.998 g/mol 5.325 mol = 0.819 M NaOH *Convert mL to L and calculate molarity 6.5 L -log(0.819 M) = 0.0866 *Use -log[OH-] to find pOH 14 - 0.0866 = 13.91 *Use pH + pOH = 14 to find pH 12. Use [H3O+] = 10^(-pH). 10^(-3.67) = 2.14 x 10^-4 M 13. Strong acids: HBr, HNO3 Weak acids: CH3COOH Strong bases: NaOH, Ba(OH)2 Weak bases: NH3 14. H2O is the acid in this case because it donates a proton and CO3 2- is the base because it gains a proton. 15. In the previous reaction, the conjugate acid is HCO3- and the conjugate base is OH-. 16. Since rainwater has a pH relatively close to neutral, it is not an acid but it is acidic. 17. Kw = [H3O+][OH-] and Kw = 1.00 x 10^-14 at standard temperature and pressure. 18. Use an ICE table and the quadratic formula. CO(g) + H2O(g) ⁶ CO2(g) + H2(g) Initial 1.0 M 2.0 M 1.0 M 0 M Change -x -x +x +x Equilibrium (1.0-x) (2.0-x) (1.0+x) x Kc = 0.64 = [1.0+x][x] [1.0-x][2.0-x] 0.64 = x + x^2 2 - 3x + x^2 0.64(2 - 3x + x^2) = x + x^2 -0.36x^2 - 2.92x + 1.28 = 0 *Standard form ax^2 + bx + c = 0 -(-2.92) ± √(-2.92)^2 - 4(-0.36)(1.28) *Quadratic formula 2(-0.36) x = -8.5, x = 0.42 *Plug in x values to see what makes sense [H2] = 0.42 M 19. Note: You do not need to do the following calculations for CaCO3 or CaO since they are solids and are not part of the equation for Kc. 26.5 g CO2 = 0.602 mol *Convert g to mol 44.01 g/mol 0.602 mol CO2 = 0.0836 M *Find molarity 7.2 L Kc = 0.084 *Use Kc = [CO2] 20. Kp = [22.30]^2 = 8.8 x 10^-4 *Calculate new Kp [33.46]^3 [15.07] 8.8 x 10^-4 < 0.0011. Since the new Kp is smaller than the given Kp, this means there are more reactants in the system, so the equilibrium will shift to the right, which will increase the pressure of NH3 and decrease the pressure of N2 and H2.

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Chapter 22, Problem 22.84 is Solved
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Textbook: Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36)
Edition: 4
Author: Randall D. Knight (Professor Emeritus)
ISBN: 9780134081496

Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36) was written by and is associated to the ISBN: 9780134081496. The answer to “The net force on the 1.0 nC charge in FIGURE P22.48 is zero. What is q?” is broken down into a number of easy to follow steps, and 16 words. This full solution covers the following key subjects: . This expansive textbook survival guide covers 42 chapters, and 4463 solutions. Since the solution to 22.84 from 22 chapter was answered, more than 424 students have viewed the full step-by-step answer. This textbook survival guide was created for the textbook: Physics for Scientists and Engineers: A Strategic Approach, Standard Edition (Chs 1-36), edition: 4. The full step-by-step solution to problem: 22.84 from chapter: 22 was answered by , our top Physics solution expert on 12/28/17, 08:06PM.

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The net force on the 1.0 nC charge in FIGURE P22.48 is