Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 18, Problem 89QRT
Interpretation Introduction
Interpretation:
Disappearance of
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Explain why acid chlorides are more reactive than amides in reactions with nucleophiles.
Calculating the pH of a weak base titrated with a strong acid
An analytical chemist is titrating 101.7 mL of a 0.3500M solution of piperidine (C5H10NH) with a 0.05700M solution of HClO4. The pK of piperidine is 2.89.
Calculate the pH of the base solution after the chemist has added 682.9 mL of the HClO solution to it.
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Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HClO solution added.
4
Round your answer to 2 decimal places.
pH = .11
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Ar
The following is a two groups (Regular tomato sauce & Salt Reduced Tomato Sauce) of data recorded by a team analysising salt content in tomato sauce using the MOHR titration method:
Regular Tomato Sauce
Salt Reduced Tomato Sauce
340.0
262.7
QUESTION: For both groups of data provide answers to the calculations attached in the image
Chapter 18 Solutions
Chemistry: The Molecular Science
Ch. 18.2 - Prob. 18.1PSPCh. 18.2 - Prob. 18.1ECh. 18.2 - Prob. 18.2PSPCh. 18.2 - Prob. 18.2ECh. 18.3 - Prob. 18.3PSPCh. 18.3 - Prob. 18.3ECh. 18.3 - Prob. 18.4CECh. 18.4 - Prob. 18.4PSPCh. 18.4 - Prob. 18.5ECh. 18.4 - Prob. 18.5PSP
Ch. 18.4 - Prob. 18.6PSPCh. 18.4 - Prob. 18.7PSPCh. 18.4 - Prob. 18.6ECh. 18.4 - Prob. 18.7CECh. 18.5 - Prob. 18.8ECh. 18.5 - Prob. 18.9CECh. 18.6 - Prob. 18.10ECh. 18.6 - Prob. 18.11ECh. 18.7 - Prob. 18.12ECh. 18.8 - Prob. 18.13ECh. 18.8 - Prob. 18.14ECh. 18.9 - Prob. 18.15ECh. 18 - Prob. 1SPCh. 18 - Prob. 2SPCh. 18 - Prob. 3SPCh. 18 - Prob. 4SPCh. 18 - Prob. 5SPCh. 18 - Prob. 1QRTCh. 18 - Prob. 2QRTCh. 18 - Prob. 3QRTCh. 18 - Prob. 4QRTCh. 18 - Prob. 5QRTCh. 18 - Prob. 6QRTCh. 18 - Prob. 7QRTCh. 18 - Prob. 8QRTCh. 18 - Prob. 9QRTCh. 18 - Complete the table.Ch. 18 - Prob. 11QRTCh. 18 - Prob. 12QRTCh. 18 - Prob. 13QRTCh. 18 - Prob. 14QRTCh. 18 - Prob. 15QRTCh. 18 - Prob. 16QRTCh. 18 - Prob. 17QRTCh. 18 - Prob. 18QRTCh. 18 - Prob. 19QRTCh. 18 - Prob. 20QRTCh. 18 - Prob. 21QRTCh. 18 - Prob. 22QRTCh. 18 - Prob. 23QRTCh. 18 - Prob. 24QRTCh. 18 - Prob. 25QRTCh. 18 - Prob. 26QRTCh. 18 - Prob. 27QRTCh. 18 - Prob. 28QRTCh. 18 - Prob. 29QRTCh. 18 - Prob. 30QRTCh. 18 - Prob. 31QRTCh. 18 - Prob. 32QRTCh. 18 - Prob. 33QRTCh. 18 - Prob. 34QRTCh. 18 - Prob. 35QRTCh. 18 - Prob. 36QRTCh. 18 - Prob. 37QRTCh. 18 - Prob. 38QRTCh. 18 - Prob. 39QRTCh. 18 - Prob. 40QRTCh. 18 - Prob. 41QRTCh. 18 - Prob. 42QRTCh. 18 - Prob. 43QRTCh. 18 - Prob. 44QRTCh. 18 - Prob. 45QRTCh. 18 - Prob. 46QRTCh. 18 - Prob. 47QRTCh. 18 - Prob. 48QRTCh. 18 - Prob. 49QRTCh. 18 - Prob. 50QRTCh. 18 - Prob. 51QRTCh. 18 - Prob. 52QRTCh. 18 - Prob. 53QRTCh. 18 - Prob. 54QRTCh. 18 - Prob. 55QRTCh. 18 - Prob. 56QRTCh. 18 - Prob. 57QRTCh. 18 - Prob. 58QRTCh. 18 - Prob. 59QRTCh. 18 - Prob. 60QRTCh. 18 - Prob. 61QRTCh. 18 - Prob. 62QRTCh. 18 - Prob. 63QRTCh. 18 - Prob. 64QRTCh. 18 - Prob. 65QRTCh. 18 - Prob. 66QRTCh. 18 - Prob. 67QRTCh. 18 - Prob. 68QRTCh. 18 - Prob. 69QRTCh. 18 - Prob. 70QRTCh. 18 - Prob. 71QRTCh. 18 - Prob. 72QRTCh. 18 - Prob. 73QRTCh. 18 - Prob. 74QRTCh. 18 - Prob. 75QRTCh. 18 - Prob. 76QRTCh. 18 - Prob. 77QRTCh. 18 - Prob. 78QRTCh. 18 - Prob. 79QRTCh. 18 - Prob. 80QRTCh. 18 - Prob. 81QRTCh. 18 - Prob. 82QRTCh. 18 - Prob. 83QRTCh. 18 - Prob. 84QRTCh. 18 - Prob. 85QRTCh. 18 - Prob. 86QRTCh. 18 - Prob. 87QRTCh. 18 - Prob. 88QRTCh. 18 - Prob. 89QRTCh. 18 - Prob. 91QRTCh. 18 - Prob. 92QRTCh. 18 - Prob. 93QRTCh. 18 - Prob. 94QRTCh. 18 - Prob. 95QRTCh. 18 - Prob. 96QRTCh. 18 - Prob. 18.ACPCh. 18 - Prob. 18.BCPCh. 18 - Prob. 18.CCPCh. 18 - Prob. 18.DCPCh. 18 - Prob. 18.ECP
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- 7. Concentration and uncertainty in the estimate of concentration (class data) Class mean for sample (Regular) |[Cl-] (mmol/L) class mean Sn za/2 95% Confidence Interval (mmol/L) [Na+] (mg/100 mL) 95% Confidence Interval (mg/100 mL)arrow_forwardThe following is a two groups (Regular tomato sauce & Salt Reduced Tomato Sauce) of data recorded by a team analysising salt content in tomato sauce using the MOHR titration method: Regular Tomato Sauce Salt Reduced Tomato Sauce 223.4 148.7 353.7 278.2 334.6 268.7 305.6 234.4 340.0 262.7 304.3 283.2 244.7 143.6 QUESTION: For both groups of data calculate the answers attached in the image.arrow_forwardGive reason(s) for six from the followings [using equations if possible] a. Addition of sodium carbonate to sulfanilic acid in the Methyl Orange preparation. b. What happened if the diazotization reaction gets warmed up by mistake. c. Addition of sodium nitrite in acidified solution in MO preparation through the diazotization d. Using sodium dithionite dihydrate in the second step for Luminol preparation. e. In nitroaniline preparation, addition of the acid mixture (nitric acid and sulfuric acid) to the product of step I. f. What is the main reason of the acylation step in nitroaniline preparation g. Heating under reflux. h. Fusion of an organic compound with sodium. HAND WRITTEN PLEASEarrow_forward
- edict the major products of the following organic reaction: u A + ? CN Some important notes: • Draw the major product, or products, of the reaction in the drawing area below. • If there aren't any products, because no reaction will take place, check the box below the drawing area instead. Be sure to use wedge and dash bonds when necessary, for example to distinguish between major products that are enantiomers. Explanation Check Click and drag to start drawing a structure. Х © 2025 McGraw Hill LLC. All Rights Reserved. Te LMUNDARYarrow_forwardSketch the intermediates for A,B,C & D.arrow_forwardCan the molecule on the right-hand side of this organic reaction be made in good yield from no more than two reactants, in one step, by moderately heating the reactants? O ? A . If your answer is yes, then draw the reactant or reactants in the drawing area below. You can draw the reactants in any arrangement you like. . If your answer is no, check the box under the drawing area instead. Explanation Check Click and drag to start drawing a structure. ㅇ 80 F5 F6 A 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Cente FIGarrow_forward
- In methyl orange preparation, if the reaction started with 0.5 mole of sulfanilic acid to form the diazonium salt of this compound and then it converted to methyl orange [0.2 mole]. If the efficiency of the second step was 50%, Calculate: A. Equation(s) of Methyl Orange synthesis: Diazotization and coupling reactions. B. How much diazonium salt was formed in this reaction? C. The efficiency percentage of the diazotization reaction D. Efficiency percentage of the whole reaction.arrow_forwardHand written equations pleasearrow_forwardHand written equations pleasearrow_forward
- > each pair of substrates below, choose the one that will react faster in a substitution reaction, assuming that: 1. the rate of substitution doesn't depend on nucleophile concentration and 2. the products are a roughly 50/50 mixture of enantiomers. Substrate A Substrate B Faster Rate X Ś CI (Choose one) (Choose one) CI Br Explanation Check Br (Choose one) © 2025 McGraw Hill LLC. All Rights Farrow_forwardNMR spectrum of ethyl acetate has signals whose chemical shifts are indicated below. Which hydrogen or set of hydrogens corresponds to the signal at 4.1 ppm? Select the single best answer. The H O HỌC—C—0—CH, CH, 2 A ethyl acetate H NMR: 1.3 ppm, 2.0 ppm, 4.1 ppm Check OA B OC ch B C Save For Later Submit Ass © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center |arrow_forwardHow many signals do you expect in the H NMR spectrum for this molecule? Br Br Write the answer below. Also, in each of the drawing areas below is a copy of the molecule, with Hs shown. In each copy, one of the H atoms is colored red. Highlight in red all other H atoms that would contribute to the same signal as the H already highlighted red Note for advanced students: In this question, any multiplet is counted as one signal. 1 Number of signals in the 'H NMR spectrum. For the molecule in the top drawing area, highlight in red any other H atoms that will contribute to the same signal as the H atom already highlighted red. If no other H atoms will contribute, check the box at right. Check For the molecule in the bottom drawing area, highlight in red any other H atoms that will contribute to the same signal as the H atom already highlighted red. If no other H atoms will contribute, check the box at right. O ✓ No additional Hs to color in top molecule ง No additional Hs to color in bottom…arrow_forward
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