Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Chapter 16, Problem 17KT
Interpretation Introduction
Interpretation:
The key term corresponding to the definition “the constant that expresses the molar equilibrium concentrations of ions in an aqueous solution for a slightly ionized acid or base” is to be stated.
Concept introduction:
In a reversible reaction, reactants and products are present in an equilibrium. The equilibrium constant is equal to the ratio of molar concentration of product to the molar concentration of reactant each raised to power of its stociometric coefficient.
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Give 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 PLEASE
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.
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© 2025 McGraw Hill LLC. All Rights Reserved. Te
LMUNDARY
Sketch the intermediates for A,B,C & D.
Chapter 16 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 16 - Prob. 1CECh. 16 - Prob. 2CECh. 16 - Prob. 3CECh. 16 - Prob. 4CECh. 16 - Prob. 5CECh. 16 - Prob. 6CECh. 16 - Prob. 7CECh. 16 - Prob. 8CECh. 16 - Prob. 9CECh. 16 - Prob. 10CE
Ch. 16 - Prob. 1KTCh. 16 - Prob. 2KTCh. 16 - Prob. 3KTCh. 16 - Prob. 4KTCh. 16 - Prob. 5KTCh. 16 - Prob. 6KTCh. 16 - Prob. 7KTCh. 16 - Prob. 8KTCh. 16 - Prob. 9KTCh. 16 - Prob. 10KTCh. 16 - Prob. 11KTCh. 16 - Prob. 12KTCh. 16 - Prob. 13KTCh. 16 - Prob. 14KTCh. 16 - Prob. 15KTCh. 16 - Prob. 16KTCh. 16 - Prob. 17KTCh. 16 - Prob. 18KTCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 1STCh. 16 - Prob. 2STCh. 16 - Prob. 3STCh. 16 - Prob. 4STCh. 16 - Prob. 5STCh. 16 - Prob. 6STCh. 16 - Prob. 7STCh. 16 - Prob. 8STCh. 16 - Prob. 9STCh. 16 - Prob. 10STCh. 16 - Prob. 11STCh. 16 - Prob. 12STCh. 16 - Prob. 13STCh. 16 - Prob. 14STCh. 16 - Prob. 15ST
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- Can 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_forwardIn 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_forward
- Hand 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_forward
- How 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_forwardin the kinetics experiment, what were the values calculated? Select all that apply.a) equilibrium constantb) pHc) order of reactiond) rate contstantarrow_forwardtrue or false, given that a 20.00 mL sample of NaOH took 24.15 mL of 0.141 M HCI to reach the endpoint in a titration, the concentration of the NaOH is 1.17 M.arrow_forward
- in the bromothymol blue experiment, pKa was measured. A closely related compound has a Ka of 2.10 x 10-5. What is the pKa?a) 7.1b) 4.7c) 2.0arrow_forwardcalculate the equilibrium concentration of H2 given that K= 0.017 at a constant temperature for this reaction. The inital concentration of HBr is 0.050 M.2HBr(g) ↔ H2(g) + Br2(g)a) 4.48 x 10-2 M b) 5.17 x 10-3 Mc) 1.03 x 10-2 Md) 1.70 x 10-2 Marrow_forwardtrue or falsegiven these two equilibria with their equilibrium constants:H2(g) + CI2(l) ↔ 2HCI(g) K= 0.006 CI2(l) ↔ CI2(g) K= 0.30The equilibrium contstant for the following reaction is 1.8H2(g) + CI2 ↔ 2HCI(g)arrow_forward
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