Chemistry for Changing Times
14th Edition
ISBN: 9780134212777
Author: John W. Hill; Terry W. McCreary
Publisher: Pearson Education (US)
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Chapter 11, Problem 67AP
Interpretation Introduction
Interpretation: We have to find out age of the wood from the tomb whose carbon-14 activities are given.
Concept introduction: Radioactive decay follows first order rate kinetics. The rate law for a radioactive decay is,
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6. Consider the following exothermic reaction below.
2Cu2+(aq) +41 (aq)2Cul(s) + 12(aq)
a. If Cul is added, there will be a shift left/shift right/no shift (circle one).
b. If Cu2+ is added, there will be a shift left/shift right/no shift (circle one).
c. If a solution of AgNO3 is added, there will be a shift left/shift right/no shift (circle one).
d. If the solvent hexane (C6H14) is added, there will be a shift left/shift right/no shift (circle
one). Hint: one of the reaction species is more soluble in hexane than in water.
e. If the reaction is cooled, there will be a shift left/shift right/no shift (circle one).
f. Which of the changes above will change the equilibrium constant, K?
Show work. don't give Ai
Show work with explanation needed. don't give Ai generated solution
Chapter 11 Solutions
Chemistry for Changing Times
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Ch. 11 - Prob. 11RQCh. 11 - Prob. 12RQCh. 11 - Prob. 13RQCh. 11 - Prob. 14RQCh. 11 - Prob. 15RQCh. 11 - Prob. 16RQCh. 11 - Prob. 17RQCh. 11 - Prob. 18PCh. 11 - Prob. 19PCh. 11 - Prob. 20PCh. 11 - Prob. 21PCh. 11 - Prob. 22PCh. 11 - Prob. 23PCh. 11 - Prob. 24PCh. 11 - Prob. 25PCh. 11 - Prob. 26PCh. 11 - Prob. 27PCh. 11 - Prob. 28PCh. 11 - Prob. 29PCh. 11 - Prob. 30PCh. 11 - Prob. 31PCh. 11 - Prob. 32PCh. 11 - Prob. 33PCh. 11 - Prob. 34PCh. 11 - Prob. 35PCh. 11 - Prob. 36PCh. 11 - Prob. 37PCh. 11 - Prob. 38PCh. 11 - Prob. 39PCh. 11 - Prob. 40PCh. 11 - Prob. 41PCh. 11 - Prob. 42PCh. 11 - Prob. 43PCh. 11 - Prob. 44PCh. 11 - Prob. 45PCh. 11 - Prob. 46PCh. 11 - Prob. 47PCh. 11 - Prob. 48PCh. 11 - Prob. 49PCh. 11 - Prob. 50PCh. 11 - Prob. 51PCh. 11 - Prob. 52PCh. 11 - Prob. 53PCh. 11 - Prob. 54PCh. 11 - Prob. 55PCh. 11 - Prob. 56PCh. 11 - Prob. 57PCh. 11 - Prob. 58APCh. 11 - Prob. 59APCh. 11 - Prob. 60APCh. 11 - Prob. 61APCh. 11 - Prob. 62APCh. 11 - Prob. 63APCh. 11 - Prob. 64APCh. 11 - Prob. 65APCh. 11 - Prob. 66APCh. 11 - Prob. 67APCh. 11 - Prob. 68APCh. 11 - Prob. 69APCh. 11 - Prob. 70APCh. 11 - Prob. 71APCh. 11 - Prob. 72APCh. 11 - Prob. 73APCh. 11 - Prob. 74APCh. 11 - Prob. 75APCh. 11 - Prob. 76APCh. 11 - Prob. 77APCh. 11 - Prob. 78APCh. 11 - Prob. 11.1CTECh. 11 - Prob. 11.2CTECh. 11 - Prob. 11.3CTECh. 11 - Prob. 11.4CTECh. 11 - Prob. 11.5CTECh. 11 - Prob. 11.6CTECh. 11 - Prob. 1CGPCh. 11 - Prob. 2CGPCh. 11 - Prob. 3CGPCh. 11 - Prob. 4CGPCh. 11 - Prob. 5CGPCh. 11 - Prob. 6CGPCh. 11 - Prob. 7CGPCh. 11 - Prob. 8CGPCh. 11 - Prob. 1CHQCh. 11 - Prob. 2CHQCh. 11 - Prob. 3CHQCh. 11 - Prob. 4CHQ
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- The following mechanism for the gas phase reaction of H2 and ICI that is consistent with the observed rate law is: step 1 step 2 slow: H2(g) +ICI(g) → HCl(g) + HI(g) fast: ICI(g) + HI(g) → HCl(g) + |2(g) (1) What is the equation for the overall reaction? Use the smallest integer coefficients possible. If a box is not needed, leave it blank. + → + (2) Which species acts as a catalyst? Enter formula. If none, leave box blank: (3) Which species acts as a reaction intermediate? Enter formula. If none, leave box blank: (4) Complete the rate law for the overall reaction that is consistent with this mechanism. (Use the form k[A][B]"..., where '1' is understood (so don't write it) for m, n etc.) Rate =arrow_forwardPlease correct answer and don't use hand rating and don't use Ai solutionarrow_forward1. For each of the following statements, indicate whether they are true of false. ⚫ the terms primary, secondary and tertiary have different meanings when applied to amines than they do when applied to alcohols. • a tertiary amine is one that is bonded to a tertiary carbon atom (one with three C atoms bonded to it). • simple five-membered heteroaromatic compounds (e.g. pyrrole) are typically more electron rich than benzene. ⚫ simple six-membered heteroaromatic compounds (e.g. pyridine) are typically more electron rich than benzene. • pyrrole is very weakly basic because protonation anywhere on the ring disrupts the aromaticity. • thiophene is more reactive than benzene toward electrophilic aromatic substitution. • pyridine is more reactive than nitrobenzene toward electrophilic aromatic substitution. • the lone pair on the nitrogen atom of pyridine is part of the pi system.arrow_forward
- The following reactions are NOT ordered in the way in which they occur. Reaction 1 PhO-OPh Reaction 2 Ph-O -CH₂ heat 2 *OPh Pho -CH2 Reaction 3 Ph-O ⚫OPh + -CH₂ Reaction 4 Pho Pho + H₂C OPh + CHOPh H₂C -CH₂ Reactions 1 and 3 Reaction 2 O Reaction 3 ○ Reactions 3 and 4 ○ Reactions 1 and 2 Reaction 4 ○ Reaction 1arrow_forwardSelect all possible products from the following reaction: NaOH H₂O a) b) ОН HO O HO HO e) ОН f) O HO g) h) + OHarrow_forward3. Draw diagrams to represent the conjugation in these molecules. Draw two types of diagram: a. Show curly arrows linking at least two different ways of representing the molecule b. Indicate with dotted lines and partial charges (where necessary) the partial double bond (and charge) distribution H₂N* H₂N -NH2arrow_forward
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