Chemistry: The Science in Context (Fourth Edition)
4th Edition
ISBN: 9780393124187
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Geoffrey Davies
Publisher: W. W. Norton & Company
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Question
Chapter 13, Problem 13.151AP
(a)
Interpretation Introduction
To determine: The two balanced equations that describe the given reactions that occur over and over again to produce a silicone
Interpretation: The two balanced equations that describe the given reactions that occur over and over again to produce a silicone polymer are to be stated.
Concept introduction: The repeated bonding of monomer units lead to the formation of polymer. They are mainly synthesized through condensation or addition reaction of monomer units.
(b)
Interpretation Introduction
To determine: An explanation corresponding to the fact that silicones are water repellent.
Interpretation: An explanation corresponding to the fact that silicones are water repellent is to be stated.
Concept introduction: The presence of alkyl or aryl group over atom provides hydrophobicity to the molecule. The long chain alkyl groups are hydrophobic in nature and repel water from the surface.
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Students have asked these similar questions
Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard
the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
For the decomposition reaction of N2O5(g): 2 N2O5(g) · 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5
-> NO2 + NO3_(K1)
NO2 + NO3 →> N2O5 (k-1)
→
NO2 + NO3
→
NO2 + O2 + NO (K2)
NO + N2O5
→ NO2 + NO2 + NO2 (K3)
Give the expression for the acceptable rate.
(A).
d[N₂O]
dt
=
-1
2k,k₂[N205]
k₁+k₂
d[N₂O5]
(B).
dt
=-k₁[N₂O₂] + k₁[NO2][NO3] - k₂[NO2]³
(C).
d[N₂O]
dt
=-k₁[N₂O] + k₁[N205] - K3 [NO] [N205]
(D).
d[N2O5]
=-k₁[NO] - K3[NO] [N₂05]
dt
A 0.10 M solution of acetic acid (CH3COOH, Ka = 1.8 x 10^-5) is titrated with a 0.0250 M solution of magnesium hydroxide (Mg(OH)2). If 10.0 mL of the acid solution is titrated with 20.0 mL of the base solution, what is the pH of the resulting solution?
For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3)
→
Give the expression for the acceptable rate.
→
→
(A).
d[N205]
dt
==
2k,k₂[N₂O₂]
k₁+k₁₂
(B).
d[N2O5]
=-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³
dt
(C).
d[N2O5]
=-k₁[N₂O] + k [NO] - k₂[NO] [NO]
d[N2O5]
(D).
=
dt
= -k₁[N2O5] - k¸[NO][N₂05]
dt
Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard
the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.
Chapter 13 Solutions
Chemistry: The Science in Context (Fourth Edition)
Ch. 13.2 - Prob. 1PECh. 13.2 - Prob. 2PECh. 13.2 - Prob. 3PECh. 13.2 - Prob. 4PECh. 13.3 - Prob. 5PECh. 13.3 - Prob. 6PECh. 13.6 - Prob. 7PECh. 13.6 - Prob. 8PECh. 13.7 - Prob. 9PECh. 13.7 - Prob. 10PE
Ch. 13.7 - Prob. 11PECh. 13.8 - Prob. 12PECh. 13.8 - Prob. 13PECh. 13 - Prob. 13.1VPCh. 13 - Prob. 13.2VPCh. 13 - Prob. 13.3VPCh. 13 - Prob. 13.4VPCh. 13 - Prob. 13.5VPCh. 13 - Prob. 13.6VPCh. 13 - Prob. 13.7VPCh. 13 - Prob. 13.8VPCh. 13 - Prob. 13.9VPCh. 13 - Prob. 13.10VPCh. 13 - Prob. 13.11VPCh. 13 - Prob. 13.12VPCh. 13 - Prob. 13.13VPCh. 13 - Prob. 13.14VPCh. 13 - Prob. 13.15VPCh. 13 - Prob. 13.16VPCh. 13 - Prob. 13.17QPCh. 13 - Prob. 13.18QPCh. 13 - Prob. 13.19QPCh. 13 - Prob. 13.20QPCh. 13 - Prob. 13.21QPCh. 13 - Prob. 13.22QPCh. 13 - Prob. 13.23QPCh. 13 - Prob. 13.24QPCh. 13 - Prob. 13.25QPCh. 13 - Prob. 13.26QPCh. 13 - Prob. 13.27QPCh. 13 - Prob. 13.28QPCh. 13 - Prob. 13.29QPCh. 13 - Prob. 13.30QPCh. 13 - Prob. 13.31QPCh. 13 - Prob. 13.32QPCh. 13 - Prob. 13.33QPCh. 13 - Prob. 13.34QPCh. 13 - Prob. 13.35QPCh. 13 - Prob. 13.36QPCh. 13 - Prob. 13.37QPCh. 13 - Prob. 13.38QPCh. 13 - Prob. 13.39QPCh. 13 - Prob. 13.40QPCh. 13 - Prob. 13.41QPCh. 13 - Prob. 13.42QPCh. 13 - Prob. 13.43QPCh. 13 - Prob. 13.44QPCh. 13 - Prob. 13.45QPCh. 13 - Prob. 13.46QPCh. 13 - Prob. 13.47QPCh. 13 - Prob. 13.48QPCh. 13 - Prob. 13.49QPCh. 13 - Prob. 13.50QPCh. 13 - Prob. 13.51QPCh. 13 - Prob. 13.52QPCh. 13 - Prob. 13.53QPCh. 13 - Prob. 13.54QPCh. 13 - Prob. 13.55QPCh. 13 - Prob. 13.56QPCh. 13 - Prob. 13.57QPCh. 13 - Prob. 13.58QPCh. 13 - Prob. 13.59QPCh. 13 - Prob. 13.60QPCh. 13 - Prob. 13.61QPCh. 13 - Prob. 13.62QPCh. 13 - Prob. 13.63QPCh. 13 - Prob. 13.64QPCh. 13 - Prob. 13.65QPCh. 13 - Prob. 13.66QPCh. 13 - Prob. 13.67QPCh. 13 - Prob. 13.68QPCh. 13 - Prob. 13.69QPCh. 13 - Prob. 13.70QPCh. 13 - Prob. 13.71QPCh. 13 - Prob. 13.72QPCh. 13 - Prob. 13.73QPCh. 13 - Prob. 13.74QPCh. 13 - Prob. 13.75QPCh. 13 - Prob. 13.76QPCh. 13 - Prob. 13.77QPCh. 13 - Prob. 13.78QPCh. 13 - Prob. 13.79QPCh. 13 - Prob. 13.80QPCh. 13 - Prob. 13.81QPCh. 13 - Prob. 13.82QPCh. 13 - Prob. 13.83QPCh. 13 - Prob. 13.84QPCh. 13 - Prob. 13.85QPCh. 13 - Prob. 13.86QPCh. 13 - Prob. 13.87QPCh. 13 - Prob. 13.88QPCh. 13 - Prob. 13.89QPCh. 13 - Prob. 13.90QPCh. 13 - Prob. 13.91QPCh. 13 - Prob. 13.92QPCh. 13 - Prob. 13.93QPCh. 13 - Prob. 13.94QPCh. 13 - Prob. 13.95QPCh. 13 - Prob. 13.96QPCh. 13 - Prob. 13.97QPCh. 13 - Prob. 13.98QPCh. 13 - Prob. 13.99QPCh. 13 - Prob. 13.100QPCh. 13 - Prob. 13.101QPCh. 13 - Prob. 13.102QPCh. 13 - Prob. 13.103QPCh. 13 - Prob. 13.104QPCh. 13 - Prob. 13.105QPCh. 13 - Prob. 13.106QPCh. 13 - Prob. 13.107QPCh. 13 - Prob. 13.108QPCh. 13 - Prob. 13.109QPCh. 13 - Prob. 13.110QPCh. 13 - Prob. 13.111QPCh. 13 - Prob. 13.112QPCh. 13 - Prob. 13.113QPCh. 13 - Prob. 13.114QPCh. 13 - Prob. 13.115QPCh. 13 - Prob. 13.116QPCh. 13 - Prob. 13.117QPCh. 13 - Prob. 13.118QPCh. 13 - Prob. 13.119QPCh. 13 - Prob. 13.120QPCh. 13 - Prob. 13.121QPCh. 13 - Prob. 13.122QPCh. 13 - Prob. 13.123QPCh. 13 - Prob. 13.124QPCh. 13 - Prob. 13.125QPCh. 13 - Prob. 13.126QPCh. 13 - Prob. 13.127QPCh. 13 - Prob. 13.128QPCh. 13 - Prob. 13.129QPCh. 13 - Prob. 13.130QPCh. 13 - Prob. 13.131QPCh. 13 - Prob. 13.132QPCh. 13 - Prob. 13.133QPCh. 13 - Prob. 13.134QPCh. 13 - Prob. 13.135APCh. 13 - Prob. 13.136APCh. 13 - Prob. 13.137APCh. 13 - Prob. 13.138APCh. 13 - Prob. 13.139APCh. 13 - Prob. 13.140APCh. 13 - Prob. 13.141APCh. 13 - Prob. 13.142APCh. 13 - Prob. 13.143APCh. 13 - Prob. 13.144APCh. 13 - Prob. 13.145APCh. 13 - Prob. 13.146APCh. 13 - Prob. 13.147APCh. 13 - Prob. 13.148APCh. 13 - Prob. 13.149APCh. 13 - Prob. 13.150APCh. 13 - Prob. 13.151AP
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- For the decomposition reaction of N2O5(g): 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed: N2O5 NO2 + NO3 (K1) | NO2 + NO3 → N2O5 (k-1) | NO2 + NO3 NO2 + O2 + NO (k2) | NO + N2O51 NO2 + NO2 + NO2 (K3) → Give the expression for the acceptable rate. → → (A). d[N205] dt == 2k,k₂[N₂O₂] k₁+k₁₂ (B). d[N2O5] =-k₁[N₂O] + k₁[NO₂] [NO3] - k₂[NO₂]³ dt (C). d[N2O5] =-k₁[N₂O] + k [NO] - k₂[NO] [NO] d[N2O5] (D). = dt = -k₁[N2O5] - k¸[NO][N₂05] dt Do not apply the calculations, based on the approximation of the stationary state, to make them perform correctly. Basta discard the 3 responses that you encounter that are obviously erroneous if you apply the formula to determine the speed of a reaction.arrow_forwardR lactam or lactone considering as weak acid or weak base and whyarrow_forward81. a. Propose a mechanism for the following reaction: OH CH2=CHCHC=N b. What is the product of the following reaction? HO H₂O N=CCH2CH2CH OH HO CH3CCH=CH2 H₂O C=N 82. Unlike a phosphonium ylide that reacts with an aldehyde or a ketone to form an alkene a sulfonium uliaarrow_forward
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