EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM
EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM
8th Edition
ISBN: 9780134554433
Author: CORWIN
Publisher: PEARSON CO
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Chapter 12, Problem 10E
Interpretation Introduction

(a)

Interpretation:

Whether the representative particle in acetone is an atom, formula unit or molecule is to be stated.

Concept introduction:

The covalent bond is formed when two non-metals are combined by sharing the valence electrons. The particle that is held together by a covalent bond is known as a molecule. This can be identified when there are two non-metals in an atom.

Interpretation Introduction

(b)

Interpretation:

Whether the representative particle in cobalt is an atom, formula unit or molecule is to be stated.

Concept introduction:

The atom is the basic unit of a particle that cannot be broken by any chemical means. The atom shows the property of an element. Every state is composed of a neutral or an ionized atom.

Interpretation Introduction

(c)

Interpretation:

Whether the representative particle in magnetite is an atom, formula unit or molecule is to be stated.

Concept introduction:

In an ionic bond, the cation is attracted to an anion. This attraction is due to the opposite charges on both the ions. The particle that is held together by an ionic bond is known as formula unit. This can be identified when the particle has a metal and a non-metal.

Interpretation Introduction

(d)

Interpretation:

Whether the representative particle in phosphorus is an atom, formula unit or molecule is to be stated.

Concept introduction:

The covalent bond is formed when two non-metals are combined by sharing the valence electrons. The particle that is held together by a covalent bond is known as a molecule. This can be identified when there are two non-metals in an atom.

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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.
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.
R lactam or lactone considering as weak acid or weak base and why

Chapter 12 Solutions

EP INTRODUCTORY CHEM.-MOD.MASTERINGCHEM

Ch. 12 - Prob. 11CECh. 12 - Prob. 12CECh. 12 - Prob. 1KTCh. 12 - Prob. 2KTCh. 12 - Prob. 3KTCh. 12 - Prob. 4KTCh. 12 - Prob. 5KTCh. 12 - Prob. 6KTCh. 12 - Prob. 7KTCh. 12 - Prob. 8KTCh. 12 - Prob. 9KTCh. 12 - Prob. 10KTCh. 12 - Prob. 11KTCh. 12 - Prob. 12KTCh. 12 - Prob. 13KTCh. 12 - Prob. 14KTCh. 12 - Prob. 15KTCh. 12 - Prob. 16KTCh. 12 - Prob. 17KTCh. 12 - Prob. 18KTCh. 12 - Prob. 19KTCh. 12 - Prob. 20KTCh. 12 - Prob. 21KTCh. 12 - Prob. 22KTCh. 12 - Prob. 23KTCh. 12 - Prob. 24KTCh. 12 - Prob. 25KTCh. 12 - Prob. 26KTCh. 12 - Prob. 27KTCh. 12 - Prob. 28KTCh. 12 - Prob. 29KTCh. 12 - Prob. 1ECh. 12 - Prob. 2ECh. 12 - Prob. 3ECh. 12 - Prob. 4ECh. 12 - Prob. 5ECh. 12 - Prob. 6ECh. 12 - Prob. 7ECh. 12 - Prob. 8ECh. 12 - Prob. 9ECh. 12 - Prob. 10ECh. 12 - Prob. 11ECh. 12 - Prob. 12ECh. 12 - Prob. 13ECh. 12 - Prob. 14ECh. 12 - Prob. 15ECh. 12 - Prob. 16ECh. 12 - Prob. 17ECh. 12 - Prob. 18ECh. 12 - Prob. 19ECh. 12 - Prob. 20ECh. 12 - Prob. 21ECh. 12 - Prob. 22ECh. 12 - Prob. 23ECh. 12 - Prob. 24ECh. 12 - Prob. 25ECh. 12 - Prob. 26ECh. 12 - Prob. 27ECh. 12 - Prob. 28ECh. 12 - Prob. 29ECh. 12 - Prob. 30ECh. 12 - Prob. 31ECh. 12 - Prob. 32ECh. 12 - Prob. 33ECh. 12 - Prob. 34ECh. 12 - Prob. 35ECh. 12 - Prob. 36ECh. 12 - Prob. 37ECh. 12 - Prob. 38ECh. 12 - Prob. 39ECh. 12 - Prob. 40ECh. 12 - Prob. 41ECh. 12 - Prob. 42ECh. 12 - Prob. 43ECh. 12 - Prob. 44ECh. 12 - Prob. 45ECh. 12 - Prob. 46ECh. 12 - Prob. 47ECh. 12 - Prob. 48ECh. 12 - Prob. 49ECh. 12 - Prob. 50ECh. 12 - Prob. 51ECh. 12 - Prob. 52ECh. 12 - Prob. 53ECh. 12 - Prob. 54ECh. 12 - Prob. 55ECh. 12 - Prob. 56ECh. 12 - Prob. 57ECh. 12 - Prob. 58ECh. 12 - Prob. 59ECh. 12 - Prob. 60ECh. 12 - Prob. 61ECh. 12 - Prob. 62ECh. 12 - Prob. 63ECh. 12 - Prob. 64ECh. 12 - Prob. 65ECh. 12 - Prob. 66ECh. 12 - Prob. 67ECh. 12 - Prob. 68ECh. 12 - Prob. 69ECh. 12 - Prob. 70ECh. 12 - Prob. 71ECh. 12 - Prob. 72ECh. 12 - Prob. 73ECh. 12 - Prob. 74ECh. 12 - Prob. 75ECh. 12 - Prob. 76ECh. 12 - Prob. 77ECh. 12 - Prob. 78ECh. 12 - Prob. 79ECh. 12 - Prob. 80ECh. 12 - Prob. 81ECh. 12 - Prob. 82ECh. 12 - Prob. 83ECh. 12 - Prob. 84ECh. 12 - Prob. 85ECh. 12 - Prob. 86ECh. 12 - Prob. 87ECh. 12 - Prob. 88ECh. 12 - Prob. 89ECh. 12 - Prob. 90ECh. 12 - Prob. 91ECh. 12 - Prob. 92ECh. 12 - Prob. 93ECh. 12 - Prob. 94ECh. 12 - Prob. 95ECh. 12 - Prob. 96ECh. 12 - Prob. 97ECh. 12 - Prob. 98ECh. 12 - Prob. 99ECh. 12 - Prob. 100ECh. 12 - Prob. 1STCh. 12 - Prob. 2STCh. 12 - Prob. 3STCh. 12 - Prob. 4STCh. 12 - Prob. 5STCh. 12 - Prob. 6STCh. 12 - Prob. 7STCh. 12 - Prob. 8STCh. 12 - Prob. 9STCh. 12 - Prob. 10STCh. 12 - Prob. 11STCh. 12 - Prob. 12STCh. 12 - Prob. 13STCh. 12 - Prob. 14STCh. 12 - Prob. 15STCh. 12 - Prob. 16STCh. 12 - Prob. 17STCh. 12 - Prob. 18ST
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