Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 19, Problem 19.32QE

(a)

Interpretation Introduction

Interpretation:

Formula of hexaaquachromium(III) hexacyanoferrate(III) complex has to be given.

Concept Introduction:

Nomenclature of coordination compounds:

Rules of Union of Pure and Applied Chemistry (IUPAC) system of nomenclature are,

  • In coordination compounds, cation name comes first then anion. Name of ligands are given in the order of alphabetical then metal cation followed by the oxidation state in Roman numeral in parentheses then the anions are named.
  • If one type of ligand is comes more times means it can be named with prefix: (1) mono- (usually omitted), (2) di-, (3) tri-, (4) tetra flowed by its name.
  • Metal in the anionic complex is named as –ate in its suffix and negative ligands are named as –ato in its suffix.

(b)

Interpretation Introduction

Interpretation:

Formula of bromochlorobis(ethylenediamine)cobalt(III) complex has to be given.

Concept Introduction:

Refer to part (a)

(c)

Interpretation Introduction

Interpretation:

Formula of carbonylpentacyanocobaltate(III) complex has to be given.

Concept Introduction:

Refer to part (a)

(d)

Interpretation Introduction

Interpretation:

Formula of (diethylenetriamine)trinitrochromium(III) complex has to be given.

Concept Introduction:

Refer to part (a)

(e)

Interpretation Introduction

Interpretation:

Formula of pentaaquathiocyanatoiron(III) complex has to be given.

Concept Introduction:

Refer to part (a)

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Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100
Part IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained. Draw the structures of these fragments.
For each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table  and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.

Chapter 19 Solutions

Chemistry: Principles and Practice

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