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Concept explainers
You are given a small bar of an unknown metal X. You find the density of the metal to be 10.5 g/cm3. An X-ray diffraction experiment measures the edge of the face-centered cubic unit cell as 4.09 Å (1 Å = 10−10 m). Identify X.
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Interpretation:
A metal has to be identified given its density and edge length of FCC unit cell.
Concept introduction:
In crystalline solids, the components are packed in regular pattern and neatly stacked. The components are imagined as spheres and closely packed. This phenomenon is called “close packing” in crystals. The two major types of close packing of the spheres in the crystal are – hexagonal close packing and cubic close packing. Cubic close packing structure has face-centered cubic (FCC) unit cell.
In face-centered cubic unit cell, each of the six corners is occupied by every single atom. Each face of the cube is occupied by one atom.
Each atom in the corner is shared by eight unit cells and each atom in the face is shared by two unit cells. Thus the number of atoms per unit cell in FCC unit cell is,
Answer to Problem 55E
Answer
The metal is identified as “Silver”.
Explanation of Solution
Explanation
Calculate the volume of unit cell of unknown metal.
Density of the titanium is given and mass of the titanium is calculated in the previous step. Substituting these two values in the equation
Calculate the mass of unit cell of unknown metal.
Density of the unit cell is given. The mass of unit cell is calculated using the equation
Calculate the atomic mass of unknown metal.
In FCC unit cell of a metal, each unit cell has 4 metal atoms. So each metal atom has an average mass of 1/4th of the mass of unit cell. The atomic mass of a metal atom corresponds to the Avogadro number of the average mass of a metal atom.
Identify the metal corresponding to the atomic mass calculated.
The identified metal is “Silver”.
Platinum is the metal that possibly matches best with the calculated atomic mass.
Conclusion
Density and edge length of FCC unit cell of a metal is given and it is identified as “Silver”.
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Chapter 10 Solutions
Chemistry with Access Code, Hybrid Edition
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- Part I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward• Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forward
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