The atomic radius of calcium in its cubic close packing structure is given and its density has to be determined. Concept introduction: In packing of atoms in a crystal structure, the atoms are imagined as spheres. 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, 8 × 1 8 atoms in corners + 6 × 1 2 atoms in faces = 1 + 3 = 4 atoms The edge length of one unit cell is given by a = 2R 2 where a = edge length of unit cell R = radius of atom
The atomic radius of calcium in its cubic close packing structure is given and its density has to be determined. Concept introduction: In packing of atoms in a crystal structure, the atoms are imagined as spheres. 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, 8 × 1 8 atoms in corners + 6 × 1 2 atoms in faces = 1 + 3 = 4 atoms The edge length of one unit cell is given by a = 2R 2 where a = edge length of unit cell R = radius of atom
Solution Summary: The author explains that the atomic radius of calcium in its cubic close packing structure is given and its density has to be determined.
The atomic radius of calcium in its cubic close packing structure is given and its density has to be determined.
Concept introduction:
In packing of atoms in a crystal structure, the atoms are imagined as spheres. 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,
8×18atomsincorners+6×12atomsinfaces=1+3=4atoms The edge length of one unit cell is given bya=2R2where a=edge length of unit cellR=radiusofatom
b) Certain cyclic compounds are known to be conformationally similar to carbohydrates, although they are not
themselves carbohydrates. One example is Compound C shown below, which could be imagined as adopting
four possible conformations. In reality, however, only one of these is particularly stable. Circle the conformation
you expect to be the most stable, and provide an explanation to justify your choice. For your explanation to be
both convincing and correct, it must contain not only words, but also "cartoon" orbital drawings contrasting the
four structures.
Compound C
Possible conformations (circle one):
Дет
Lab Data
The distance entered is out of the expected range.
Check your calculations and conversion factors.
Verify your distance. Will the gas cloud be closer to the cotton ball with HCI or NH3?
Did you report your data to the correct number of significant figures?
- X
Experimental Set-up
HCI-NH3
NH3-HCI
Longer Tube
Time elapsed (min)
5 (exact)
5 (exact)
Distance between cotton balls (cm)
24.30
24.40
Distance to cloud (cm)
9.70
14.16
Distance traveled by HCI (cm)
9.70
9.80
Distance traveled by NH3 (cm)
14.60
14.50
Diffusion rate of HCI (cm/hr)
116
118
Diffusion rate of NH3 (cm/hr)
175.2
175.2
How to measure distance and calculate rate
For the titration of a divalent metal ion (M2+) with EDTA, the stoichiometry of the reaction is typically:
1:1 (one mole of EDTA per mole of metal ion)
2:1 (two moles of EDTA per mole of metal ion)
1:2 (one mole of EDTA per two moles of metal ion)
None of the above
Chapter 9 Solutions
Bundle: Chemistry: An Atoms First Approach, Loose-leaf Version, 2nd + OWLv2 with Student Solutions Manual, 4 terms (24 months) Printed Access Card
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Unit Cell Chemistry Simple Cubic, Body Centered Cubic, Face Centered Cubic Crystal Lattice Structu; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=HCWwRh5CXYU;License: Standard YouTube License, CC-BY