
Concept explainers
Interpretation:
The angles of diffraction of X rays having
Concept introduction:
A unit cell of the crystal is the three-dimensional arrangement of the atoms present in the crystal. The unit cell is the smallest and simplest unit of the crystal which on repetition forms an entire crystal. Unit cell can be a cubic unit cell or hexagonal unit cell. The classification of a unit cell depends on the lattice site occupied by the atoms.

Answer to Problem 21.38E
The angles of diffraction of X rays having
Explanation of Solution
The
Where,
•
•
The Bragg’s equation for diffraction of X rays is given by an expression as shown below.
Where,
•
•
•
•
Substitute equation (1) in equation (2).
The above formula can be written as follows:
The structure of
Substitute
Substitute
Substitute
Substitute
Substitute
Substitute
Thus, the angles of diffraction of X rays at different planes are shown below.
The angles of diffraction of X rays having
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Chapter 21 Solutions
PHYSICAL CHEMISTRY-STUDENT SOLN.MAN.
- 9 7 8 C 9 8 200 190 B 5 A -197.72 9 8 7 15 4 3 0: ང་ 200 190 180 147.52 134.98 170 160 150 140 130 120 110 100 90 90 OH 10 4 3 1 2 -143.04 140. 180 170 160 150 140 130 120 110 100 90 CI 3 5 1 2 141.89 140.07 200 190 180 170 160 150 140 130 120 110 100 ៖- 90 129. 126.25 80 70 60 -60 50 40 10 125.19 -129.21 80 70 3.0 20 20 -8 60 50 10 ppm -20 40 128.31 80 80 70 60 50 40 40 -70.27 3.0 20 10 ppm 00˚0-- 77.17 30 20 20 -45.36 10 ppm -0.00 26.48 22.32 ―30.10 ―-0.00arrow_forwardAssign all the carbonsarrow_forwardC 5 4 3 CI 2 the Righ B A 5 4 3 The Lich. OH 10 4 5 3 1 LOOP- -147.52 T 77.17 -45.36 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm B -126.25 77.03 200 190 180 170 160 150 140 130 120 110 100 90 80 70 60 50 40 30 20 10 ppm 200 190 180 170 160 150 140 130 120 110 100 90 80 TO LL <-50.00 70 60 50 40 30 20 10 ppm 45.06 30.18 -26.45 22.36 --0.00 45.07 7.5 1.93 2.05 -30.24 -22.36 C A 7 8 5 ° 4 3 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 8 5 4 3 ཡི་ OH 10 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 5 4 3 2 that th 7 I 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 115 2.21 4.00 1.0 ppm 6.96 2.76 5.01 1.0 ppm 6.30 1.00arrow_forward
- Curved arrows were used to generate the significant resonance structure and labeled the most significant contribute. What are the errors in these resonance mechanisms. Draw out the correct resonance mechanisms with an brief explanation.arrow_forwardWhat are the: нсе * Moles of Hice while given: a) 10.0 ml 2.7M ? 6) 10.ome 12M ?arrow_forwardYou are asked to use curved arrows to generate the significant resonance structures for the following series of compounds and to label the most significant contributor. Identify the errors that would occur if you do not expand the Lewis structures or double-check the mechanisms. Also provide the correct answers.arrow_forward
- how to get limiting reactant and % yield based off this data Compound Mass 6) Volume(mL Ben zaphone-5008 ne Acetic Acid 1. Sam L 2-propanot 8.00 Benzopin- a col 030445 Benzopin a Colone 0.06743 Results Compound Melting Point (°c) Benzopin acol 172°c - 175.8 °c Benzoping to lone 1797-180.9arrow_forwardAssign ALL signals for the proton and carbon NMR spectra on the following pages.arrow_forward7.5 1.93 2.05 C B A 4 3 5 The Joh. 9 7 8 1 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 7 8 0.86 OH 10 4 3 5 1 2 7.5 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 1.5 1.0 ppm 9 7 8 CI 4 3 5 1 2 7.0 6.5 6.0 5.5 5.0 4.5 4.0 3.5 3.0 2.5 2.0 2.21 4.00 1.5 2.00 2.07 1.0 ppm 2.76arrow_forward
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