Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Explore thousands of questions and answers from knowledgeable experts in various fields on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
The unit cell volume of the crystal is [tex]V_c = 9.9084*10^-^2^3 cm^3 / unit cell[/tex] and the density of titanium is calculated as 4.71g/cm^3 while the literature value is 4.5 g/cm^3
Data;
- radius = 0.1445 nm
- c/a = 1.58
- A = 46.88 g/mol
Unit Cell Volume
The unit cell volume can be calculated as
[tex]V_c = 6R^2c\sqrt{3}\\[/tex]
let's substitute the values into the formula
[tex]V_c = 6 * (1.445*10^-^8)^2 * 1.58 8 * a * \sqrt{3} \\a = 2R\\V_c = 6 * (1.445*10^-^8)^2 * 1.58* 2 * 1.445*10^-^8 * \sqrt{3}\\ V_c = 9.984*10^-^2^3 cm^3 / unit cell[/tex]
The unit cell volume of the crystal is [tex]V_c = 9.9084*10^-^2^3 cm^3 / unit cell[/tex]
Density of Ti
The density of titanium can be calculated as
[tex]\rho = \frac{nA}{V_c N_a}[/tex]
- n = 6 for hcp
- A = 46.88 g/mol
- Na = Avogadro's number
let's substitute the values into the formula
[tex]\rho = \frac{nA}{V_c Na}\\ \rho = \frac{6*46.88}{9.9084*10^-^2^3* 6.023*10^2^3} \\\rho = 4.71 g/cm^3[/tex]
The density of titanium is calculated as 4.71g/cm^3 while the literature value is 4.5 g/cm^3
Learn more on crystal lattice here;
https://brainly.com/question/6610542
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.