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Science of Materials

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TALLER N° 2
1. El papel aluminio que se usa para almacenar alimentos pesa unos 0.3g por pulgada cuadrada. ¿Cuántos átomos de aluminio contiene una pulgada cuadrada de ese papel?
 Masa aluminio: 0.3 g
 Peso atómico: 26.98 g/mol n° moles= 0.3/26.98 = 0.011 moles n° átomos= (0.011*6.023x10^23) = 6.697x10^21 átomos
Rta: una pulgada cuadrada de ese papel aluminio contiene 6.69x10^21 átomos
2. Calcule la cantidad de átomos de hierro en una tonelada corta (2000lb) de hierro.
 Masa de hierro: 2000 lb= 907184,74 g
 Peso atómico Fe:55,845 g/mol n° moles= 907184,74/55,845 = 16244,69 moles n° átomos= (16244,69*6,023x10^23) = 9,7842x10^27 átomos
Rta: una tonelada corta de hierro tiene 9,7842x10^27 átomos
3. Para niquelar una parte de acero de 200in², de superficie, con una capa de 0.002in de espesor de níquel: ¿Cuántos átomos de Níquel se requieren? ¿Cuántos moles de Níquel se requieren?
 Volumen: 0,4 in³ =6,55 cm³
 ρNi: 8,9 g/cm³
 peso atómico Ni: 58,69 g/mol
Gramos de Ni= 6,55 cm³*8,9 g/cm³ = 58,295 g n° moles= 58,295/58,69 = 0,9933 moles n° átomos= (0,9933*6,023x10^23) = 5,9825x10^23 átomos
Rta: se requieren 5,9825x10^23 átomos y 0,9933 moles
4. Un alambre de Oro tiene 0.70mm de diámetro y 8.0cm de largo, la densidad del
Oro es 19.3g/cm³. ¿Cuántos átomos contiene este alambre?
 Peso atómico Au: 196,96 g/mol
 ρAu= 19,3 g/cm³
V= s*h
S= (0,035)²*(3,1416)= 3,8485x10^-3 cm² v= 3,8485x10^-3*(8) = 0,03079 cm³
Gramos de Au = 19,3*0,0379 =0,5942 g n° moles= 0,5942/196,96 =3,0169x10^-3 moles n° átomos= (3,0169x10^-3*6,023x10^23) = 1,3171^21 átomos
Rta: El alambre contiene 1,3171^21 átomos
5. Escriba los números cuánticos de un elemento que tiene valencia 2 y número atómico 27. ¿Cuántos electrones debe haber en el subnivel 3d?
Configuración electrónica z= 27: 1s2 2s2 2p6 3s2 3p6 4s2 3d7
Configuración Electrón Nivel [n] Subnivel [l] Orbital [ml] Spin [ms]
1s2 1 1 0 0 ½
2 1 0 0 -½
2s2 3 2 0 0 ½
4 2 0 0 -½
2p6 5 2 1 -1 ½
6 2 1 -1 -½
7 2 1 0 ½
8 2 1 0 -½
9 2 1 1 ½
10 2 1 1 -½
3s2 11 3 0 0 ½
12 3 0 0 -½
3p6 13 3 1 -1 ½
14 3 1 -1 -½
15 3 1 0 ½
16 3 1 0 -½
17 3 1 1 ½
18 3 1 1 -½
3d7 19 3 2 -2 ½
20 3 2 -2 -½
21 3 2 -1 ½
22 3 2 -1 -½
23 3 2 0 ½
24 3 2 1 ½
25 3 2 2 ½
4s2 26 4 0 0 ½
27 4 0 0 -½
Rta: debe haber 7 electrones en el subnivel 3d.
6. Escriba la configuración electrónica del Indio Z=49. ¿Cuántos niveles de energía tiene y cuál es su valencia?
Configuración electrónica: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p1
Rta: El Indio tiene 5 niveles de energía y si valencia es 3.
7. Escriba los números cuánticos de cada uno de los electrones del Cobre en la capa
M.
Z Cu=29 c.e.= 1s2 2s2 2p6 3s2 3p6 4s2 3d9
8. En los metales la carga eléctrica se transfiere con el movimiento de los electrones de valencia. ¿Cuántos portadores potenciales de carga hay en un alambre de
Aluminio de 1mm de diámetro y 100m de longitud?
 V alambre: 78.5398 cm3
 ρ Al: 2.6 g/cm3
Configuración Electrón Nivel [n] Subnivel [l] Orbital [ml] Spin [ms]
3s2 11 3 0 0 ½
12 3 0 0 -½
3p6 13 3 1 -1 ½
14 3 1 -1 -½
15 3 1 0 ½
16 3 1 0 -½
17 3 1 1 ½
18 3 1 1 -½
3d9 19 3 2 -2 ½
20 3 2 -2 -½
21 3 2 -1 ½
22 3 2 -1 -½
23 3 2 0 ½
24 3 2 0 -½
25 3 2 1 ½
26 3 2 1 -½
27 3 2 2 ½
 Peso atómico: 26,9815 g/mol
 Z= 13
Portadores potenciales de carga son los electrones de valencia.
C.e.= 1s2 2s2 sp6 3s2 3p1 3 electrones de valencia.
Gramos de Al= 78.5398*2.6 = 204.2035 g nº átomos= (204.2035*6.023x1023)/26.9815 =4.5584x1024 átomos e- Valencia =(4.5584x1024)(3) =1.3675x1025 e- de Valencia.
Rta: Hay 1.3675x1025portadores potenciales de carga en el alambre.
9. Si el radio atómico del plomo vale 0.175 nm, calcular el volumen de la celdilla unidad en metros cúbicos.
 Estructura de Pb: FCC
Radio atómico [m]= 0.175*(1x10-9) =1.75x10-10 m
V FCC =
=1.2127x10-28 m3
Rta: El volumen de La celda unitaria de Pb es 1.2127x10-28 m3
10. Demuestre que el factor de empaquetamiento atómico para la celda BCC es 0,68.
Factor empaquetamiento=
Factor empaquetamiento=
=
=0.68
11. Demuestre que el factor de empaquetamiento atómico para la celda FCC es 0,74
Factor empaquetamiento=
=
=0.74
12. Calcular el radio de un átomo de Paladio (Pd), estructura FCC, con densidad de
12 gr/cm3, y un peso atómico de 106,4 gr/mol.
 Volumen de celda =
Rta: El radio atómico del paladio es
13. En el diagrama se muestran el peso atómico, la densidad y el radio atómico de tres hipotéticas aleaciones. Determinar para cada una si su estructura cristalina es
FCC, BCC o CS y justifíquelo.
Densidad =
A. Densidad
= 6.40 g/cm3
B. Densidad
= 12.3 g/cm3
C. Densidad
= 9.6 g/cm3
Rta: La estructura de la aleación A es BCC, la estructura de la aleación B es CS y la estructura de la aleación C es BCC.
14. Determinar los índices de las direcciones mostradas en la siguiente celdilla unidad de tipo cúbico:
_
 A= (0,1,1)-(1,0,1)= (-1,0,1) =[1 1 0]
 B= (0,½,½)-(0,0,0) =(0,½, ½)*2= [0 1 1]
_ _
 C= (1,0,0)-(1,½,1)= (0,-½,-1)*2=[0 1 2]
_
 D=(1,0,½)-(½,1,0 )= (½,-1,½)*2= [1 2 1]
15. Determinar los índices de los planos mostrados en la siguiente celdilla unidad de tipo cúbico:
 A: x=½, y=½, z= ∞ plano ( 2 2 0)
 B: x=1, y=½, z= ½ plano (1 2 2)
16. Calcular y comparar las densidades lineales en las direcciones [100], [110], [111] en una celda FCC ρ =
 ρ[1 0 0] =
 ρ[1 1 0] =
 ρ[1 1 1] =
17. Calcular y comparar las densidades planares en las direcciones (100), (110) en una celda FCC
Plano ρ=
=

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