If a star contains a certain element, it can be determined by studying the star’s

Calculate the unit cell edge length for an 79 wt% Ag- 21 wt% Pd alloy. All of the palladium is in solid solution, and the crysta

ankoles [38]

Answer:

The edge length is 0.4036 nm

Solution:

As per the question:

Density of Ag, $\rho = 10.49 g/cm^{3}$

Density of Pd, $\rho = 12.02 g/cm^{3}$

Atomic weight of Ag, A = 107.87 g/mol

Atomic weight of Pd, A' = 106.4 g/mol

Now,

The average density, $\rho_{a} = \frac{n A_{avg}} {V_{c}\times N_{A}}$

where

$V_{c} = a^{3}$ = Volume of crystal lattice

a = edge length

n = 4 = no. of atoms in FCC

Therefore,

$\rho_{a} = = \frac{n A_{avg}} {V_{c}\times N_{A}}$

Therefore, the length of the unit cell is given as:

$a = (\frac{nA_{avg}}{\rho_{a}\times N_{a}})^{1/3}$ (1)

Average atomic weight is given as:

$A_{avg} = \frac{100}{\frac{C_{Ag}}{A_{Ag}} + \frac{C_{Pd}}{A_{Pd}}}$

where

$C_{Ag}$ = 79 %

$A_{Ag}$ = 107

$C_{Pd}$ = 21%

$A_{Pd}$ = 106

Therefore,

$A_{avg} = \frac{100}{\frac{79}{107} + \frac{21}{106}} = 106.78$

In the similar way, average density is given as:

$\rho_{a} = \frac{100}{\frac{C_{Ag}}{\rho_{Ag}} + \frac{C_{Pd}}{\rho_{Pd}}}$

$\rho_{a} = \frac{100}{\frac{79}{10.49} + \frac{21}{12.02}} = 10.78 g/cm^{3}$

Therefore, edge length is given by eqn (1) as:

$a = (\frac{4\times 106.78}{10.78\times 6.023 X 10^23})^{1/3} = 4.036\times 10^{- 8} cm = 0.4036\times 10^{- 9} m = 0.4036 nm$

5 0

2 years ago

A 4.6-kg block of ice originally at 263 K is placed in thermal contact with a 15.7-kg block of silver (cAg = 233 J/kg-K) which i

Viktor [21]

Answer:

temperature at 326.44 K system achieve equilibrium

Explanation:

given data

mass of block of ice = 4.6 kg

temperature = 263 K

thermal contact = 15.7-kg

specific heat of silver cAg = 233 J/kg-K

initially temperature = 1052 K

to find out

what temperature will the system achieve equilibrium

solution

first we consider final temperature of the system is T

we know that specific heat of water (C w) = 4186 J(kg K)

and

specific heat of ice ( C i ) = 2030 J/(kg K)

and

latent heat of fusion of ice ( Lf ) = 3.33 × $10^{5}$ J/kg

and we know that system is insulated

so heat lost by silver = heat generated by ice .................1

so we can say

mass of silver × specific heat of silver × ( initial temp - final temp ) = mass of ice × specific heat of ice × ( ice temp ) + mass of ice × latent heat of fusion of ice + mass of ice × specific heat of water × (final temperature )

put here value we get

mass of silver × specific heat of silver × ( initial temp - final temp ) = mass of ice × specific heat of ice × ( ice temp ) + mass of ice × latent heat of fusion of ice + mass of ice × specific heat of water × (final temperature )

15.7 × 233 × ( 1052 - T ) = 4.6 × 2030 × 10 + 4.6 × 3.33 × $10^{5}$ + 4.6 × 4286 × ( T - 273 )

solve we get

T = 326.44 K

so temperature at 326.44 K system achieve equilibrium

7 0

2 years ago

Which state(s) of matter have no definite shape or volume? liquids and solids liquids and gasses gasses solids

Alex777 [14]

Answer:

Liquid has a defined volume but undefined shape

Explanation:

Explanation: