Solid State

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the answers

? H =? U +? n_gRT

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

Since, atom B forms CCP structure. Therefore, there will be 4-B atoms.

Now, atom A occupies 1/3 of tetrahedral voids.

Hence, number of A atoms = 1/3 * 8 = 8/3

The correct formula of the compound = A? /? B?

= A? B?

x + y = 2 + 3 = 5

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2 months ago

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Vishal Baghel

Contributor-Level 10

n (O? ) = 4/32 = 1/8 mol
n (H? ) = 2/2 = 1 mol
n (Total) = n (O? ) + n (H? ) = 1/8 + 1 = 9/8 mol
PV = nRT
P (Total) * 1 = (9/8) * 0.082 * 273
P (Total) = 25.18 atm

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2 months ago

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Vishal Baghel

Contributor-Level 10

Λ°m (NaCl) = 126.45Scm² mol? ¹
Λ°m (HCl) = 426.16Scm² mol? ¹
Λ°m (CH? COONa) = 91Scm² mol? ¹
Λ°m (CH? COOH) =Λ°m (CH? COONa) +Λ°m (HCl)−Λ°m (NaCl)
= 91 + 426.16 – 126.45
= 391.72Scm² mol? ¹

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New answer posted

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V
Vishal Baghel

Contributor-Level 10

Both statements are correct independently but the reason is not the correct reason of the assertion.

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R
Raj Pandey

Contributor-Level 9

Edge length in bcc, a? = 27 Å
Let, Edge length in fcc be a? Å
Now, the same element crystallises in bcc as well as fcc.
For bcc: 4r = √3 a? ⇒ r = (√3 / 4) a?
For fcc: 4r = √2 a? ⇒ r = a? / (2√2)
So, (√3 / 4) a? = a? / (2√2)
(√3 / 4) * 27 = a? / (2√2)
a? = 33.13 Å
The nearest integer is 33.

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2 months ago

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A
alok kumar singh

Contributor-Level 10

First, the number of unit cells in the given mass is determined. In an HCP structure, there are 6 atoms and 18 total voids (6 octahedral + 12 tetrahedral) per unit cell. Multiplying the number of unit cells by 18 gives the total number of voids. The result is14.9 x 10²¹, which is rounded.

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2 months ago

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Raj Pandey

Contributor-Level 9

Covalent solids have high melting point due to strong covalent bonds, also they are insulator in both solid as well as in molten state.

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R
Raj Pandey

Contributor-Level 9

For an acidic buffer solution, pH = pKa + log ( [Base]/ [Acid]).

Given pH = 5.74 and pKa = 4.74.

5.74 = 4.74 + log ( [Base]/1).

1 = log ( [Base]).

[Base] = 10M.

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