The density of solid ag is 10.5 g/cm3. how many atoms are present per cubic centimeter of ag?

as a solid, ag adopts a face-centered cubic unit cell. how many unit cells are present per cubic centimeter of ag?

what is the volume of a unit cell of this metal?

what is the edge length of a unit cell of ag?

break it down to me step-by-step, so i can understand it.

Unit cells which are present per cubic centimeter of Ag =

Volume =

Edge length =

Explanation:

(a)

Given that:-

The density of the solid Ag = 10.5 g/cm³

Molar mass of silver = 107.8682 g/mol

So, Moles present per cm³ of Ag = =0.0973 mol/cm³

Also, 1 mole = atoms.

So,

Atoms present per cm³ of Ag =

Thus, answer =

In FCC, the number of atoms  in the unit cell = 4 unit cells

So,

Unit cells which are present per cubic centimeter of Ag =

Unit cells which are present per cubic centimeter of Ag =

(b)

The reciprocal of the unit cell/cm³ is the volume of the unit cell.

So,

Volume =

(c)

Also, Volume =

Thus, edge length = =

Edge length =

it should be all of them because it asks if you included all of those in your answer

explanation:

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol