Lead crystallizes in a face centered cubic unit cell structure and has a densiy of 11.40 g/cm3 . Calculate the atomic radius of lead in picometers. Molar mass of lead is 207.2 g/mol

Atomic radius of lead = R = (1.75 × 10⁻¹⁰) m = 175 pm

Explanation:

For a crystalline structure,

ρ = (nA)/(V꜀Nₐ)

where

ρ = density = 11.40 g/cm³ = 11400 kg/m³ for Lead

n = number of atoms per unit cell = 4 atoms (for an FCC crystal)

A = Atomic number = 207.2 g/mol = 0.2072 kg/mol

V꜀ = Volume of the unit cell = ?

Nₐ = Avogadro's constant = (6.022 × 10²³) atoms per mol

V꜀ = (nA)/(ρNₐ)

V꜀ = (4×0.2072)/(11400×6.022×10²³)

V꜀ = (1.207 × 10⁻²⁸) m³

But, V꜀ = (edge length of the unit cell)³ = a³

a³ = (1.207 × 10⁻²⁸)

a = (4.9424058 × 10⁻¹⁰) m

But for an FCC crystal, the radius of atom is related to the edge length through,

a = 2R√2

R = a/(2√2)

R = (4.9424058 × 10⁻¹⁰)/(2√2)

R = (1.7474043 × 10⁻¹⁰) m = (1.75 × 10⁻¹⁰) m = 175 pm

the answer is a

photosynthesis

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.