Given the standard heats of reaction Reaction ∆H 0

M(s) + 2 X2(g) → MX4(g) −98.7 kJ/mol

X2(g) → 2 X(g) +327.3 kJ/mol

M(g) → M(s) −20.1 kJ/mol

calculate the average bond energy for a single

M-X bond.

Answer in units of kJ/mol.

Explanation:

M(s) → M (g ) + 20.1 kJ --- ( 1 )

X₂ ( g ) → 2X (g ) + 327.3 kJ ( 2 )

M( s) + 2 X₂(g) → M X₄ (g ) - 98.7 kJ ( 3 )

( 3 ) - 2 x ( 2 ) - ( 1 )

M( s) + 2 X₂(g) - 2 X₂ ( g ) - M(s)  → M X₄ (g ) - 98.7 kJ -  2 [ 2X (g ) + 327.3 kJ ] - M (g ) - 20.1 kJ

0 = M X₄ (g ) - 4 X (g ) - M (g ) - 773.4 kJ

4 X (g ) +  M (g ) =  M X₄ (g ) - 773.4kJ

heat of formation of M X₄ (g ) is - 773.4 kJ

Bond energy of one M - X bond =  773.4 / 4 =  193.4 kJ / mole

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3

ok whats the question if not i cannot you i need question and need a thing to answer