When the enthalpy value is given, we can calculate how much heat is use or produces in a given equation.
67.6 kCal ---> 67.6 kCal= 1 mol of reaction
1 mol of reaction= 1 mol of CO (based on the coefficient)
so 1 mole of CO gives us 67.6 kCal of heat.
calculation:
1 mol CO
<span>Xe = VIII = 8 valence electrons
F = VII = 4 (7 ve) = 28 valence electrons</span>
total ve = 8 + 28 = 36 ve
<span>36 - 4(2) = 28 ve
(there are 2 electrons in each bond x 4 bonds)</span>
<span>28 - 4(6) = 4
(We assign the remaining electrons to F atoms)</span>
<span>4 - 2(2) = 0
(Therefore 4 electrons left => we have 2 lone pairs)</span>
The steric number = No. of
σ bonds + #lone pairs
= 4 σ bonds + 2 lone pairs
= 6 => d²sp³ (6 hybrid orbitals)
<span>4 bonds + 2 lone pairs
=> square planar</span>
Note that
The heating value of standard coal is about 30,080 kJ/kg
1 L of water has a mass of 1.0 kg
The mass of 15 L of water = 15 kg.
The latent heat of vaporization of water is about 2260 kJ/kg,
The energy required to boil 15 L of water is
(2260 kJ/kg)*(15 kg) = 33900 kJ
The mass of coal required to provide this energy is
(33900 kJ)/(30080 kJ/kg) = 1.127 kg
Because 1 kg = 2.205 lb, the mass of coal required is
(1.127 kg)*(2.205 lb/kg) = 2.485 lb
Answer: 2.49 lb (nearest hundredth)