Answer:
Heating water until it boils. The water particles would gain thermal energy from the heat source and there would be an increase in the kinetic energy between the water particles. During this stage, the temperature of the water particles remains the same as the thermal energy is used to break the strong bonds of attraction between the water particles so that they can be further apart and transition from the liquid state to the gas state.
Density=mass/volume so mass=density/volume mass=500ml/1.43g/ml mass=0.00286
Here are the given:
<span>ΔHf° = –423 kJ/mol
</span> ΔHsub = 119 kJ/mol
IE = 469 kJ/mol
ΔHEA = –301 kJ/mol
BE = 161 kJ/mol
The lattice energy of the compound is solved using the formula:
U = <span>ΔHf° - </span>ΔHsub - BE - IE - ΔHEA
U = -423 - 119 - 161 - 469 - (-301)
U = -871 kJ/mol
Therefore, the lattice energy is 871 kJ/mol (released).
Answer:
39.3%
Explanation:
CaF2 + H2SO4 --> CaSO4 + 2HF
We must first determine the limiting reactant, the limiting reactant is the reactant that yields the least number of moles of products. The question explicitly says that H2SO4 is in excess so CaF2 is the limiting reactant hence:
For CaF2;
Number of moles reacted= mass/molar mass
Molar mass of CaF2= 78.07 g/mol
Number of moles reacted= 11g/78.07 g/mol = 0.14 moles of Calcium flouride
Since 1 mole of calcium fluoride yields two moles of 2 moles hydrogen fluoride
0.14 moles of calcium fluoride will yield 0.14×2= 0.28 moles of hydrogen fluoride
Mass of hydrogen fluoride formed (theoretical yield) = number of moles× molar mass
Molar mass of hydrogen fluoride= 20.01 g/mol
Mass of HF= 0.28 moles × 20.01 g/mol= 5.6 g ( theoretical yield of HF)
Actual yield of HF was given in the question as 2.2g
% yield of HF= actual yield/ theoretical yield ×100
%yield of HF= 2.2/5.6 ×100
% yield of HF= 39.3%