MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g)
Using the standard enthalpies of formation given in the source below:
(−601.24 kJ) + (2 x −92.30 kJ) − (−641.8 kJ) − (−285.8 kJ) = +141.76 kJ
So:
MgCl2(s) + H2O(l) → MgO(s) + 2 HCl(g), ΔH = +141.76 kJ
Answer:
500g because 300g is less dense than 500g its kinda like a weight
Explanation:
i
Basis: 1 L of the substance.
(1.202 g/mL) x (1000 mL) = 1202 g
mass solute = (1202 g) x 0.2 = 240.2 g
mass solvent = 1202 g x 0.8 = 961.6 g
moles KI = (240.2 g) x (1 mole / 166 g) = 1.45 moles
moles water = (961.6 g) x (1 mole / 18 g) = 53.42 moles
1. Molality = moles solute / kg solvent
= 1.45 moles / 0.9616 kg = 1.5 m
2. Molarity = moles solute / L solution
= 1.45 moles / 1 L solution = 1.45 M
3. molar mass = mole solute / total moles
= 1.45 moles / (1.45 moles + 53.42 moles) = 0.0264
