Answer:
0.369 moles of H₂O.
Explanation:
The molar mass of the hydrate salt is equal to the molar mass of the anhydrous salt plus the molar mass of water times the moles of water.
M(MgSO₄·7H₂O) = M(MgSO₄) + 7 × M(H₂O)
M(MgSO₄·7H₂O) = 168 g/mol + 7 × 18 g/mol
M(MgSO₄·7H₂O) = 294 g/mol
Every 294 g of MgSO₄·7H₂O there are 126 g of H₂O. So, for 15.5 g of MgSO₄·7H₂O,

Answer:
The freezing point of the solution is -4.46 °C
Explanation:
Step 1: Data given
Number of moles MgBr2 = 0.80 moles
Mass of water = 1.00 kg
Water has a freezing point depression constant of 1.86°C.kg/mol
Freezing point of water = 0°C
Step 2: Calculate the freezing point of the solution
ΔT = i * Kf * m
⇒ΔT = the freezing point depression = TO BE DETERMINED
⇒with i = the can't Hoff factor of MgBr2 = 3
⇒with Kf = the freezing point depression constant of 1.86°C/molal
⇒with m = the molality = 0.80 moles / / 1.00 kg = 0.80 molal
ΔT = 3 * 1.86 °C/molal * 0.80 molal
ΔT = 4.46 °C
Step 3: Calculate the freezing point of the solution
0°C - 4.46 °C = -4.46 °C
The freezing point of the solution is -4.46 °C
The ions charge comes from the number of valence electrons. All elements excluding transition metals want to have 8 valence electrons. The first 3 groups will try to lose electrons to get to 8 where as the 5, 6, and 7 ones will try to gain electrons. Hydrogen is to small to get 8 electrons so it settles for 2 like helium because helium is a noble gas. oxygen needs 2 electrons and hydrogen gives one thats hy H is positive H3O is positive and OH is nwgitive.