Answer is: unsaturated.
Solubility of potassium chlorate on 70°C is approximately 30 grams in 100 grams of water.
Solubility of potassium chlorate on 70°C is approximately 10 grams in 100 grams of water.
So if dissolve 15 g of potassium chlorate in 201 g of water, there is less salt than it solubility and solution is unsaturated.
For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:
ΔTf<span> = (K</span>f)(<span>m)(i)
</span>where:
ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
m = molality of solution (mol solute/kg solvent) = ?
i = ions in solution = 3
Computing for molality:
Molar mass of MgBr2 = 184.113 g/mol
m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg
For the problem,
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf
Tf = -1.59 celsius
I believe is different in pressure
Answer:
50.76 mol H2O.
Explanation:
The photosynthesis follows the equation:
6CO2 + 6H2O ---> C6H12O6 + 6O2
This means that 6 mol of H2O are needed to obtain 1 mol of C6H12O6 (see the numbers that precedes every molecule to know how many mols are in game).
So we can say that:
1 mol C6H12O6 --------- 6 mol H2O
8.46 mol C6H12O6 -----x= 8.46 x 6 : 1 = 50.76 mol H20
Answer:
compound is formed .............
