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
Answer:
Sucrose: glucose and fructose
Explanation:
<em>What monosaccharides will result from the hydrolysis of sucrose?</em>
<em>Sucrose</em> is a <em>disaccharide</em> composed of 2 different <em>monosaccharides</em>: glucose and fructose joining by a 1 ⇒ 2 bond. These monosaccharides will be released upon the hydrolysis of sucrose.
<em>What monosaccharide will result from the hydrolysis of starch?</em>
<em>Starch</em> is a <em>polysaccharide</em> composed of numerous glucose monomers joined by glycosidic bonds (1 ⇒ 4 and 1 ⇒ 6). These monosaccharides will be released upon the hydrolysis of starch.
Answer:
An exothermic process releases heat, causing the temperature of the immediate surroundings to rise. An endothermic process absorbs heat and cools the surroundings.
