Answer is: <span>he boiling point of a 1.5 m aqueous solution of fructose is </span>100.7725°C.
The boiling point
elevation is directly proportional to the molality of the solution
according to the equation: ΔTb = Kb · b.<span>
ΔTb - the boiling point
elevation.
Kb - the ebullioscopic
constant. of water.
b - molality of the solution.
Kb = 0.515</span>°C/m.
b = 1.5 m.
ΔTb = 0.515°C/m · 1.5 m.
ΔTb = 0.7725°C.
Tb(solution) = Tb(water) + ΔTb.
Tb(solution) = 100°C + 0.7725°C = 100.7725°C.
Oxygen and carbon dioxide
Answer:
The mass of water is 36 g.
Explanation:
Mass of hydrogen = 4 g
Mass of water = ?
Solution:
First of all we will write the balance chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of hydrogen = mass / molar mass
Number of moles of hydrogen = 4 g/ 2 g/mol
Number of moles of hydrogen = 2 mol
Now we compare the moles of water with hydrogen from balance chemical equation.
H₂ : H₂O
2 : 2
Mass of water = moles × molar mass
Mass of water = 2 mol × 18 g/mol
Mass of water = 36 g
If the water oxygen is in excess than mass of water would be 36 g.
Answer:
this is a oxidation reaction
product is magnesium oxide
the balanced equation is
2Mg + O2 -> 2MgO
