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.
Answer:
c. 15 g Kr
Explanation:
The amount of a gas (Moles) is directely proportional to its pressure. That means the higher amount of moles, the highest pressure and vice versa.
Using molar mass of the compounds (Ne=20.2g/mol, Ar = 39.9g/mol, Kr = 83.8g/mol, CO₂ = 44 g/mol and F₂ = 38.0g/mol), moles of 15.0g of each gas are:
Ne = 15g ₓ (1mol / 20.2g) = <em>0.74 moles of Ne</em>
Ar = 15g ₓ (1mol / 39.9g) = <em>0.38 moles of Ar</em>
Kr = 15g ₓ (1mol / 83.8g) = <em>0.18 moles of Kr</em>
CO₂ = 15g ₓ (1mol / 44g) = <em>0.34 moles of CO₂</em>
F₂ = 15g ₓ (1mol / 38g) = <em>0.39 moles of F₂</em>
<h3>As 15g of Kr contains the less quantity of moles, this sample will con have the lowest pressure</h3>
Answer:
because of the location of the 6 the answer is 6 cent
Copper chloride ...............
