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:
0.21 kg
Step-by-step explanation:
Volume of cola = 2.0 L × (33.8 fl oz/1L)
= 67.6 fl oz
Grams of sugar = 67.6 fl oz × (25 g sugar/8.0 fl oz)
= 210 g sugar
Kilograms of sugar = 210 g × (1 kg/1000 g)
= 0.21 kg
According to the periodic table, carbon's molar mass is 12.011 grams per mole (that's the small number under the element). So, just multiply like this to get the answer:

So, there are approximately 0.208 grams in 2.5 moles of carbon.
Answer:
Hexose category and ketohexose category
Explanation:
The classification of the Carbohydrate tagatose by carbonyl group is that it is a monosaccharide and has a hexose structure hence it belongs to the Hexose category
Based on the number of carbon atoms the structure has a ketofunctionality hence it is classified under the ketohexose category
Attached below is the remaining part of the solution
D=m/v ⇒ m=d*v
d=density
m=mass
v=volume
d(ether)=0.71 gr/cm³=0.71 gr/ ml
v=130 ml
m=d*v
m=0.71 gr/ml*(130 ml)=92.3 g
Solution: m=92.3 g