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:
Frequency of spinal trait in the population = 0.11
Explanation:
Given:
Total spineless trait = 600
Number of spinal trait = 80
Find:
Frequency of spinal trait in the population
Computation:
Total population = Total spineless trait + Number of spinal trait
Total population = 600 + 80
Total population = 680
Frequency of spinal trait in the population = Number of spinal trait / Total population
Frequency of spinal trait in the population = 80 / 680
Frequency of spinal trait in the population = 0.11
Answer:
Explanation:
Hello,
In this since one mole equals 6.022x10²³ particles of silver by means of the Avogadro's number, we can compute the moles in 4.27x10²² particles as shown below:
Best regards.
Answer:
409 g/mol
Explanation:
you divide the mass by the moles to get the molar mass
Answer:
Find the domain and the range of the following:
x y
3 2
5 7
1 4
9 2
3 7
Explanation:
