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:
75603.86473 K
Explanation:
Given that:
The 1st excited electronic energy level of He atom = 3.13 × 10⁻¹⁸ J
The objective of this question is to estimate the temperature at which the ratio of the population will be 5.0 between the first excited state to the ground state.
The formula for estimating the ratio of population in 1st excited state to the ground state can be computed as:

From the above equation:
Δ E = energy difference = 3.13 × 10⁻¹⁸ J
k = Boltzmann constant = 1.38 × 10⁻²³ J/K

Thus:





T = 75603.86473 K
The 4:00 time because later that afternoon would be later in the day
Nitrogen N I think would be the answer
Hi.
I did some digging and I think I found what you're looking for.
I found this on Q(uizlet)
basic or acidic conditions and the reactants must be heated.
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