Chemistry, 24.10.2020 16:00 theo39

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Answer from: Syaralopez

I believe the boiling point is 102 degrees Celsius and the freezing point is -2 degrees Celsius (so C).

Explanation:

Gas has a higher entropy than liquid, who has a higher entropy than solid. If a solid melts or a liquid vaporizes, the entropy increases. If a solute is present, it also contributes to the entropy of the mixture. The liquid already has a higher entropy (molecule randomness or disorder) than the pure liquid, so there's less likelihood of forming gas. This, therefore, means that it has a higher boiling point. This randomness also opposes freezing. A lower temperature has to reached before balance between solid and solution reached. So, the freezing point is lower.

Please let me know if I'm incorrect :)

Answer from: Quest

remember that hf is only the energy of one photon. not two. not three. one. 1) 5.12 * 10^3 photons find the energy of one photon, then determine how many photons are needed to reach the sensitivity. e = hf = (6.626 * 10^-34 js)(5.95 * 10^19 hz) = 3.94247 * 10^-14 j per photon (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons round to three sig figs. 2) 5.77 * 10^-17 j energy of pulse = energy per photon * number of photons = hfn = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250) = 5.774559 * 10^-17 j round to three sig figs. 3) 4.07 * 10^8 kj/mol 10^12 pm = 1 m c = fλ f = c/λ e = hf = hc/λ = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol) = 40702571.4 kj/mol round to three sig figs. 4) 9.94 * 10^-1 m e = hc/λ λ = hc / e = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j) = 0.9939 m round to three sig figs. 5) 9.06 * 10^3 photons energy of pulse = energy per photon * number of photons number of photons = energy of pulse / energy per photon = energy of pulse / (hf) = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)] = 9055.23695 photons round to three sig figs. 6) 2.78 * 10^-15 j this is problem 1, except in reverse. sensitivity = minumum number of photons * energy per photon = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz) = 2.78070758 × 10^-15 j round to three sig figs.

Answer from: Quest

here are some of the ways in which chemical and nuclear reactions differ.

explanation:

$\begin{array}{ll}\qquad\textbf{chemical} & \qquad\textbf{nuclear} \\\text{energy changes $\approx 10^{2}$ kj} & \text{energy changes$\approx 10^{8}$ kj ($10^{6}$ times more}) \\\text{mass is conserved} & \text{mass changes to energy} \\\text{involves electrons} & \text{involves protons and neutrons} \\\text{atoms do not change} & \text{atoms transmute} \\\end{array}$

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