Chemistry, 10.09.2019 01:30 edwardordonez66

In a laboratory experiment, a fermenting aqueous solution of glucose and yeast produces carbon dioxide gas and ethanol. the solution was heated by burning natural gas in a bunsen burner to distill the ethanol that formed in the flask. during the distillation, the ethanol evaporated and then condensed in the receiving flask. the flame of the burner was kept too close to the bottom of the flask and some of the glucose decomposed into a black carbon deposit on the inside of the flask. during this experiment the following changes occurred. which of these changes involved a physical change and not a chemical change? check all that apply. check all that apply.
1-condensation of ethanol
2-evaporation of ethanol
3- formation of carbon dioxide gas from glucose burning of natural gas
4-formation of ethanol from glucose by yeast
5-formation of a carbon deposit inside the flask

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Answer from: deeknuk

Two answers apply to physical change:

1-Condensation of Ethanol

2-Evaporation of Ethanol

Explanation:

Physical and Chemical changes are two very different types of change. However, both processes are possible because of the same thing: the investment of energy.

Physical change requires a lower investment of energy and this is the reason why these kind of changes can be reverted and the molecular structure of the matter does not change.

For example: I can easily change the state of water with fire or a fridge, and go from ice to liquid to vapor and back as many times I want. This is possible because I'm not giving enough energy to break or modify bonding forces inside the molecule of water, but enough energy to modify the interaction between two different molecules of water.

This is why Ethanol can be evaporated and then condensated into liquid again, representing this physical changes.

Chemical change, on the other hand, requires a higher investment of energy and because of this is that the bonds between atoms inside a molecule can be affected (bonds between atoms in the molecule are stronger). Therefore, when the molecules themselves are modified, the matter is transformed from a compound into a new one. This is why glucose becomes carbon dioxide, ethanol or carbon. These three compounds appear when the bonds between Carbons in the glucose are broken.

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

Ithink the answer is d

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