Chemistry, 22.09.2019 02:20 crookdamian21

In this experiment, 1-2 ml of saturated sodium chloride is used to transfer the crude product after the initial distillation. why is saturated sodium chloride, rather than pure water, used for this procedure?

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Saturated sodium chloride is used to transfer the product rather than water since it is not polar and rinsing the product with water would revert any 4-methylcyclohexene back to 4-methylcyclohexanol in the Hickman Head and thus lowering the percent yield; using water would shift the equilibrium towards the reactants. Also sodium chloride removes the small amount of phosphoric acid and also a small amount of water. If one were to add water, both 4-methylcyclohexene and phosphoric acid are partially soluble making difficult to remove the water later; sodium chloride makes the water less reactive so easier to remove by making the aqueous later more polar.

Answer from: taterbugwallen7944

Because it helps to remove water from the system.

Explanation:

The saturated sodium chloride solution has a strong affinity for water molecules and there is the possibility of changing the saturated solution to a dilute solution in the presence of pure water. Because of these reasons, the saturated sodium chloride solution removes water molecules from the system to become a diluted solution. That is the reason why the saturated solution was used instead of pure water.

Answer from: Quest

answer: i dont know the answer but this might

explanation:

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.