Chemistry, 05.07.2019 20:00 tytybruce2

Construct a flow chart to demonstrate how you could separate a mixture of 1,4- dichlorobenzene, 4-chlorobenzoic acid, and 4-chloroaniline using an extraction procedure. (see the last section of technique 10.1 for a short discussion on flow charts) 6. deduce

Answers: 1

Show answers

Answers

Answer from: derbraz6770

Explanation:

Hello,

On the attached picture, you will find the corresponding flowchart for the requested extraction.

In this case, the first step consists on the addition of ether and diluted hydrochloric acid in order to ionize the 4-chloroaniline into an aqueous-like phase to subsequently reform it to neutral. On side of it, the other phase contains both 4-chlorobenzoic acid and 1,4- dichlorobenzene and is separated again with ether but now in the presence of sodium bicarbonate in order to form aqueous-like sodium 4-chlorobenzoate to subsequently take it back to 4-chlorobenzoic acid (second phase 1 sub process). At the time, the 1,4- dichlorobenzene is separated through the second organic phase 2.

This extraction is attained by forming non-organic intermediates which are immiscible with the additional organic fraction in each step, in order to achieve the process successfully by gravity separations (separating funnel).

Best regards.

Answer from: Quest

answerleaving cheese out to get moldy;

Answer from: Quest

Note that 2 mol dm⁻³ = 2 mol/(dm³)outline: we're given the concentration of hcl, which is 2 mol dm⁻³ (cubic meters in the denominator can represent volume, so i know this is concentration). we're an amount of calcium carbonate caco₃, which is 5 g. we're presented with a chemical equation which includes caco₃ and hcl. we are told to find the volume of hcl. from what we see so far, this seems to be a stoichiometry problem. we'll have to use the given information about calcium carbonate to find the amount of moles of hcl needed to fully react with calcium carbonate. then we'll have information about the moles of hcl and the concentration of hcl; that will be enough to figure out volume according to the formula concentration = moles / volume.======all the answer choices are not in dm.convert the given concentration so that it uses cm⁻³.we can use conversion factors. we have to cube our conversion factors so that dimensional analysis works.we can convert to meters first, using the fact that 10 dm = 1 m ⇔ 10³ dm³ = 1 m³then we use the fact that 10² cm = 1 m ⇔ (10² cm)³ = (1 m)³ ⇔ 10⁶ cm³ = 1 m³so converting: $\dfrac{\text{2 mol}}{\text{dm}^{3}} \cdot \dfrac{\text{10}^3\text{ dm}^3}{\text{1 m}^3} \cdot \dfrac{\text{1 m}^3}{10^6\text{ cm}^3} \\ \\ = \dfrac{\text{2 mol}}{1} \cdot \dfrac{\text{10}^3}{1} \cdot \dfrac{1}{10^6\text{ cm}^3} \qquad \text{\footnotesize cancelling units} \\ \\ = \dfrac{2\text{ mol}}{10^3 \text{ cm}^3} \\ \\ = 0.002\text{ mol/cm}^3$ now we have to find the moles of hydrochloric acid neededfind the moles of calcium carbonate. we have to find the molar mass of calcium carbonate, caco₃, which can be found by molar mass = (mass ca) + (mass c) + 3(mass o) = 40 + 12 + 3(16) = 100 g/molconvert the given 5 g to moles: mol caco₃ = 5 g caco₃ * 1 mol/(100 g) = 0.05 mol caco₃use stoichiometry (ratios) to figure out the moles of hcl needed. ratio wise, there's 2 hcls for every caco₃, so use dimensional analysis mol hcl = mol caco₃ * (2 mol hcl) / (1 mol caco₃) = 0.05 mol hcl * 2 = 0.1 mol hcl we have 0.1 mol hcl and concentration of hcl (0.002 mol/cm³).we need to find volumesince concentration = moles / volume, rearrange to get volume = moles / concentration volume = 0.1 mol / (0.002 mol/cm³) volume = 50 cm³