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Dvinal [7]
3 years ago
14

Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide wh

at lithium diorganocopper (Gilman) reagent is needed to convert 1-bromopropane into propylcyclohexane. Draw the structure of the organolithium compound that is used to prepare Gilman.
Chemistry
1 answer:
Ann [662]3 years ago
7 0

Answer:

See explanation and image attached

Explanation:

The  Gilman reagent is a lithium and copper (diorganocopper) reagent with a general formula R2CuLi.  R is an alkyl or aryl group.

They are useful in the synthesis of alkanes because they react with organic halides to replace the halide group with an R group.

In this particular instance, we intend to synthesize propylcyclohexane. The structure of the  lithium diorganocopper (Gilman) reagent required is shown in the image attached to this answer.

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When 18.0 g H20 is mixed with 33.5 g Fe, which is the limiting reactant?
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Answer:

si.mple fe

Explanation:

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3 years ago
Hydrobromic acid, HBr, is a strong acid. Hydrofluoric acid, HF, is a weak acid. If you prepared equal
snow_tiger [21]
HBr and HF are both monoprotic Arrhenius acids—that is, in aqueous solution, they dissociate and ionize to give hydrogen ions. A strong acid ionizes completely; a weak acid ionizes partially.

In this case, HBr, being a strong acid, would ionize completely in water to yield H+ and Br- ions. However, HF, being a weak acid, would ionize only to a limited extent: some of the HF molecules will ionize into H+ and F- ions, but most of the HF will remain undissociated.

pH is, by definition, a measurement of the concentration of hydrogen ions in solution (pH = -log[H+]). A higher concentration of hydrogen ions gives a lower pH, while a lower concentration of hydrogen ions gives a higher pH. At 25 °C, a pH of 7 indicates a neutral solution; a pH less than 7 indicates an acidic solution; and a pH greater than 7 indicates a basic solution.

If we have equal concentrations of HBr and HF, then the HBr solution will have a greater concentration of hydrogen ions in solution than the HF solution. Consequently, the pH of the HBr solution will be less than the pH of the HF solution.

Choice A is incorrect: Strong acids like HBr dissociate completely, not partially.

Choice B is incorrect: While the initial concentration of HBr and HF are the same, the H+ concentration in the HBr solution is greater. Since pH is a function of H+ concentration, the pH of the two solutions cannot be the same.

Choice C is correct: A greater H+ concentration gives a lower pH value. The HBr solution has the greater H+ concentration. Thus, the pH of the HBr solution would be less than that of the HF solution.

Choice D is incorrect for the reason why choice C is correct.

4 0
3 years ago
Calculate the molar solubility of PbBr2 (Ksp = 4.67x10-6) in 0.10M NaBr solution.
Softa [21]

Explanation:

PbBr_{2} will dissociate into ions as follows.

         PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)

Hence, K_{sp} for this reaction will be as follows.

                   K_{sp} = [Pb^{2+}][Br^{-}]^{2}

We take x as the molar solubility of PbBr_{2} when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

               [Pb^{2+}] = [Pb^{2+}]_{o} + x

               [Br^{-}]^{2} = [Br^{-}]_{o} + 2x

So, equilibrium solubility expression will be as follows.

            K_{sp} = ([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains [Br^{-}]_{o} = 0.10 and there will be no lead ions. So, [Pb^{2+}]_{o} = 0

So, [Br^{-}]_{o} + 2x will approximately equals to [Br^{-}]_{o}.

Hence, K_{sp} = x[Br^{-}]^{2}_{o}

            4.67 \times 10^{-6} = x \times (0.10)^{2}

                        x = 4.67 \times 10^{-4} M

Thus, we can conclude that molar solubility of PbBr_{2} is 4.67 \times 10^{-4} M.

5 0
3 years ago
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The answer is B as isotopes are different versions of the same chemical element containing the same amount of protons and electrons but different amounts of neutrons.

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How many significant digits in 0.23100
Reil [10]

Answer:

There are 5 significant digits in 0.23100.

Explanation:

This is because all non-zero digits are considered significant and zeros after decimal points are considered significant.

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