What would be the effect of carrying out the sodium iodide in acetone reaction with the alkyhalides using an iodide solution half as concentrated?
A reaction with alkyl halides in NaI with acetone is by the Sn2 mechanism. The rate for an Sn2 mechanism is directly proportional to the concentration of the nucleophile:
rate = k[nucleophile][alkylhalide]
If the iodine solution (the nucleophile) is half as concentrated, then the rate will also be halved.
rate = k [nucleophile]/2 [alkyl halide]
If the KOH is in a moderately concentrated aqueous solution, the net reaction can be: 2 Al + 6 H2O => 2 Al(OH)3 + 3 H2. In this instance, the KOH does not undergo any net reaction; instead it catalyzes the reaction between aluminum and water by preventing the solid aluminum from maintaining a passivation layer on its surface. If the KOH is in a still more concentrated aqueous solution, the reaction can be: 2…
No, a concentrated solution need not be saturated always. Concentrated simply implies the presence of a particular solute in a solution in high percentile. Saturation implies that the addition of even a very small amount of a solute will result in a change of phase. Concentrated solution is a solution that contains a large amount of solute relative to the amount that could dissolve.
A solution is a solute dissolved in a solvent. A concentrated solution is all the solute that be dissolved in a solvent at normal temperature. A super-concentrated solution is all the solute that can be dissolved in a solution after mixing in the solute during high temperature / pressure. The concentration after cooling to normal temperature / pressure is greater than a regular concentrated solution.
The reaction results vastly varies from the concentration of the hydrochloric acid. If copper is added to a solution of the dilute acid, no reaction would take place. If it is concentrated acid, it would form the complex tetrachlorocuprate(II) ion. The medium concentrated acids give off the intermediate products when forming the above ion.