When the concentration of H₃O⁺ becomes equal to the concentration of OH⁻ ions, it is called the Neutral Solution.
Example:
A neutral water with pH 7 contains Hydronium and Hydroxyl ions in following number,
[H₃O⁺] = 1.0 × 10⁻⁷ M
[OH⁻] = 1.0 × 10⁻⁷ M
Answer:
Yes this works.
Explanation:
This works because when the sugar is crushed, more of the surface area of the sugar gets exposed to the solvent, allowing the solvent to dissolve it a lot quicker than when it wasn't crushed.
Stirring also helps the sugar particles dissolve faster. This is because the particles are now agitated, which increases their kinetic energy. This increase in Kinetic energy is reflected by increase in temperature, which helps the particles dissolve faster.
The problem ask on which of the following reactions and subsequent values and the value is given on you question. Base on your data and also to my formulation, solution and further simplification about the problem and the given values, i came up with an answer of <span>∆Hrxn = 5∆Hf(O2(g)) + 2∆Hf(C2H2(g)) - 2∆Hf(H2O(g)) - 4∆Hf(CO2(g)).</span>
Here, the rate of reaction would be: rate = k[A]
In short, Your Answer would be Option D
Hope this helps!
Answer:
See explanation
Explanation:
In order to do this, we need to use 3 reagents to get the final product.
The first one, and logic is the halogenation of the alkene. Doing this, with Br2/CCl4, we'll get an alkane with two bromines, one in carbon 2 and the other in carbon 3.
Then, the next step is to eliminate one bromine of the reactant. The best way to do this, is using sodium ethoxide in ethanol. This is because sodium ethoxide is a relatively strong base, and it will promove the product of elimination in major proportions rather than the sustitution product. If we use NaOH is a really strong base, and it will form another product.
When the sodium ethoxide react, it will form a double bond between carbon 1 and 2 (The carbon where one bromine was with the methyl, changes priority and it's now carbon 3).
The final step, is now use acid medium, such H3O+/H2O or H2SO4/H2O. You can use any of them. This will form an carbocation in carbon 2 (it's a secondary carbocation, so it's more stable that in carbon 1), and then, the water molecule will add to this carbon to form the alcohol.
See the attached picture for the mechanism of this.