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.
Answer:
If your lab has litmus paper, you can use it to determine your solution's pH. When you place a drop of a solution on the litmus paper, the paper changes color based on the pH of the solution. Once the color changes, you can compare it to the color chart on the paper's package to find the pH.
Explanation:
A solution's pH will be a number between 0 and 14. A solution with a pH of 7 is classified as neutral. If the pH is lower than 7, the solution is acidic. When pH is higher than 7, the solution is basic. These numbers describe the concentration of hydrogen ions in the solution and increase on a negative logarithmic scale.
For example, If Solution A has a pH of 3 and Solution B has a pH of 1, then Solution B has 100 times as many hydrogen ions than A and is therefore 100 times more acidic.
Answer: There are 0.00269 moles of acetic acid in buffer.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
Molarity of acetic acid solution = 0.0880 M
Volume of solution = 30.6 mL
Putting values in equation 1, we get:
Thus there are 0.00269 moles of acetic acid in buffer.
The colour of copper sulphate solution is blue
