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.
0.114 mol/l
The equilibrium equation will be:
Kc = ([Br2][Cl2])/[BrCl]^2
The square factor for BrCl is due to the 2 coefficient on that side of the equation.
Now solve for BrCl, substitute the known values and calculate.
Kc = ([Br2][Cl2])/[BrCl]^2
[BrCl]^2 * Kc = ([Br2][Cl2])
[BrCl]^2 = ([Br2][Cl2])/Kc
[BrCl] = sqrt(([Br2][Cl2])/Kc)
[BrCl] = sqrt(0.043 mol/l * 0.043 mol/l / 0.142)
[BrCl] = sqrt(0.001849 mol^2/l^2 / 0.142)
[BrCl] = sqrt(0.013021127 mol^2/l^2)
[BrCl] = 0.114110152 mol/l
Rounding to 3 significant figures gives 0.114 mol/l
Explanation:
(a) potassium oxide with water

According to reaction,1 mole of potassium oxide reacts with 1 mole of water to give 1 mole of potassium hydroxide.
(b) diphosphorus trioxide with water

According to reaction,1 mole of diphosphorus trioxide reacts with 2 moles of water to give 2 moles of phosphorus acid.
(c) chromium(III) oxide with dilute hydrochloric acid,

According to reaction,1 mole of chromium(III) oxide reacts with 6 moles of hydrochloric acid to give 2 moles of chromium(III) chloride and 3 moles of water.
(d) selenium dioxide with aqueous potassium hydroxide

According to reaction,1 mole of selenium dioxide reacts with 2 moles of potassium hydroxide to give 1 mole of potassium selenite and 1 mole of water.