Answer: The hydroboration of an alkene occurs in TWO CONCERTED STEP which places the boron of the borane on the LESS SUBSTITUTED carbon of the double bond. The oxidizing agent then acts as a nucleophile, attacking the electrophilic BORON and resulting in the placement of a hydroxyl group on the attached carbon. Thus, the major product of the hydroboration oxidation reaction DOES NOT follow Markovnikov's rule.
Explanation:
Hydroboration is defined as the process which allows boron to attain the octet structure. This involves a two steps pathway which leads to the production of alcohol.
--> The first step: this involves the initiation of the addittion of borane to the alkene and this proceeds as a concerted reaction because bond breaking and bond formation occurs at the same time.
--> The second step: this involves the addition of boron which DOES NOT follow Markovnikov's rule( that is, Anti Markovnikov addition of Boron). This is so because the boron adds to the less substituted carbon of the alkene, which then places the hydrogen on the more substituted carbon.
Note: The Markovnikov rule in organic chemistry states that in alkene addition reactions, the electron-rich component of the reagent adds to the carbon atom with fewer hydrogen atoms bonded to it, while the electron-deficient component adds to the carbon atom with more hydrogen atoms bonded to it.
Answer:
B. 0.971 g
Explanation:
When MgCl₂(aq) reacts with Pb(NO₃)₂(aq), PbCl₂(s) and Mg(NO₃)₂(aq) are produced:
MgCl₂(aq) + Pb(NO₃)₂(aq) →, PbCl₂(s) + Mg(NO₃)₂(aq)
Thus, we need to find imiting reactant finding moles of each reactant:
<em>Moles MgCl₂:</em>
15.5mL = 0.0155L * (0.225 mol / L) = 3.49x10⁻³ moles
<em>Moles Pb(NO₃)₂:</em>
37.5mL = 0.0375L * (0.250mol / L) = 9.38x10⁻³ moles
As the ratio of the reactants is 1:1, the moles of PbCl₂ are 3.48x10⁻³ moles.
We need to convert thes moles to mass using molar mass of PbCl₂ (278.1g/mol), thus:
3.48x10⁻³ moles * (278.1g/mol) =
0.968g of PbCl₂ are precipitate
Thus, right answer is:
<h3>B. 0.971 g</h3>
Answer:
(i) Change in colour (ii) Change in temperature (iii) Formation of precipitate
Explanation:
(i) Change in colour: Reaction between lead nitrate solution and potassium iodide solution. Pb(NO3)2(aq)+2KI → PbI2(s)+2KNO3(aq) In this reaction, colour changes from colourless to yellow. (ii)Change in temperature: Action of dilute sulphuric acid on zinc. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2 In this reaction, heat is evolved (iii) Formation of precipitate: Action of barium chloride on sodium sulphate. BaCl2(aq) +Na2SO4(aq) → BaSO4(s) +2NaCl(aq) BaSO4(s)
Answer:
pH = 7.30
Explanation:
You can find the pH using the following equation:
pH = -log[H⁺]
In this equation, [H⁺] represents the hydrogen ion concentration. You can plug the given value into the equation and solve to find the pH.
pH = -log[H⁺]
pH = -log[5 x 10⁻⁸]
pH = 7.30
