Answer:
5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O
Explanation:
Fe⁺² + MnO₄⁻ + H⁺ => Mn⁺² + Fe⁺³ + H₂O
5(Fe⁺² => Fe⁺³ + 1e⁻) => 5Fe⁺² => 5Fe⁺³ + 5e⁻
<u>MnO₄⁻ + 5e⁻ => Mn⁺² => MnO₄⁻ + 8H⁺ + 5e⁻ => Mn⁺² + 4H₂O</u>
=> 5Fe⁺² + MnO₄⁻ + 8H⁺ => 5Fe⁺³ + Mn⁺² + 4H₂O
Answer:
50 g Sucrose
Explanation:
Step 1: Given data
- Concentration of the solution: 2.5%
Step 2: Calculate the mass of sucrose needed to prepare the solution
The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:
2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose
The hydrocarbon is used in excess.
<h3><u>Explanation</u>:</h3>
The bromination of an arene is not simple as bromination of an alkane. This is because the carbocation or free radicle formation in benzene is a very energy consuming process. This is why a lewis base like aluminium bromide or ferric bromide is used. The ferric bromide takes in the bromine radicle and forms the brominium cation which helps in the formation of electrophile. Now this electrophile brominium cation attacks the benzene ring and forms a temporary sp3 hybrid carbon intermediate. Then the hydrogen is taken by the FeBr4- forming HBr and regenerating the FeBr3 as well as Aromaticity of the arene species at the same time. Here hydrocarbon is used in excess just to prevent the chances of multiple substitution in the same arene molecule.
