The hydrolysis of esters in base is called saponification .
So, option C is correct one.
The saponification is the process that involves conversion of fats , oils , lipids into soap and water in the presence of alkaline medium. Saponification is the process of making soap.
During the saponification process, the mixture has an acidity, which tells that it's not safe for usage. After the saponification process is complete, the pH should be a base.The process of formation of carboxylic salt and water by hydrolysis of ester in base is called saponification.
learn more about saponification
brainly.com/question/2263502
#SPJ4
Answer:
53.18 grams of H₂SO₄ are needed to react with 78.86 grams of Al(NO₃)₃.
Explanation:
The balanced reaction between Al(NO₃)₃ and H₂SO₄ is:
2 Al(NO₃)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 HNO₃
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of reagents are required:
- Al(NO₃)₃: 2 moles
- H₂SO₄: 3 moles
Being the molar mass of the elements:
- Al: 27 g/mole
- N: 14 g/mole
- O: 16 g/mole
- H: 1 g/mole
- S: 32 g/mole
then the molar mass of the reactants are:
- Al(NO₃)₃: 27 g/mole + 3*(14 g/mole + 3*16 g/mole)= 213 g/mole
- H₂SO₄: 2*1 g/mole + 32 g/mole +4*16 g/mole= 98 g/mole
Then, by reaction stoichiometry, the following reagent mass amounts are required:
- Al(NO₃)₃: 2 moles* 213 g/mole= 436 g
- H₂SO₄: 3 moles*98 g/mole= 294 g
Then you can apply the following rule of three: If by stoichiometry 436 g of Al(NO₃)₃ react with 294 g of H₂SO₄, 78.86 g of Al(NO₃)₃ with how much mass of H₂SO₄ will it react?
mass of H₂SO₄=53.18 g
<u><em>53.18 grams of H₂SO₄ are needed to react with 78.86 grams of Al(NO₃)₃.</em></u>
Answer:
Some of KCl will precipitate out of the solution.
Explanation:
Gases solubility increases with decrease in temperature and therefore nitrogen gas will not bubble out is the temperature is dropped from 75 deg C to room temperature.
On the other hand, KCL is a salt in solid state and the solubility of solids decreases with decrease in temperature. Therefore some of the KCl will precipitate out of the solution when temperature drops from 75 deg C to room temperature.
By not shaking it up or leaving it in the heat.
Answer : The heat of the reaction is, 1.27 kJ/mole
Explanation :
First we have to calculate the heat released.
Formula used :
or,
where,
Q = heat = ?
m = mass of sample = 1.50 g
c = specific heat of water =
= initial temperature =
= final temperature =
Now put all the given value in the above formula, we get:
Now we have to calculate the heat of the reaction in kJ/mol.
where,
= enthalpy change = ?
Q = heat released = 0.0238 kJ
n = number of moles NH₄NO₃ =
Therefore, the heat of the reaction is, 1.27 kJ/mole