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3 years ago

Raw materials used in the manufacturing of indigenous soap

Chemistry

1 answer:

Harrizon [31]3 years ago

5 0

Answer:

Fat

Alkali

Explanation:

Fat and alkali are the two primary raw materials needed to manufacture soap.

Sodium hydroxide or potassium hydroxide is generally used as an alkali. The use of alkali depends on the intended application of the soap.

Raw animal fat was used in the past but these days, processed fat is used in the soap manufacturing process. Vegetable fats ( e.g, palm oil, olive oil, coconut oil) are also being used in soap manufacturing.

Additives are also used to enrich the color and texture of the soap.

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Explanation:

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Answer:

The correct option is: c. petroleum jelly, d. Polyethylene glycol 4000/600 mixture

Explanation:

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3 years ago

Assuming the volumes are additive, what is the [Cl−] in a solution obtained by mixing 297 mL of 0.675 M KCl and 664 mL of 0.338

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Answer: The concentration of chloride ions in the solution obtained is 0.674 M

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KCl:

Molarity of KCl solution = 0.675 M

Volume of solution = 297 mL

Putting values in equation 1, we get:

$0.675=\frac{\text{Moles of KCl}\times 1000}{297}\\\\\text{Moles of KCl}=\frac{(0.675mol/L\times 297)}{1000}=0.200mol$

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Moles of chloride ions in KCl = 0.200 moles

For magnesium chloride:

Molarity of magnesium chloride solution = 0.338 M

Volume of solution = 664 mL

Putting values in equation 1, we get:

$0.338=\frac{\text{Moles of }MgCl_2\times 1000}{664}\\\\\text{Moles of }MgCl_2=\frac{(0.338mol/L\times 664)}{1000}=0.224mol$

1 mole of magnesium chloride produces 2 moles of chloride ions and 1 mole of magnesium ion

Moles of chloride ions in magnesium chloride = $(2\times 0.224)=0.448mol$

Calculating the chloride ion concentration, we use equation 1:

Total moles of chloride ions in the solution = (0.200 + 0.448) moles = 0.648 moles

Total volume of the solution = (297 + 664) mL = 961 mL

Putting values in equation 1, we get:

$\text{Concentration of chloride ions}=\frac{0.648mol\times 1000}{961}\\\\\text{Concentration of chloride ions}=0.674M$

Hence, the concentration of chloride ions in the solution obtained is 0.674 M

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