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3 years ago

What are the common uses of acetone and formalin?

2 answers:

3 0

Acetone It is good for plastics and some synthetic fibers. It is used for cleaning tools. it is used for thinning polyester resin.Formalin is used as an antiseptic, and preservative. It is also used to treat the fungal diseases of fish. Formaldehyde solutions can be used as disinfectants and germicides, as they quickly kill bacteria and other potentially harmful microorganisms

lorasvet [3.4K]3 years ago

3 0

Answer:

cetone is mainly used in the manufacturing of many different chemicals and also used as solvent in items such as remover for nail polish, paint, lacquers and finishes, cleaners, etc.

Formaline is used as antibacterial, preservative and disinfectant. It's the pungent gas formaldehyde's 37% aqueous solution and also has the formula HCHO.

Explanation:

Acetone is widely used to wash nail however formalin, often identified as formaldehyde, is often used in embalming.

Acetone is mainly used in the manufacturing of many different chemicals and also used as solvent in items such as remover for nail polish, paint, lacquers and finishes, cleaners, etc.

Formaline is used as antibacterial, preservative and disinfectant. It's the pungent gas formaldehyde's 37% aqueous solution and also has the formula HCHO.

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A chemistry teacher adds 50.0 mL of 1.50 M H2SO4 solution to 200 mL of water. What is the concentration of the final solution? A

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CaO + H2O -> Ca(OH)2

yawa3891 [41]

The % yield of Ca(OH)₂ : 62.98%

<h3>Further eplanation </h3>

Percent yield is the compare of the amount of product obtained from a reaction with the amount you calculated

General formula:

Percent yield = (Actual yield / theoretical yield )x 100%

An actual yield is the amount of product actually produced by the reaction. A theoretical yield is the amount of product that you calculate from the reaction equation according to the product and reactant coefficients

Reaction

CaO + H₂O ⇒ Ca(OH)₂

mass CaO= 4.2 g

mol CaO(MW=56,0774 g/mol) :

$\tt mol=\dfrac{mass}{MW}\\\\mol=\dfrac{4.2}{56,0774 g/mol}\\\\mol=0.075$

mol Ca(OH)₂ based on mol CaO

mol ratio CaO : Ca(OH)₂,= 1 : 1, so mol Ca(OH)₂ = 0.075

mass Ca(OH)₂(MW=74,093 g/mol) ⇒ theoretical

$\tt mass=mol\times MW\\\\mass=0.075\times 74,093 g/mol\\\\mass=5.557~g$

% yield :

$\tt =\dfrac{actual}{theoretical}\times 100\%\\\\=\dfrac{3.5}{5.557}\times 100\%\\\\=62.98\%$

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2 years ago