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alukav5142 [94]
3 years ago
7

1.5 What is the purpose of adding sodium carbonate to the reaction mixture? I

Chemistry
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer: As sodium carbonate is added, bubbles of carbon dioxide form. When all the acid has been neutralised, adding sodium carbonate will no longer produce bubbles of carbon dioxide gas. The final rinsing with water should remove traces of the water soluble salt formed.

Explanation:

plzz mark as brainliest

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On treatement of 2 -with hydrogen gas and lindlar catalyst, the major product obtained is cis-2butene.

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A tablet of Pain Be Gone Aspirin, which had a mass of 1.213 g, was pulverized and 1.159 g were dissolved in 10.0 mL of ethyl alc
Elan Coil [88]

Answer:

a. Moles of NaOH  = 0.001643 moles

b. 0.296 g

c. 0.3098 g

d. Not acceptable

Explanation:

a.

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For NaOH :

Molarity = 0.1052 M

Volume = 15.62 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 15.62×10⁻³ L

Thus, moles of NaOH :

Moles=0.1052 \times {15.62\times 10^{-3}}\ moles

Moles of NaOH  = 0.001643 moles

b.

The reaction of NaOH with the acetylsalicylic acid is in the ratio of 1:1.

Thus, Moles of NaOH = Moles of acetylsalicylic acid = 0.001643 moles

Molar mass of acetylsalicylic acid = 180.16 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Mass = Moles * Molar mass = 0.001643 moles * 180.16 g/mol = 0.296 g

c.

1.159 g of sample contains 0.296 g of acetylsalicylic acid

1.213 g of sample contains \frac{0.296}{1.159}\times 1.213 g of acetylsalicylic acid

Mass of acetylsalicylic acid = 0.3098 g = 309.8 mg

d. Sample contains = 309.8 mg

Manufacturer claiming = 315 mg to 335 mg

Thus , it is not acceptable.

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3 years ago
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