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Andrews [41]
3 years ago
13

Example. A Student wants to increase the maximum if sugar that can dissolve in water..She crushes the sugar and then stirs it in

to the water .Does This work?
Chemistry
2 answers:
Advocard [28]3 years ago
8 0

Answer:

Yes this works.

Explanation:

This works because when the sugar is crushed, more of the surface area of the sugar gets exposed to the solvent, allowing the solvent to dissolve it a lot quicker than when it wasn't crushed.

Stirring also helps the sugar particles dissolve faster. This is because the particles are now agitated, which increases their kinetic energy.  This increase in Kinetic energy is reflected by increase in temperature, which helps the particles dissolve faster.

Delvig [45]3 years ago
6 0

Answer: yes it does work.

Explanation:

When the sugar is crushed by the student, an increase in the number of molecules of sugar occurs. Meaning that more number of the solute (sugar molecules) are able to interact with the solvent (the water).

Therefore, there will be more number of collisions taking place between the solute and solvent molecules.

Hence, we can say it works because this leads to more interaction or collisions between the solute and solvent particles, as a result of this increased interaction between solute and the solvent, sugar will dissolve readily into the water.

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Compound C with molar mass 205gmol-1 , contains 3.758g of carbon , 0.316g of hydrogen and 1.251g of oxygen. Determine the molecu
labwork [276]

Answer:

Molecular formula => C₁₂H₁₂O₃

Explanation:

From the question given above, the following data were obtained:

Molar mass of compound = 205gmol¯¹

Mass of Carbon (C) = 3.758 g

Mass of Hydrogen (H) = 0.316 g

Mass of Oxygen (O) = 1.251 g

Molecular formula =?

We'll begin by determining the empirical formula of the compound. This can be obtained as follow:

C = 3.758 g

H = 0.316 g

O = 1.251 g

Divide by their molar masses

C = 3.758 /12 = 0.313

H = 0.316 /1 = 0.316

O = 1.251 /16 = 0.078

Divide by the smallest.

C = 0.313 / 0.078 = 4

H = 0.316 / 0.078 = 4

O = 0.078 / 0.078 = 1

Thus, the empirical formula of the compound is C₄H₄O

Finally, we shall determine the molecular formula of the compound. This can be obtained as follow:

Molar mass of compound = 205gmol¯¹

Empirical formula => C₄H₄O

Molecular formula =>?

[C₄H₄O]ₙ = 205

[(12×4) + (4×1) +16]n = 205

[48 + 4 +16]n = 205

68n = 205

Divide both side by n

n = 205 / 68

n = 3

Molecular formula => [C₄H₄O]ₙ

Molecular formula => [C₄H₄O]₃

Molecular formula => C₁₂H₁₂O₃

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