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tekilochka [14]
2 years ago
5

You make an experiment to determine the percent mass of carbon in sugar. You find out that there is 46.63g of carbon in 75.00g o

f sugar. What is the percent mass of carbon?
Chemistry
1 answer:
Aleksandr [31]2 years ago
3 0

Taking into account the definition of percentage composition, the percent composition of carbon in this sample is 62.17%.

<h3>Percentage composition</h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

Then, the percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

percentageby mass= \frac{mass of solute}{mass of solution}x100

<h3>This case</h3>

In this case, you know:

  • mass of solute= 46.63 g
  • mass of solution= 75 g

Replacing:

percentageby mass= \frac{46.63 g}{75 g}x100

Solving:

<u><em>percentage by mass= 62.17 %</em></u>

Finally, the percent composition of carbon in this sample is 62.17%.

Learn more about percent composition:

<u>brainly.com/question/24201923?referrer=searchResults</u>

<u>brainly.com/question/9779410?referrer=searchResults</u>

<u>brainly.com/question/17030163?referrer=searchResults</u>

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Answer:

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Explanation:

To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.

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<u>Solubility in 0.0120 M CoBr₂ (S')</u>

First, we will consider the ionization of CoBr₂, a strong electrolyte.

CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)

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    CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)

I                       0           0.0240

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E                       S'            0.0240 + S'

Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')

In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.

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