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2. The diagram shows an experiment in which a model cell was placed in a sucrose solution. At the start of the experiment the model cell weighed 25 mathrm(~g) and at the end it weighed 30 mathrm(~g) . What was the percentage increase in mass? A 5-0 x B 16-7 x C 20-0 x D 83-3 x A B C D

Question

2. The diagram shows an experiment in which a model cell was placed in a sucrose solution.
At the start of the experiment the model cell weighed 25 mathrm(~g) and at the end it weighed 30 mathrm(~g) . What was the percentage increase in mass?
A 5-0 x 
B 16-7 x 
C 20-0 x 
D 83-3 x 
A
B
C
D

2. The diagram shows an experiment in which a model cell was placed in a sucrose solution. At the start of the experiment the model cell weighed 25 mathrm(~g) and at the end it weighed 30 mathrm(~g) . What was the percentage increase in mass? A 5-0 x B 16-7 x C 20-0 x D 83-3 x A B C D

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UnaMaster · Tutor for 5 years

Answer

The percentage increase in mass is \(20\%\).

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## Step1: Identify the initial and final masses of the model cell. The initial mass is \(25 \, \mathrm{g}\) and the final mass is \(30 \, \mathrm{g}\).<br /><br />## Step2: Understand the formula for percentage increase. The percentage increase can be calculated using the formula:<br />### \(\frac{{\text{Final} - \text{Initial}}}{\text{Initial}} \times 100\%\)<br /><br />## Step3: Substitute the given values into the formula. The calculation will be:<br />### \(\frac{{30 \, \mathrm{g} - 25 \, \mathrm{g}}}{25 \, \mathrm{g}} \times 100\%\)<br /><br />## Step4: Perform the subtraction in the numerator:<br />### \(\frac{{5 \, \mathrm{g}}}{25 \, \mathrm{g}} \times 100\%\)<br /><br />## Step5: Simplify the fraction:<br />### \(\frac{1}{5} \times 100\%\)<br /><br />## Step6: Multiply to find the percentage:<br />### \(0.2 \times 100\% = 20\%\)
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