Using the calibration curve below (equation of the line: y=0.0039x + 0.9981) to determine the percentage of sugar in a sugar water solution with a density of 1.034 g/mL. Density versus Percent Sugar 1.07 1.06 y = 0.0039x + 0.9981 R= 0.9998 1.05 1.04 1.03 1.02 1.01 1 0.99 4. 6 8 10 12 14 16 Percent Sugar (% m/m ) 8.5% 1.0% 9.2% 4.2 % 9.9% Density (g/mL)
Using the calibration curve below (equation of the line: y=0.0039x + 0.9981) to determine the percentage of sugar in a sugar water solution with a density of 1.034 g/mL. Density versus Percent Sugar 1.07 1.06 y = 0.0039x + 0.9981 R= 0.9998 1.05 1.04 1.03 1.02 1.01 1 0.99 4. 6 8 10 12 14 16 Percent Sugar (% m/m ) 8.5% 1.0% 9.2% 4.2 % 9.9% Density (g/mL)
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![## Determining the Percentage of Sugar in a Sugar Water Solution
### Using the Calibration Curve
To determine the percentage of sugar in a sugar water solution with a known density of 1.034 g/mL, we can use the provided calibration curve. The equation of the line depicted in the calibration curve is:
\[ y = 0.0039x + 0.9981 \]
\[ R^2 = 0.9998 \]
Where:
- \( y \) represents the density (g/mL)
- \( x \) represents the percent sugar (% m/m)
### Graph Description: Density versus Percent Sugar
The graph "Density versus Percent Sugar" plots density (in g/mL) on the y-axis and the percent sugar (in % m/m) on the x-axis. Four marked points on the graph indicate measured data, and a trend line is fitted to these points with the given linear equation.
- The graph ranges from 0% to 15% sugar on the x-axis.
- On the y-axis, density ranges from 0.99 g/mL to 1.07 g/mL.
- The plotted data points lie close to the trend line, illustrating a direct linear relationship between density and sugar concentration, hence a high \( R^2 \) value of 0.9998, indicating an excellent fit of the line to the data.
### Calculation Example
Given a solution density of 1.034 g/mL, we can use the equation to find the corresponding percentage of sugar:
\[ 1.034 = 0.0039x + 0.9981 \]
Solving for \( x \):
1. Subtract 0.9981 from both sides:
\[ 1.034 - 0.9981 = 0.0039x \]
\[ 0.0359 = 0.0039x \]
2. Divide both sides by 0.0039:
\[ x = \frac{0.0359}{0.0039} \]
\[ x \approx 9.21 \]
Thus, the percentage of sugar in the sugar water solution with a density of 1.034 g/mL is approximately **9.2%**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7eb32d09-79f1-492f-b21e-e525fbfadae0%2Fccef7383-bc01-49cc-9428-344d97213b75%2Fktqbpzs_processed.png&w=3840&q=75)
Transcribed Image Text:## Determining the Percentage of Sugar in a Sugar Water Solution
### Using the Calibration Curve
To determine the percentage of sugar in a sugar water solution with a known density of 1.034 g/mL, we can use the provided calibration curve. The equation of the line depicted in the calibration curve is:
\[ y = 0.0039x + 0.9981 \]
\[ R^2 = 0.9998 \]
Where:
- \( y \) represents the density (g/mL)
- \( x \) represents the percent sugar (% m/m)
### Graph Description: Density versus Percent Sugar
The graph "Density versus Percent Sugar" plots density (in g/mL) on the y-axis and the percent sugar (in % m/m) on the x-axis. Four marked points on the graph indicate measured data, and a trend line is fitted to these points with the given linear equation.
- The graph ranges from 0% to 15% sugar on the x-axis.
- On the y-axis, density ranges from 0.99 g/mL to 1.07 g/mL.
- The plotted data points lie close to the trend line, illustrating a direct linear relationship between density and sugar concentration, hence a high \( R^2 \) value of 0.9998, indicating an excellent fit of the line to the data.
### Calculation Example
Given a solution density of 1.034 g/mL, we can use the equation to find the corresponding percentage of sugar:
\[ 1.034 = 0.0039x + 0.9981 \]
Solving for \( x \):
1. Subtract 0.9981 from both sides:
\[ 1.034 - 0.9981 = 0.0039x \]
\[ 0.0359 = 0.0039x \]
2. Divide both sides by 0.0039:
\[ x = \frac{0.0359}{0.0039} \]
\[ x \approx 9.21 \]
Thus, the percentage of sugar in the sugar water solution with a density of 1.034 g/mL is approximately **9.2%**.
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