What happens to the glucose molecule during the process of cellular respiration?
2 answers:
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Answer:
1. density = 0.89 g/cm3
2. Yes is possible to identify the liquid
3. ethanolamine
Explanation:
<u>Data:</u>
mass = 682 g
volume = 0.767 L = 767 mL or cm3
1.
To calculate the density of the liquid it is necessary to know that the density formula is:
The data obtained is replaced in the formula:
2.
With the given data it is possible to identify the liquid, this because the density value is a basic property of each liquid.
3.
It is possible to determine what liquid it is, since when comparing the value obtained with those reported in the collection of Material Safety Data Sheets (MSOS), the value that agrees is that of ethanolamine.
Density is sensitive to temperature for gases and liquids, although not much for liquids. We use the data in the picture. Using linear interpolation, we determine the densities at 14°C and 20°C.
@20°C: Density = 0.99823 g/cm³ or g/mL
@14°C:
(10 - 14)/(10 - 20) = (0.99973 - Density)/(0.99973 - 0.99823)
Solving for density:
Density = 0.99913 g/cm³ or g/mL
Mass @ 20°C = 50 mL * 0.99823 g/mL = 49.9115 g
Mass @ 14°C = 50 mL * 0.99913 g/mL = 49.9565 g
Difference of Masses = |49.9115 g - 49.9565 g| = 0.045 g
@20°C: Density = 0.99823 g/cm³ or g/mL
@14°C:
(10 - 14)/(10 - 20) = (0.99973 - Density)/(0.99973 - 0.99823)
Solving for density:
Density = 0.99913 g/cm³ or g/mL
Mass @ 20°C = 50 mL * 0.99823 g/mL = 49.9115 g
Mass @ 14°C = 50 mL * 0.99913 g/mL = 49.9565 g
Difference of Masses = |49.9115 g - 49.9565 g| = 0.045 g