Finding the Densities of an Unknown Solid and Liquid in Order to Determine What they Are
Purpose: To find the densities of an unknown solid and liquid in order
to determine what they are.
Materials: Toluene(V) and Zinc(E)
Equipment: A stoppered flask, and an analytical balance
Safety: To begin the experiment, we wore our safety goggles. We were
careful with the chemicals as to not get it on our hands, or clothes.
When we were finished with our chemicals, we poured our “unknown
liquid” into a container, specifically for that, located under the
hood. We put our metal in the designated place, rinsed our flasks and
the stoppers, and left them to dry.
Procedure: In the 1st part of the experiment, we were trying to find
the density of an unknown liquid. To perform this experiment, we
found the mass of the empty stoppered flask by weighing it on the
analytical balance. We then found the mass of the stoppered flask
plus water. By subtracting the mass of the empty stoppered flask from
the mass of the stoppered flask filled with water, we found the mass
of the water. Next, we found the mass of the stoppered flask plus an
“unknown liquid.” (V) By subtracting the mass of the empty stoppered
flask from the mass of the stoppered flask filled with an unknown
liquid, we found the mass of the unknown liquid. Next, through
dimensional analysis, we found the volume of the flask, which is also
equal the volume of the substance which it can hold. Having found
this information, we were able to take the mass of the “unknown
liquid” and divide it by the volume of the “unknown liquid” to get the
density of the “unknown liquid.
In the 2nd part of the experiment, we were trying to find the density
of an unknown metal. First, we found the mass of the stoppered flask
plus about 50 grams of the “unknown metal.” (E) Next, we added water
to the flask- and found the mass of the stoppered flask plus water and
Finding Out the Changes of Mass in Potato Chips Due to Osmosis Aim: We have been asked to investigate the effect that osmosis has on potato chips. Prediction: I predict that the potato cylinder in the lowest sugar solution (water) will gain the most mass through osmosis, whereas the 80% sugar concentration will lose the most mass through osmosis. I predict this because I think that the potato chips will try to gain and lose water between the two sides of its cell wall in order to reach an equilibrium between them. Because the 80% sugar solution has a high amount of sugar molecules, which can not diffuse through the membrane, I think that the potato will lose water in order to make the solution on the outside of its membrane equal to the water concentration on the inside (thus losing mass). I also predict that the other concentrations (60%, 40%, and 20%) will lose mass, but in lower numbers compared to the 80% sugar solution (the lower the concentration, the higher the mass).
HYPOTHESIS: If I use the Metric Measurement and become more familiar with it, then I will be able to convert other measurements and learn to use it and understand it better since other countries used the Metric Measurement.
For the Identification of a Volatile Liquid Data Analysis I had the unknown number A53826. There are many ways that I used to determine the identification of the liquid. Some characteristics of the liquid that will help further in my research are that the liquid was colorless with a harsh strong odor. Before I did my research there were many components I did to help with my research. I first started by performing an experiment that helped determine the mass of my sample. The mass of my sample is one of the key components that will help me determine the moles of my sample. By using the ideal gas law (PV=nRT to n= (PV)/RT) using the mass of my sample in grams, the temperature of my sample after it evaporated during the double-boiling lab, the atmospheric pressure in the room using a barometer, and the volume of the flask I found the moles of the sample (approximately 0.00530 moles). Using the moles I found the molecular weight (mass of sample/ moles of sample). The average molecular weight was 94.9 g/mol. I first started my research by calculating the empirical formula by doing a combustion analysis for 2.0000 grams of my sample unknown (CxHyCLz) yielding 1.7787 grams of carbon dioxide and 0.7294 grams of water. By finding grams of carbon from the amount of grams in carbon dioxide, and finding grams of hydrogen from the amount of grams in water I added the two together (grams of carbon and hydrogen) and then subtracting it from 2.0000 grams of the sample to get my grams of chloride. I then converted the grams to moles and divided it by the smallest amount of mole ratio to get my empirical formula, CH_2 Cl. By using the empirical formula mass, 49.477 grams, and the molecular weight of my sample, 94.9 g/mol, I was able to find my mol...
To undertake titration and colorimetry to determine the concentration of solutions By carrying out titrations and colorimetry, the aim of this investigations was to use these methods such that the concentrations of different solutions used can be identified, and to help find the concentration of the unknown solution that were given. Using Titration and colorimetry the concentrations of different solutions in general can be determined and this helps to identify solutions with unknown concentrations. In this assignment I was asked to carry out two different scientific techniques and find the concentration of different solutions.
The oil glass bottle is made up of three trapezium prisms and a half sphere. So to find the volume of the oil glass bottle you would need to find the volume of all three truncated cones and the volume of the half sphere and plus them all together. To find the volume of the truncated cones, you need to use the formula V = π [s (R + r) + R2 + r2], this will give you the volume for all the truncated cones. Also to find the volume of the half sphere is two-thirds times pi times radius3 (2/3 x π x r3). Once these volumes have been found and plus together, you would receive the volume of the oil glass bottle.
To wrap things up, the lab was made for us to really comprehend about the physical properties of substances. As a result, salt, sand, and iron was used for the purpose. Then our jobs were to figure out how to separate an unknown mixture and to revive them back away from one another. By doing this lab, I got to understand the differences between pure substances, compounds, elements, and mixtures.The reason why was because we observed substances by themselves and did not use them any further such as salt and iron, this hinted these were elements due to the fact that they could not be broken down any further since they had nothing else added to them. Plus, substances such the combination of salt and iron or sand and salt were mixtures since several
to be done. This was to find out what amount of each liquid would be
Once I have found the volume and mass of the marshmallow pies, I will then find their density: taking all three marshmallow pies, I will weigh all three on the triple beam balance, and after doing so, I will identify their volume by using the length, width, and height method; after, I’ll divide to get the result on what amount of density the marshmallow pies have, to get my final results for my
Summary: In this experiment my group and I performed a lipid chromatography and a lipid emulsion test. The lipid chromatography would separate the molecules found in the lipid by polarity. The solvent we used in this lab moved the most nonpolar molecules farther away from the point of origin. Rf describes the relationship between the distance moved by pigment and the distance from origin to solvent front. The lipid chromatography taught me that the chromatography spots that are closer to the point of origin (which is also the least nonpolar molecule) were typically phospholipids. The next spot toward the top is likely to be cholesterol, followed by fatty acids and then triglycerides (which is the most nonpolar molecule). The larger the Rf value the less polar the molecule is and the father it is from the origin point. Next we performed an emulsification test to test if if lipids were present in various samples. An emulsification test validates the presence of
Independent: 20 mL of a substance (fat- free milk, fat- free lactose- free milk, vanilla ice cream, and plain yogurt)
So we can work out through this method that the volume of a box with
After finishing the trials, our group subtracted the mass of the glassware without water from the mass of the glassware with water in order to find the mass of the water in grams. Then, we divided the mass of the water by the density(g/cm^3) of the water in order to find the volume (mL). An example calculation from the 5.00mL pipet is: (4.9285mL+4.8839mL+4.9367mL+4.9265mL+4.9134mL)/5 = 4.9178. In most cases, the temperature of the water was around 23 degrees celsius, making the density about .998408 g/cm^3 for many of the trials. The densities we used were found online. The next calculations we performed were to determine the average volume of the water in each person’s five trials by adding up all of the volumes(mL) and dividing that number by five. Using the average volume, we then calculated the
Moles Volume HCl Volume Water 2 M 10 cm 3 0 cm 3 1.5 M 7.5 cm 3 2.5 cm 3 1 M 5 cm 3 5 cm 3 0.5 M 2.5 cm 3 7.5 cm 3
The form of Density is partitioned into three sections: A measures 1 - 23, B measures 18 - 40, A1 measures 41 - end. The first A section can be broken into two parts: Aa mm1-14 and Ab mm.15-23. The B section may be broken into two smaller parts, the first Ba from measures 24-29 and the second Bb from measures 32-36 with the omitted portions (mm. 29-32 and mm. 36-40) functioning as transitional material. The unmistakable return of A occurs in measure 41.
In density base approach continuity equations are used to solve the density while pressure field is find from equation of state.