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The effects temperature has on reaction rate
The effects temperature has on reaction rate
The effects temperature has on reaction rate
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Identifying features and key components of unknown chemical mixtures are an essential part of chemistry. In many cases it is necessary to analyses and understand the chemical makeup of pharmaceuticals. Qualitative analysis is a method of testing a chemical mixture to determine its chemical makeup, such as its functional group. Identifying functional groups of an unknown compound are important, as it helps in understanding how a chemical acts under certain conditions, such as when changing physical states. Depending on the functional group, the boiling point will be higher or lower, due to the strong or weak intermolecular forces. In this experiment qualitative analysis is used to identify two unknown samples, a liquid and a solid (L21 and …show more content…
Classification Test:
Jones Test:
One drop of the liquid (L21) was placed in a test tube. But for the solid unknown (S21) 5mg was added to the test tube. 0.5mL of acetone was added with 1 drop of jones reagent. The results were recorded. The test tubes were then discarded in the hazardous waste container.
TCICA:
0.5mL of TICICA in acetonitrile, 1 drop of 1M HCL, and 1 drop of unknown liquid (L21) or 10mg of solid unknown (S21) was added to a test tube. A timer was set up and started after the solutions were mixed together in the test tube. Then the test tube was flicked several times and then placed in a test tube rack holder. The reaction stopped when a precipitate formed. The speed of the reaction was recorded and the results of the test were identified. The NaHSO_3 and water were added to the test tubes and disposed of in the waste container. Iodoform:
Four drops of the unknown liquid (L21) or 50mg of solid unknown (S21) is mixed with 2mL of water in a test tube. 2mL of 3M sodium Hydroxide was then added. 3mL of iodine solution was added. The results of the reaction were
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Since it is a primary alcohol, this confirms the conjecture that was made. Looking at the H-NMR, it is obvious that there are four types of hydrogen. One type is at 3.65ppm which is the OH bond. The second type at 1.6ppm is the CH2 next to the OH bond, because the splitting at 3.65ppm is 2, meaning it is next to a CH2. The third type is at 1.3ppm which indicates the rest of the CH2’s in the chain. Lastly the fourth type is at 0.9ppm, indicating a CH3 at the end of the structure. The splitting is 2, which means it is next to a CH2. Just looking at the H-NMR, this supports that the unknown is 1-octadecanol. Looking at the C-NMR next, there are many carbon peaks. The 63ppm identifies the CO bond. The peak at 17ppm identifies the CH3 at the end of the chain. The rest of the peaks fall within the CH2 range, this proves the rest of the carbons in the
The purpose for this experiment was to determine why it was not possible to obtain a high percent yield when Calcium Nitrate Ca(〖NO_3)〗_2 with a concentration of 0.101 M was mixed with Potassium Iodate KIO_3 with concentration of 0.100 M at varying volumes yielding Calcium Iodate precipitate and Potassium Nitrate. Filtration was used to filter the precipitates of the solutions. The percent yield for solution 1 was 87.7%, and the percent yield for solution 2 was 70.8%. It was not possible to obtain a high percent yield because Calcium Iodate is not completely soluble and some of the precipitates might have been rinsed back to the filtrates when ethanol was used to remove water molecules in the precipitate.
...icted α-methyl-2-naphthalenemethanol. Probably the most obvious clue that corresponded to this secondary alcohol was the seven integrated hydrogens within the aromatic region of 7.5-7.9 ppm. This compound was the only one that had seven hydrogens belonging to naphthalene. The other two secondary alcohols 3-methoxy-α-methylbenzyl alcohol and 4-bromo-α-methylbenzyl alcohol have only four aromatic hydrogens.
Solid A was identified to be sodium chloride, solid B was identified to be sucrose, and Solid C was identified to be corn starch. Within the Information Chart – Mystery White Solid Lab there are results that distinguishes itself from the other 4 experimental results within each test. Such as: the high conductivity and high melting point of sodium chloride, and the iodine reaction of corn starch. Solid A is an ionic compound due to its high melting point and high electrical conductivity (7), within the Information Chart – Mystery White Solid Lab there is only one ionic compound which is sodium chloride, with the test results of Solid A, it can be concluded that is a sodium chloride. Solid B was identified as sucrose due to its low electrical
The purpose of the Unknown White Compound Lab was to identify the unknown compound by performing several experiments. Conducting a solubility test, flame test, pH paper test, ion test, pH probe test, conductivity probe test, and synthesizing the compound will accurately identified the unknown compound. In order to narrow down the possible compounds, the solubility test was used to determine that the compound was soluble in water. Next, the flame test was used to compare the unknown compound to other known compounds such as potassium chloride, sodium chloride, and calcium carbonate. The flame test concluded that the cation in the unknown compound was potassium. Following, pH paper was used to determine the compound to be neutral and slightly
At this point the identity of the unknown compound was hypothesized to be calcium nitrate. In order to test this hypothesis, both the unknown compound and known compound were reacted with five different compounds and the results of those reactions were compared. It was important to compare the known and unknown compounds quantitatively as well to ensure that they were indeed the same compound. This was accomplished by reacting them both with a third compound which would produce an insoluble salt that could be filte...
Furthermore, you will compare their reactions with water, universal indicator, vinegar and iodine solution to discover what the mystery powder is. Purpose: To identify the mystery powder based on its physical and
While the solution is being stirred, an air condenser is attached to the vial and 12.2 mg of NaBH4 is added in 3 portions through the condenser. The condenser is capped with a drying tube containing calcium chloride and cotton. After thirty minutes a TLC analysis is taken of the reaction to see how many compounds are present. Three compounds were identified on the TLC meaning that the reaction did not go to completion and the mixture was placed back on a hot plate to react further for ten minutes longer. The TLC showed the starting product 4-tert-butylcyclohexanone and the cis-/trans- forms of 4-tert-butylcyclohexanol; the cis isomer is more polar and will appear above the trans
The C-H (sp3) hydrogens from our product displayed at wavelength 2959 cm-1 correlates to the methyl groups located on the ends of isopentyl acetate4. A really prominent, strong peak located at 1742 cm-1 shows that a C=O ester stretch is located in the product, along with at 1244 cm-1 the spectrum shows a strong peak representing the C(=O)-O stretch that is crucial to the structure of isopentyl acetate. Shown in my IR spectrum is a weak O-H (H-bonded) peak at 3464 cm-1 which shows that I have an impurity of isopentyl alcohol in my product. Isopentyl alcohol has similar boiling points and density as my product so the impurity could have easily boiled out with the isopentyl acetate during distillation. The isopentyl alcohol was also present in my 1H-NMR spectrum backing up the impurity peak at 3464
Refer to Chemistry Lab # 2 – Investigating Changes. No changes have been made in this experiment. Methods = == ==
By doing this experiment, I can know the physical and chemical properties of these samples. After I get my results about the physical and chemical properties of these samples, I can compare my results with the information given by the past student and identify the 5 unknown samples, finding out which sample is which substance. Hypothesis = ==
The conical vial was placed in a small beaker and allowed to cool to room temperature. The mixture was Cooled thoroughly in an ice bath for 15-20 minutes and crystals collected by vacuum filtration on a Hirsch funnel. The vial was rinsed with about 5 mL of ice water and transferred into to the Hirsch funnel and again washed with two additional 5mL portions of ice water. Crystals were dried for 5-10 minutes by allowing air to be drawn through them while they remained on the Hirsch funnel. The product was transferred to a watch glass plate and allow the crystals to dry in air. Crude acetaminophen product was weighed and set aside a small sample for a melting point determination and a color comparison after the next step. Calculation of the percentage yield of crude acetaminophen (MW = 151.2). was done and recorded in the lab notebook.
If the solid dissolved in the solvent at room temperature, then it was too soluble and that solvent could be eliminated. The acetanilide is completely dissolved in ethanol and dichloromethane, therefore eliminating them from being the suitable solvent. If the solid did not dissolve at room temperature then it was placed in the sand bath and left to boil. If the solid dissolved, it was placed in the ice bath and if crystals were observed coming out of the solution then the suitable solvent was found. The suitable solvent was water as the crystals came out once placed in the ice bath.
Chemistry 1A03 is a discussion of chemical fundamentals, including bonding, structure, reactivity and energetics, with emphasis on applications to health, energy and the environment. The typical week for chemistry starts on Mondays where I watch online modules covering the information in the weekly lectures. Before the lectures but after watching the modules, an online quiz is due at midnight every Tuesday must be completed. Some may find this structure for learning odd however, it helps hone my time management skills as the deadline for the weekly quizzes is rather rigid. Furthermore, the quizzes force me to learn the material and mitigates the effects of procrastination. Procrastination is the bane of my success and leads me to cram my learning.
Firstly, an amount of 40.90 g of NaCl was weighed using electronic balance (Adventurer™, Ohaus) and later was placed in a 500 ml beaker. Then, 6.05 g of Tris base, followed by 10.00 g of CTAB and 3.70 g of EDTA were added into the beaker. After that, 400 ml of sterilized distilled water, sdH2O was poured into the beaker to dissolve the substances. Then, the solution was stirred using the magnetic stirrer until the solution become crystal clear for about 3 hours on a hotplate stirrer (Lab Tech® LMS-1003). After the solution become clear, it was cool down to room temperature. Later, the solution was poured into 500 ml sterilized bottle. The bottle then was fully wrapped with aluminium foil to avoid from light. Next, 1 mL of 2-mercaptoethanol-β-mercapto was added into fully covered bottle. Lastly, the volume of the solution in the bottle was added with sdH2O until it reaches 500 ml. The bottle was labelled accordingly and was stored on chemical working bench.
In this experiment three different equations were used and they are the Stoichiometry of Titration Reaction, Converting mL to L, and Calculating the Molarity of NaOH and HCl (Lab Guide pg. 142 and 143).