Analysis of a Commercial Bleach Essay

In this Chemistry Lab the main objective is to perform accurate chemical analysis for the quantity of elements and compounds in a sample. There will be a compound made then synthesized. The methods used were acid-base titrations, redox titrations, gravity filtration, and distillation. General conclusions included

Mole of chlorine : 1.0217g - .221g - .3946 g = .4061 g of chlorine

H. How would you prepare 10 mL of a 0.25M HCl solution if 1M HCl was available? How much

(0.074 mol HCl x 1 mol NaOH) / 1 mol HCl = 0.074 mol NaOH

The concentration or molarity of a solution can be seen through the ratio of solute amount to solution volume. This means that when you take the solute amount of a solution in moles and divide it by its volumes in liters you are able to find the solution’s overall concentration or molarity. This is proven in my data table below labeled, Molarity Simulation Data, and it helps to show that when you get the moles of a solute and the volume of the solvent you can easily and accurately figure out the molarity of the solution you are investigating. For example, when there were .75 moles of an energy drink power, and only .25 liters of water to mix it in, as shown in #3 of my data table, I was able to identify the molarity as 3 M through the process of division The way I executed this experiment lets me to see the impressionable relationship between the solute amount in moles and the solution volume in liters and how they affect molarity of a solution. The data table allows me to demonstrate this pattern. My argument is, the molarity of an aqueous solution can be found if you take the moles found in a given solute and divide them by the volume of the solvent. The data presented in this paper proves that the given formula shows the dependent, mathematical relationship that all three of these factors

deionized water to make a 250 mL solution, the molarity of that solution would be 0.00935 M.

Molar Mass is basically in grams per mole for example salt has a molar mass of 5.8g per mole. The mass of an element refers to 6.0221415x1023 molecules of the substance. Molar Mass helps us figure out how many grams per mole we need to do this project and it helps us know how much we need to put into the solution. If you don’t Find

3.6.3. 2, 4 – D (2, 4–Dichloro phenoxy acetic acid) stock solution (1mg/ml): 10.0mg of 2.4-D being weighed and dissolved completely in 1N NaOH to a final total volume

Our purpose in this lab was to create four solutions of the same components and different concentration. I found the concentration in Molarity (mol/liter). I would dilute a solution to create new solutions, using the formula M1xV1=M2xV2 where M=molarity and V=volume. In this lab, there were four solutions, Solution A, Solution B, Solution C, and Solution D. The solute or substance that dissolves was sugar and the solvent or substance that the solute dissolves in was water. For Solution A, I weighed 3.42 g sugar and diluted to 100 mL to get a concentration of 0.10 M and the color of bright red. For Solution B, I weighed 0.342 g sugar and diluted to 100 mL to get a concentration of 0.010 M and the color of pink. For Solution C, I measured 10

Molarity – the concentration of a solution expressed in moles of solute per liter of solution

The objective of this lab was first to convert the mass of a compound to the number of moles and number of molecules and then determine the concentration of salt and its component. The first thing we did was get the mass of an empty container by using a scale and it came out to be 16.87g. Next thing we did was pick a substance which in this case it was Potassium Chloride and placed it on the scale to get a total mass of 31.20g. The container the Potassium Chloride was in only had a mass of 16.87g which means that the mass of the substance was 14.33g. To convert the mass to the number of moles we took the amount of the substance 14.33g and divided it by the mass of Potassium Chloride 74.55g and figured out that the number of moles was 0.192.

Upon the addition of water, it was noted that a layer separation occurred and the water layer remained on top, with the 2-methylphenol layer on the bottom layer. Then, conversion calculations were performed to determine the appropriate amount of 3M NaOH to be added to the 2-methylphenol solution. From the calculations, it was determined that 1.08 mL were to be added. 3M NaOH itself was a cloudy solution in appearance and upon the addition of 3M NaOH to the 2-methylphenol solution, it was noted a color change occurred and it became a yellow-green solution. Following this, the same calculations used previously, were used to determine the appropriate amount of sodium chloroacetate, which was found to be 0.38 g (3.26 mmol). Sodium choloroacetate was a white, crushed solid that was then combined with 1 mL of water and was swirled until the sodium chloroacetate completely dissolved. This sodium choloracetate solution was then transferred to the 2-methylphenol solution by the use of a medicine

First of all, we positioned all our substances together. We got our 6-250 ml beakers, 1-100 ml graduated cylinder, 1 knife, 1 weight boat, 1 scale, white potato, 1 Cork borer, 1 forceps, labels for beakers, water, 0.5 inventory of urea pay attention, and paper towels. First, we started by means of putting in the beakers, we categorized every beaker with the aid of the extraordinary molarities of the solution. There has been a total of six beakers with a total volume of 100 ml, containing a very last attention solution of the following- 0.0, 0.1, 0.2, 0.3, 0.4, 0.5 M % of Urea, with an inventory solution of 0.5M. Next, we calculated every solution the use of the Diluting answers method (C1V1=C2V2), and then we took the final volume and subtracted the final inventory method to determine the amount of water wanted to complete

Experiment to investigate the amount of sodium hydroxide needed to neutralize the solution of vinegar

Molar Consentration or Molarity: Concentration of a solution measured as the number of moles of solute per liter of solution. For example, a 6 M HCl solution contains 6 moles of HCl per liter of solution.