It is known that chlorine is more electronegative than bromine, and thus chlorine is more reactive, and less discriminatory as to what it will react with, thus making bromine more “selective”. Another pertinent piece of information to look at would be stability. The stability of a free radical increases as the number of carbon substituents increases. Therefore, primary is the least stable and tertiary is the most stable. Also, the more stable the free radical that is left behind, the weaker its C-H bond strength will be.
Abdul Gadoush Period 2 Chemistry Honors 1-16-14 Metal Gizmo 1. Purpose/Problem: The whole purpose to this experiment was to identify which metal of the following: Copper, Magnesium or zinc had the most occurring reactions in the constant chemicals that they were inserted in. The metal that had the least amount of reactions in each chemical would be the least reactant, and the metal that had the most occurring reactions in the chemical would be the most reactant. For this process to occur (make metal gizmos), Naugatuck Metal Works needs one highly reactive metal and another that isn't a reactive metal. - signs of chemical change = changes the shape and color, creates a gas, distributes heat, Etc.
The structure on the left gives N three bonds which is good, the structure on the right puts the formal negative charge on the more electronegative element. My guess is that the left structure is more important because N really likes to have 3 bonds. O O C O C H O O O C O C H O 2.) These are equally important. N F S F F 3.)
The purpose of the lab was to determine which reactant was the limiting reactant, and to see how much of the other reactant was used. The true molarity of a compound can be defined as the amount of moles per liter of that substance. The equation of this single displacement chemical reaction done during this lab is 2Al(s) + 3CuCl(aq) → 3Cu (s) + 2AlCl2 (aq). In the reaction, the solid Aluminum replaces the Copper in Copper (II) Chloride to produce solid copper, and Aluminum Chloride. In order to find which reactant is the limiting reactant, an equation based on the molarity of the Copper (II) Chloride may be used, or the products of the reaction may be observed.
In the experiment, toluene was alkylated with 2-chloropropane to synthesize isomers of isopropyltoluene in 47% yield, weighing 2.66g. The final product had a purity of 91.8% while the appearance was a colourless clear liquid with a strong odour. The results suggest that although the product yield was low, the experimental design led to the production of isopropyltoluene in high purity. REFERENCES 1. Smith R, McKee J, Zanger M. The electrophilic bromination of toluene: Determination of the ortho, meta, and para ratios by quantitative FTIR spectrometry.
Focus Question Should chemists prepare Zinc Iodide, from its elements or from a Double Replacement Reaction between Barium Iodide and Zinc Sulfate? -To put things into perspective in terms of cost efficiency, zinc granules cost about $62.50 per kilogram or $.0625 per gram. Also, Iodine chips cost about $.1498, Barium Iodide Dihydrate costs about $.886 per gram, and Zinc Sulfate Heptahydrate costs about $.0405 per gram. According the chemical reaction listed below, the double replacement reaction creates a solid Barium Sulfate precipitate This means that the ZnI2 is less pure than it would be from just the Zinc and the Iodine. With the elemental reaction you get ZnI2 as your only product thus indicated that it is “pure” Zinc Iodide.
The half reactions for this system are: Oxidation of 〖Fe〗^(2+): 〖Fe〗^(2+)→ 〖Fe〗^(3+)+1e^- Reduction of 〖MnO〗_4^-: 〖MnO〗_4^-+8H_3 O^++5e^-→ 〖Mn〗^(2+)+12H_2 O Which produces the following overall equation: 〖MnO〗_4^-+8H_3 O^++5〖Fe〗^(2+)→5〖Fe〗^(3+)+〖Mn〗^(2+)+12H_2 O Equilibrium is initially obtained at a very slow rate, therefore the titration is carried out in the presence of excess sulphuric acid (H_2 〖SO〗_4) at a high temperature; in order to drastically increase the rate at which equilibrium is attained. Potassium permanganate acts as its own satisfactory indicator since the reagent 〖MnO〗_4^- anion appears to be an intense purple colour while the product 〖Mn〗^(2+) cation has a colourless appearance. However, the end point must be read quickly as the permanganate end point gradually fades due to the 〖MnO〗_4^- reacting with the 〖Mn〗^(2+) that was formed during the titration. When performing the titration, one must be cautious as side reactions can occur and these side reactions must be prevented using appropriate chemical measures. If an insufficient amount of acid was
Halides Lab: Background information: Halide ions are reactive and useful. Salts are positively charged ions (metals) combined with any negative ions (nonmetal), and when placed in a solution (water) it separates into the cations and anions that made it up. The Purpose of this lab is to find out how the Halides react with the indicators, and to determine the identity of the two unknown solutions (A and B). Color of solutions prior to experiment: NaF | NaCI | KBr | KI | Unknown A | Unknown B | clear | clear | clear | clear | clear | clear | Color of indicator prior to experiment: 5% Bleach (NaOCI) | 0.2 M Na2S2O3 | 0.1 M AgNO3 | 0.5 M Ca(NO3)2 | clear | clear | clear | clear | Halide solutions | NaF | NaCI | KBr | KI | unknown A | unknown B | Test 1: Ca(NO3)2 | Cloudy White (Nothing) | Clear | Nothing | light yellow (Nothing) | Nothing | Nothing | Test 2, Part A: AgNO3 | clear (Nothing) | Milky White | Gold (Cloudy yellow) | milky green (Cloudy yellow) | turned white, film developed on top layer | Milky | Test 2, Part B: add Na2S2O3 to test tube from part A | Dark Orange/brown | Clear | Dark Gold(precipitate yellow then clear) | milky green (no change) | white precipitation, settled on bottom | Milky | Test 3: NaOCI (Bleach) | Clear (Nothing) | Nothing | Nothing | Orange (Clear) | Nothing | Orange | Unknown A is identified as NaCI (Sodium Chloride), because in test#1 the solution turned a cloudy white color when Ca(NO3)2 (Calcium nitrate) was added. In the first part of test#2, when AgNO3 (Silver nitrate) is added, the solution turned white, with a thin layer of film developing on the surface.
This particular reaction is a strong acid and a strong base which means that when the reaction reaches the equivalence point, the moles of the acid and the base are equal and the solution is neutral so the pH should be around 7.0 depending on the final volume of each solution. To get this data, we will titrate an HCl solution with NaOH solution of which is a known concentration. We will record the initial and final reading of the NaOH while we record the pH of the titrated solution in the beaker. We will repeat this process with a solution of acetic acid which is a weak acid with NaOH and record the initial and final reading of NaOH and the pH of the solution in the beaker. Procedure Preview Calibrate the pH meter.
When the hydrate is heated, it easily loses water molecules attached and becomes an anhydrous salt. The corresponding chemical reaction for hydrated magnesium sulfate can be written as |MgSO4•7H2O ( MgSO4 + 7H2O |(1) | where MgSO4 is the anhydrous salt. Usually,