chemistry 1a. Urea [(NH2)2CO] is produced by reacting ammonia with carbon dioxide with the release of water. In one process 637.2g of ammonia is treated with 1142g of carbon dioxide A. Write a balance equation for the above reaction B. Which of the 2 reactants is the limiting reagent C. Calculate the mass of urea formed D. How much excess reagent is left at the end of the reaction E. If the actual yield of urea formed was 980g what is its yield % a. Ans.
The mass of the products in a chemical reaction depends on the amount of A. excess reactant B. limiting reactant C. oxidizing agent D. reducing subtannce 14. What is the formula mass of CaCl2? (Ca = 40.1, Cl = 35.45) A. 75.6 amu B. 110.9 amu C. 111.1 amu D. 756 amu 15.
2. Acetylation of p-methylaniline to create p-methylacetanilide. Here acetic anhydride will be used to prepare the nitrogen substituted acetamide. This is conducted to make the amine group, which is a very stong ortho and para directing activator on an aromatic ring, a lesser strength activator so that the oxidation of the methyl group can proceed. 3.
The main reason is that these buffers neutralize the solution to maintain the PH. This is when the solution is either too acidic or basic. They also neutralize a solution if it is too acidic or basic which means if they are low on the pH scale it brings them up and vice versa on the base side. Organic cells need to maintain a very specific pH level for them to function properly and survive, which is exactly why cells need buffering agents. 44.
Results | | | | Trial #1 | Trial #2 | Volume of NaOH at equivalence point | 9.5 mL | 13.7 mL | Volume of NaOH at one-half the equivalence point | 4.75 mL | 6.9 mL | pH at half equivalence point | 5.42 | 5.3 | pKa of unknown acid | 5.42 | 5.3 | Average pKa | 5.36 | 5.36 | Average Ka | 4.4 * 10-6 | 4.4 * 10-6 | Moles of unknown acid | 1.30 * 10-3 moles | 1.87 * 10-3 moles | M of unknown acid | 0.0650 M | 0.0805 M | Average M of unknown acid | 0.07275 M | 0.07275 M | 2. The data table containing all your raw and manipulated data (like the one found on page 47 of your lab manual). Be sure to include your lab partner’s name and unknown number. Unknown acid buret | | | | Trial #1 | Trial #2 | Initial buret reading | 20.5 mL | 21.8 mL | Final buret reading | 40.5 mL | 45.1 mL | Volume of acid added | 20.0
4 Calculate the molar mass of these compounds: (relative atomic masses: H = 1; N = 14; O = 16; S = 32; Cu = 64; Br = 80; Pb = 207) (a) copper nitrate, Ca(NO3)2 (b) lead bromate, Pb(BrO3)2 (c) ammonium sulfate, (NH4)2SO4 5 The equation for the complete combustion of methane is shown below. CH4 + 2O2 → CO2 + 2H2O When 80 g of methane is completely combusted, 220 g of carbon dioxide and 180 g of water are formed. (a) Why is the mass of carbon dioxide formed greater than the mass of methane burnt? (b) Calculate the mass of oxygen that reacted. (c) Calculate the mass of water formed when 20 g of methane undergoes complete combustion.
Atom Arrangement (shape) of the electron pairs around the atom Hybridisation of the atom Geometry around the atom N-1 Trigonal planar sp2 Angular/bent/V-shaped N-2 Tetrahedron sp3 Trigonal pyramidal C-3 Trigonal planar sp2 Trigonal planar C-4 Tetrahedron sp3 Tetrahedron The pKa of (the conjugate acid form at) N-1 is 3.12 and the pKa of (the conjugate acid form at) N-2 is 8.02. Draw the structure of the predominant form of nicotine that exists in the human body at pH 7.4. The pH is on the base side of 3.12 and so the conjugate base will predominate and the
The lab manual recommended that 0.5mL of waste be removed during distillation; otherwise it would affect the purity of the product. During distillation, a plastic pipet was used. It was difficult to retrieve the waste because the pipet could not reach without the apparatus at a certain angle. By the time I switched to a glass pipet, there may have been excess of 0.5mL. In attempt to remove all the waste, I likely removed from 0.5mL of product.
To find out how much acid was neutralized by single doses of each antacid we have to subtract the answers from question one from the moles of HCl (.004mol). Maalox: .004 -.001205= .002795 mol Tums: .004 - .00112 = .002880 mol Mylanta: .004 - .001 = .003 mol CVS brand: .004 - .000995 = .003005 mol Rennies: .004 - .00122 = .002780 mol Divide the number of moles of acid neutralized in each case by that antacid's mass. This gives you the moles of acid neutralized per gram of antacid, which will allow you to judge the strongest and weakest antacids on a per weight basis. Maalox: 24.1ml/20.0g = 1.21ml/g .002795 mol base / 20.0g = .00013975 mol base/pill Tums: 22.4ml/21.0g = 1.07ml/g .002880 mol base / 21.0g = .0001371 mol base/pill Mylanta: 20.0ml/18.0g = 1.11ml/g .003 mol base / 18.0g = .0001666 mol base/pill CVS brand: 19.9ml/18.3g = 1.09ml/g (.003005 mol base)/ (18.3g pill) = .0001642 mol base/pill Rennies: 24.4ml/ 17.5g = 1.39ml/g (.002780 mol base)/ (17.5g pill) = .0001588 mol base/pill
Buffers work by having molecules bind to free H+( acid) or OH- ( base) to reduce the acidity or basicity of a solution. Why are we using buffers for the pHs of 7 and 9? Buffers are being used for the pH of 7 and 9 in order to maintain a constant pH through the reactions and ensure proper functioning of the enzymes. Without buffers the pH would change and we would not get correct results for our simulated digestive system reactions. Benedicts Solution Iodine Litmus Cream Biuret Reagent What do these