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Acc15gaseschallsol1. Nitroglycerin, C3H5O9N3, explosively decomposes to produce carbon dioxide, oxygen, nitrogen and water vapor. The density of nitroglycerin is 1.59 g/mL. If a 5.00 mL vial is filled with nitroglycerin and the nitroglycerin is detonated, assuming standard temperature and that the container does not break with exploding, what is the final pressure in the container in ATMs? A. 4C3H5O9N3 ------> 12CO2 + O2 + 6N2 + 10H2O 4 mols of Nitroglycerin ----> 12 + 1 + 6 + 10 = 29 mols of gas. 5 mL X 1.59 g X 1 mol X 29 mol gas = 0.254 mol gas P = 0.254 mol * 0.0821 (atm•L/mol•K) * 273K 2. Aluminum and iron each react with hydrochloric acid in single replacement reactions (note: Iron(III) product). If a 7.00 grams of a mixture of aluminum and iron react to produce 6.96 Liters of hydrogen gas at 798 mm Hg pressure and 27°C, what is the percent of aluminum in the original mixture. n = (1.05 ATM x 6.96 L)/(0.0821 ATM*L/mol*K x 300K) = 0.297 moles of H2 produced Therefore, each element contributes equal number of moles of H2 gas. Therefore, there must have been equal number of moles of each element to begin with. Stoichiometry => 0.297 moles of H2 X 2 mol/3 mol = 0.198 mols reacted Let mass of Al = X; and Let Fe = (7.00 – X) Therefore, Xg * 1 mol/27 g + (7.00g – X) * 1 mol/55.8 g = 0.198 mols 0.0370X mol + 0.125 mol – 0.0179X mol = 0.198 mols 3.82 g/7.00 g = 0.5459985 * 100% = 54.6% Al 3. Nitrogen and an unknown gas at equal temperature and pressure, diffuse through an equal number of identical pinhole openings. It takes 84 seconds for 1.00 liter of the unknown gas to diffuse and 32 seconds for an equal volume of nitrogen to diffuse. What is the approximate molecular mass of the unknown gas? A. Let A = Nitrogen and B = unknown - - - > ta = 32 sec rate x time = distance - - - > ra x ta = Da ra x ta = rb x tb - - - - > ra / rb = tb / ta - - - -> ra / rb = 84 sec / 32 sec squaring both sides ==>> 6.891 = x/28 4. An internal combustion engine in a car operates by injecting fuel vapor and air into a cylinder, compressing the gas with a piston, and igniting the pressurized gas. Suppose an engine cylinder has a volume of 4.20 x 10 nm when the gas is injected, a temperature of 32.0°C and a pressure of 92.0 kPa. When the cylinder fires igniting the fuel, the volume changes to 49.4 cm and the temperature has changed to 400.0°C. What is the new pressure of the gas? A. 4.20 x 10 nm x (1 cm/1 x 10 nm) = 4.20 x 10 cm P2 = (92.0 kPa x 4.20 x 10 cm x 673.0°C) / (305.0 K x 49.4 cm ) 5. Samples of neon and helium are placed in separate containers connected by a pinched rubber tube; the neon is in a 5.00 L container at 634 mm Hg of pressure; the helium is in a 3.00-L container at 522 mm Hg pressure. When the clamp is removed from the rubber tube the gases are allowed to mix. What are the final partial pressures of each gas and the total pressure of the gas mixture? Assume the volume of the connecting tube is negligible. Ne -> 5.00 L * 634 mm Hg = 8.00 L * P = 396 mm Hg He = 3.00 L * 522 mm Hg = 8.00 L * P = 196 mm Hg P = P + P = 396 mm Hg + 196 mm Hg = 592 mm Hg 6. You go away skiing for spring break. When you return all you think about is soaking away your sore muscles in your hot tub. Unfortunately, while you were gone it was real cold and the power went out causing the water in your hot tub to cool. It is currently at a temperature of 18.0°C, yet you desire a hot tub temperature of 35.0°C. Your hot tub when completely filled with water has a depth of 0.750 meters and a diameter of 1.50 meters. You heat your hot tub by burning Propane. If the propane is stored in a tank in your back yard at a temperature of 22.0°C and a pressure of 475.7 kPa. How many liters of propane must you burn to heat your hot tub to your desired temperature? Heat released by burning propane = 2072.8 kJ/mol V = Π r h = 3.14159 * (75.0 cm) * 75.0 cm = 1.3253 x 10 cm = 1.32 x 10 g q = m * C * delta T = 1.32 x 10 g * 4.184 J/(g * °C) * 17.0°C = 9.3888 - > 9.39 x 10 J q = 9.42659 - > 9.43 x 10 J ==>> 9.43 x 10 kJ (If you carry till you round here) Step 3 - Calculate the number of moles of propane needed C3H8 + 5 O2 ----> 3 CO2 + 4 H2O + 2072.8 kJ 9.43 x 10 kJ / 2072.8 kJ/mol = 45.5 moles propane needed P = 475.7 kPa x (1 ATM/101.325 kPa) = 4.695 ATM Step 5 - Finally calculate the number of liters of propane needed PV = nRT ==>> V = nRT/P = (45.5 mol x 0.0821 (ATM x L/mol x K) x 295.0K)/4.695 ATM = 234.7 = 235 L 7. During a diffusion experiment, it took 75 seconds for a certain number of moles of an unknown gas to pass through a tiny hole. Under the same conditions, the same number of moles of oxygen gas passed through the hole in 30 seconds. What is the molar mass of the unknown gas? A. Let A = Oxygen and B = unknown - - - > ta = 30 sec rate x time = distance - - - > ra x ta = Da ra x ta = rb x tb - - - - > ra / rb = tb / ta - - - -> ra / rb = 75 sec / 30 sec squaring both sides ==>> 6.25 = mB/32 g/mol
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