UNIVERSITI TUNKU ABDUL RAHMAN Faculty Course Year/ Semester Session : : : : Faculty of Science Bac. of Science (Hons) Chemistry Year 3 Semester 1 201105 or May 2011 Unit Code Unit Title Lecturer Assignment : : : : UDEC3254 Spectroscopy and Chromatography II Dr. Lim Chan Kiang 2 Answer all questions 1. A mixture of steroid hormones consisting of cortisone, cortexone and corticosterone was separated using a HPLC silica column eluted with a binary mixture of acetone and hexane (20:80). (i) What is the order of elution time for the above compounds from the HPLC column? (3 marks) (ii) What is the chromatographic mechanism involved?
Where it resulted to values of 3.990602 x 10-3 s-1, 4.653278 x 10-3 s-1, 5.944044 x 10-3 s-1, 7.499958 x 10-3 s-1, 7.499958 x 10-3 s-1, 9.84554 x 10-3 s-1, for flasks 4, 5, 6, 7, 8. Then the Ionic strength was calculated using the equation µ=½∑ ci zi2 . Which resulted in 4.5392142 x 10-2, 6.4999998 x 10-2, 1.23823528 x 10-1, 2.21862742 x 10-1, and 4.17941175 x 10-1. It
IB HL Chemistry Energetics - End of Topic Test 35 min 37 marks Answer all the questions in the spaces provided 1. What energy changes occur when chemical bonds are formed and broken? A. Energy is absorbed when bonds are formed and when they are broken. B.
5. 5. II. Mass Defect and Nuclear Binding Energy Consider a helium-4 atom: (_______ protons, _________ neutrons, and __________ electrons) Recall: 1 proton = 1.007277 u, 1 neutron = 1.008665 u, and 1 electron = 0.000548 u We would expect the mass of a helium-4 atom to be as follows: 2(1.007277 u) + 2(1.008665) + 2(0.000548) = 4.032980 u However the observed mass of a helium-4 atom is 4.002596 u!!!!!! There is a difference of 0.030384 u between the observed mass and the calculated mass.
AS Applied Science (Chemistry) Unit 3: Finding out about substances Part 1: Qualitative analysis of compounds Introduction ____________________________________________________________ __________ Description of the investigation..................................................................................................................... Page 2 Description and importance of flame tests..................................................................................................... Page2 Background ____________________________________________________________ __________ The Composition of matter or materials…………………………………………………............................... Page 3 Description of the sub-particles of an atom …………………………………………………………………. Page 4 Example and definition of an ion ………………………………………………………………………….... Page 4 List of some metals with symbols ………………………………………………………………………........
Main – group metals usually for one cation (positive ion). In a binary ionic compound the metal (cation) is named first. Then the nonmetal (anion) is named, and the suffix -ide is added. To create the formula, you switch the charges, and that tells you how many of each element you will need. For example: Cation Anion Formula Name of Compound Ba2+ I- Ba2+ I- Barium Ion Iodide Ion BaI2 Barium Iodide Type 2: Binary Ionic Type 2 Binary Ionic compounds consist of a metal and a nonmetal.
When the equation of mass of a photon turns to unification, this equation is applicable to all fields from the particle to the universe as a hole. Here is the example of energy emission that in what way the fission be possible. To find this, we should clarify first the mass of a photon and the equation of unification physics. Key words: mass of a photon, unification of physics, value of Pi at excited state, unknown weight of radioactive elements, fission reaction
For example, when a carbon-14 nuclide decays by emitting a beta particle, it becomes nitrogen-14 (which is stable). [pic]C ( [pic]N + [pic]e Therefore, unlike a chemical reaction, nuclear reactions alter the identity of the atom. Types of Radioactive Decay Alpha Decays There are several different types of radioactive decay. One frequently observed decay process involves production of an alpha (α) particle, which is a helium nucleus and assigned the [pic]α. This is the common mode of decay for heavy radioactive nuclides (those with atomic number, Z > 83).
(Absolute Astronomy, 2009). The structure of Manganese(III) acetylacetonate is shown as below:(Source: Tcieurope.com) The equation is as follow:MnCl2 + 4H2O Mn(H2O)4Cl2 Mn(H2O)4] Cl + 2HC5H7O2 + NaC2H3O2 Mn(C5H7O2)2 + NaCl + HC2H2O2 4Mn(C5H7O2)2 + KMnO4 + 7HC5H7O2 + HC2H3O2 5Mn(C5H7O2)3 + KC2H3O2 + 4H2O Furthermore, bis(acetylacetonato)oxovanadium(IV) is also known as Vanadyl acetylacetonate, VO(acac)2. As we know, it is a blue green complex. bis(acetylacetonato)oxovanadium(IV) has a vanadyl group, VO2+. The vanadyl group is bonded to 2 acetylacetonate anions and the structure of the compound is as follow:This complex can be made from vanadium(IV) or vanadium(V).
Experiment 7 Formula of a Complex Ion by the Continuous Variation Method Objective: To determine the formula of a complex ion by the continuous variation method. Procedure: Refer to laboratory manual. Results and Calculations: 1. Calculate the molarity of both CuCl2 solution and the ethylenediamine solution. Number of moles of CuCl2 used = [pic] =[pic] = 0.127 moles Therefore, Molarity of CuCl2 = [pic] = [pic] = 0.063 mol/dm3 Number of moles of ethylenediamine used = [pic] =[pic] = [pic] = 0.1 mole Therefore, Molarity of ethylenediamine = [pic] = [pic] = 0.05 mol/dm3 2.