Empirical formula: CH5N Steps for molecular formula: 1- Calculate the molar mass of the empirical formula. 2- Divide the known (given) molar mass by the calculated empirical formula molar mass to get a whole number 3- Multiply that whole number through subscripts of the empirical formula to obtain the molecular formula. Example CH5N 12.01 g C x 1 C= 12.01 g/mol 1.008 g H x 5 H = 5.040
Determine the percent yield of this reaction, showing all steps of your calculation. (3 points) heoretical yield of H2 gas: (1.156 x 10^-3 moles)(1 mole H2/ 1 mole Mg) = 1.156 x 10^-3 moles Theoretical mass: (1.156 x 10^-3 moles)(2.02 g/mole) = 2.335 x 10^-3 g Using ideal gas law: P = (1.1 atm)(760 torr / 1 atm) - 19.8 torr = 816.2 torr V = 0.026 L T = 295 K Solve for n: n = PV/(RT) n = (816.2 torr)(0.026
2 marks 4 Draw the structural formula of Compound G. 1 mark 5 Using the chemical shift correlation for 13C NMR, predict the number of peaks for Compound G and draw in the position of the peaks on the blank spectrum below, annotating each peak with its corresponding structure. (2 marks) 6 Draw the structural formula for 2-chloro but-2-ene. Below this draw a structural formula of an isomer of 2-chloro but-2-ene and name this substance.
The chemical reaction used to find this constant is as follows: MgC2O4 (s) ↔Mg(aq)2++ C2O4 (aq)2- Kc= Mg2+[C2O42-][MgC2O4] Ksp=Mg2+[C2O42-] The solid salt magnesium oxalate is prepared through the following precipitation reaction: Mg(SO4)(aq)+NaC2O4 (aq) → MgC2O4 (s)+NaSO4 (aq) Next, the concentration of the Mg2+ and C2O42- ions is found through a redox titration. This redox titration uses a standardized potassium permanganate solution. The potassium permanganate solution is standardized by titrating it with samples of iron(II)ammonium sulfate hexahydrate . The end point is reached when the solution has turned light purple which is a result of excess amounts of MNO4-. This reaction can be summed up using the following formula: 5Fe2++8H++MnO4- →5Fe3++Mn2++4H2O After standardization, the potassium permanganate solution is then titrated with 3 different magnesium oxalate solutions.
Part B: The graduated pipet’s average density at 22.3 °C was determined to be 0.9785g/mL with a percentage error of 1.89% shows the graduated pipet to be more accurate and precise. Part C: Density of an unknown NaCl solution was measured and a calibration curve used to determine the percentage of NaCl by mass in the solution. y=0.007x + 0.998 which concluded that the concentration of the sodium chloride solution was 3.14%. INTRODUCTION Anything that you can see, touch, taste or smell, occupies space and has mass, it is called matter. Matter can be a gas, a liquid,
Objectives: The purpose of this lab is to observe the reaction of crystal violet and sodium hydroxide by looking at the relationship between concentration and time elapsed of the crystal violet. CV+ + OH- CVOH To quantitatively observe this reaction of crystal violet, the rate law is used. The rate law tells us that the rate is equal to a rate constant (k) multiplied by the concentration of crystal violet to the power of its reaction order ([CV+]p) and the concentration of hydroxide to the power of its reaction order ([OH-]q). Rate = k[CV+]p[OH-]q To fully understand the rate law, concentrations of the substances must be looked at first. The concentration is measured in molarity.
Aim : a) To determine reduction potentials of several redox couples. b) To determine the effect of concentration changes on cell potential. c) To determine the molar concentration of Cu2+ in the unknown using Nernst equation. Procedure : Please refer to the laboratory manual page 77 – 79. Results : Reduction Potentials of Several Redox Couples Galvanic Cell Measured Ecell Anode Equation for Anode Reaction Cathode Equation for Cathode Reaction Cu-Zn +1.19 V Zn Zn → Zn2+ + 2e- Cu Cu2+ + 2e- → Cu Cu-Mg +3.23 V Mg Mg → Mg2+ + 2e- Cu Cu2+ + 2e- → Cu Cu-Fe +0.89 V Fe Fe → Fe2+ + 2e- Cu Cu2+ + 2e- → Cu Zn-Mg +2.00 V Mg Mg → Mg2+ + 2e- Zn Zn2+ + 2e- → Zn Fe-Mg +1.54 V Mg Mg → Mg2+ + 2e Fe Fe2+ + 2e → Fe Zn-Fe +0.29 V Zn Zn → Zn2+ + 2e- Fe Fe2+ + 2e- → Fe Balanced net reaction Cu-Zn Zn (s) + Cu2+ (aq) → Zn2+ (aq) + Cu (s) Cu-Mg Mg (s) + Cu2+ (aq) → Mg2+ (aq) + Cu (s) Cu-Fe Fe (s) + Cu2+ (aq) → Fe2+ (aq) + Cu (s) Zn-Mg Mg (s) + Zn2+ (aq) → Mg2+ (aq) + Zn (s) Fe-Mg Mg (s) + Fe2+ (aq) → Mg2+ (aq) + Fe (s) Zn-Fe Zn (s) + Fe2+ (aq) → Zn2+ (aq) + Fe (s) Zn-Mg = + 2.00 V, Mg-Cu = + 3.23 V The cell potential of Zn-Cu = 3.23 V – 2.00 V = 1.23 V The measured cell potential of Zn-Cu = +1.19 V The value of the sum of the Zn-Mg and Zn-Cu cell potentials are nearly the same as the Cu-Mg cell potential.
Born- Haber cycle 1. The lattice enthalpy of magnesium chloride, MgCl2, can be determined using a Born-Haber cycle and the following enthalpy changes. |Name of process |Enthalpy change/kJ mol-1 | |Enthalpy change of formation of MgCl2 |-641 | |Enthalpy change of atomization of magnesium |+148 | |First ionization energy of magnesium |+738 | |Second ionization energy of magnesium |+1451 | |Enthalpy change of atomization of chlorine |+123 | |Electron affinity of chlorine |-349 | a) Define, using an equation with MgCl2 as an example, what is meant by the term lattice enthalpy. b) Construct a Born-Haber cycle for MgCl2, including state symbols, and calculate the lattice enthalpy of MgCl2. c) Explain why the lattice enthalpy of NaBr is much less exothermic than that of MgCl2.
The absorption spectrum is measured using a spectrophotometer and the data is graphed in Excel. The peak of the line is used to find Vmax of Fe2+. Vmax is used to find the moles of Fe2+ and ligand. The unknown n is a ratio of moles ligand divided by moles Fe2+. Results and Discussion For the first part of the experiment (Part A), five different 100 mL volumetric flasks were each filled with 1,2,3,4 and 5 mL of iron (II) solution.
Show all your work. 10. Calculate the atomic mass of the element X. Then use the periodic table to identify the Mass contribution ( ؍mass)(percent abundance) 69Ga: (68.9257 amu)(60.12%) 44.14 ؍amu 71Ga: (70.9249 amu)(39.88%) 82.82 ؍amu Atomic mass of Ga 44.14 ؍amu ؉ 28.28 amu 27.96 ؍amu has a relative abundance of 60.12% and an atomic mass of 68.9257 amu. 71Ga has a relative abundance of 39.88% and an atomic mass of 70.9249 amu.