of mixture Metal C 25.605g 24.6mL 25.2°C 100.5°C 28.7°C Calculations: Show your work and write a short explanation with each calculation. Part I: 1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J/ (g · °C), and the density of water is 1.00 g/mL. (4 points) q = m × c × Δt Given: q=?
Ijose Eghaghe Lab, Week 1 BALANCE LAB Introduction Learning to use the balances is very important in chemistry. We use the direct weighing and weighing by difference methods to find weight and mass of chemicals and compounds. The purpose of this lab is to become familiar with the different methods used in finding the weight and mass of chemicals and compounds in the Chemistry lab. The lab is performed by using centigram and electronic balance, and the direct weighing and weighing by difference methods. Procedure Part I.
Calculate the mass of water present in 5.00 ml of the solution. DETERMINATION OF THE EQUILIBRIUM CONSTANT FOR THE HYDROLYSIS OF AN ESTER I. Theory: In order to determine the equilibrium constant for a reaction, it is necessary to know the initial concentrations of the reactants and products and to be able to determine the equilibrium concentration of one of the compounds. Using the stoichiometry of the reaction and the change in concentration of that compound, the changes in concentration and the equilibrium concentrations of all the other reactants and products can be determined. From these equilibrium concentrations, the equilibrium constant for the reaction can be determined.
Approximately 3mL of Clear Liquid 2 was carefully poured into the graduated cylinder. The graduated cylinder was placed on the scale again (with the added liquid) and the mass was measured in grams to the nearest .01 gram. The graduated cylinder was then placed on the tabletop and the volume was read to the nearest tenth of a
Using this information, we can calculate the amount of the block that should be underwater using the following, mblackg=Fbouyancy=mdisplaced waterg therefore, ρblockVblock=ρwaterVdisplaced water →ρblockh=ρwaterx. Experimental Overview The purpose of this lab is to measure volumes and masses to compute density, and measure displaced volumes and buoyancy forces. First, we measured the radius of the
From the Chemicals window that appears, use the dropdown menu and select Copper Slug. Change the quantity from 0 to 1. Record the mass, in grams, of the copper slug. Step 2: Centigram Balance: Weighing by Difference Obtain the centigram balance. Obtain a 100 mL beaker, and record the mass.
Controlled The controlled variables of the experiment were: A. The volume and concentration of the Hydrochloric acid. B. The concentration of the Sodium Hydroxide. Equipment List * Boiling Tube * 10 cm3 1mol dm-3 Hydrochloric Acid (HCL) * 15 cm3 1mol dm-3 Sodium Hydroxide (NaOH) * pH and Temperature Probes * Data Logger * Measuring Cylinder ‘ * Boiling Tube * Teat Pipette Method * Add 10ml of Hydrochloric acid, measured in a measuring cylinder, into a boiling tube.
Chuong Nguyen U84536721 Post Lab Report Experiment 6 – Part B (100 points) 1. (15 points) Prepare a table listing the mass of the starting materials used and the mass of the dried product you collected. Calculate the percent yield for each compound. Part 1 | | Part 2 | | Material | Amount | Material | Amount | KMnO4 | 3.7580 g | [NiCl2(PPh3)2] | 0.6026 g | C5H8O2 | 16.6 g | NH4SCN | 0.8024 g | Product | | PPh3 | 2.0016 g | Mn(acac)3 | 6.119 g | Product | | | | [Ni(NCS)2(PPh3)2] | 1.56 g | %yield= mass of product/ mass of reactant x 100% Part I 4 KMnO4 + 12 C5H8O2 4 Mn(C5H8O2)3 + 2 K2O + 7 O2 Theoretical yield 0.027 mol KMnO4 x 352.26 g/mol Mn(C5H8O2)3 (1:1 ratio) = 9.51 g Mn(C5H8O2)3 0.166 mol C5H8O2 x 352.26 g/mol Mn(C5H8O2)3 (3:1 ratio) = 19.47 g Mn(C5H8O2)3 Limiting reactant KMnO4 Mn(acac)3 %yield = 6.110 g /9.51 g KMnO4 x 100% = 64.25% Part II [NiCl2(PPh3)2] + 2NH4SCN [Ni(NCS)2(PPh3)2] + 2 NH4Cl Theoretical yield .6026 g /259.54g/mol[NiCl2(PPh3)2] x 304.80 g/mol [Ni(NCS)2(PPh3)2] (1:1 ratio) = 0.7076 g [Ni(NCS)2(PPh3)2] 0.8024 g/ 76.122g/mol NH4SCN x 304.80 g/mol [Ni(NCS)2(PPh3)2] (2:1 ratio) = 1.606 g Limiting reactant [NiCl2(PPh3)2] [Ni(NCS)2(PPh3)2] % yield = 1.56g / 0.7076 g x 100% = 220.46 % (Excess triphenylphosphine need to account in chemical reaction and product yield) 2. (15 points): Prepare a table listing all the experimental magnetic data you obtained for the three compounds you measured.
Show your work and write a short explanation with each calculation. Part I: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation qwater = m × c × ΔT. We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. The water has absorbed the heat of the metal. So, qwater = qmetal Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal.
First, we had to calculate how many grams of copper (II) sulfate we needed to form 100 mL of a 0.200 M solution of copper (II) sulfate. We determined that we needed to use 4.994g of copper (II) sulfate to make the solution. We added distilled water to the 4.994g of copper (II) sulfate in a beaker until it reached 100 mL. Then we put the beaker on a hot plate and added a magnetic stirrer. We determined that the mass of zinc necessary to completely react with the copper (II) ions in the solution was 1.308g.