The solution was then titrated with AgNO3 and volume used to generate the colour change was recorded. From the volume of AgNO3 used for the titration, the moles of Ag+ were calculated and found to be 0.005 moles of Ag+. Silver ions and iodine ions react in mole ratio of 1:1 to form the precipitate AgI (s). From the mole ratio it was calculated that 0.005 moles of I- reacted and thus 100% of I- was
Due to water’s polar structure, ions in some compounds attract and form bonds with water molecules, forming hydrates. A hydrate is a salt that has water molecules trapped within its crystals. Every hydrate has a certain number of water molecules weakly bonded to the salt as follows: salt • number of water molecules Anhydrous salts are salts that can form hydrates but which have had all the water driven off, usually by heat. By heating the Copper (II) sulphate hydrate until its color changes from blue to white, the compound can be decomposed into CuSO4, a white crystal, and H2O gas, represented as follows: CuSO4 • xH2O(s) ←⎯→ CuSO4(s) + xH2O(g) Because of the mass conservation law and the fixed proportion of water molecules, the mole of H2O can be calculated by calculating the mass difference of the substance before and after the reaction: Mole of H2O = (Mass of substance before the reaction) (Mass
9. Theoretical yield = (150.22g/mol)(3.5 x 10^-3 mol of nucleophile) = 0.525 g Actual yield = 0.441 g, Percent Yield = (0.441g/0.525g) x 100% = 84% 10. Percent recovery from recrystallization = (0.172g/0.441g) x 100% = 38% 11. The data table provided below obtained melting point data for crude product, pure product, and mixture of the pure and 4-tert-butylbenzyl. 12.
The reason behind this discrepancy is because sodium hydrogen carbonate does not melt or change into a liquid, it decomposes, separating into carbon dioxide (gaseous state), dihydrogen oxide (water vapour, gaseous state), and sodium carbonate (solid state). Furthermore, the reason why
Abstract In this experiment, the reaction kinetics of the hydrolysis of t-butyl chloride, (CH3)3CCl, was studied. The experiment was to determine the rate constant of the reaction, as well as the effects of solvent composition on the rate of reaction. A 50/50 V/V isopropanol/water solvent mixture was prepared and 1cm3 of (CH3)3CCl was added. At specific instances, aliquots of the reaction mixture were withdrawn and quenched with acetone. In addition, phenolphthalein was added as an indicator.
What is the mass of a 22.3 cm3 sample? Let’s rearrange the density equation to solve for mass: d= mass volume ⇒d×volume=mass⇒22.59 g c m 3 ×22.3 c m 3 =504g d=massvolume⇒d×volume=mass⇒22.59gcm3×22.3cm3=504g By rearranging the equation, cm3 gets cancelled out, leaving g as a result. Quick Check: 2.8.b What is the volume of a 5.98 g sample of sodium chloride? Sodium chloride has a density of 2.16 g/cm3. 3.00 g of a sample was placed in a graduated cylinder with 10.0 mL of water.
Hydrogen gas was generated during the reaction which was seen when bubbles were formed as the penny was dissolved into the beaker. An error that could have been present during the lab includes not letting the zinc react completely with the chloride ions by removing the penny too early from the solution. For instance, the percent error of this lab was 45.6%, which was determined by the subtraction of the theoretical percent of Cu 2.5% and the experimental percent of Cu 3.64% and dividing by the theoretical percent of Cu 2.5%. This experiment showed how reactants react with one another in a solution to drive a chemical reaction and the products that result from the
Substrate concentration basically means the amount used for the substrate. The substrate in our experiment was 0.1% hydrogen peroxide. The 0.1% is the concentration amount. Just like temperature and pH, substrate concentration can speed the reaction only up to a certain limit. When we mixed pH 3 enzyme tube with substrate tube, we used 0.3 mL of hydrogen peroxide, but if we were to increase the amount, then the experiment would have been faster.
(a) Formation of calcium chloride (b) Formation of magnesium oxide (c ) formation of bonds in sodium fluoride Checking Up: 2 1. For each of the following ionic bonds: Sodium + Chlorine, Magnesium + Iodine, Sodium + Oxygen, Calcium + Chlorine and Aluminium + Chlorine a) Write the symbols for each element. b) Draw a Lewis Dot structure for the valence shell of each element. c) Draw an arrow (or more if needed) to show the transfer of electrons to the new element. d) Write the resulting chemical formula.
He proved that 6.022 x 1023 particles are present in 1 mole of a gas. which means if there is 1 mole of Hilum atom, then there are 6.022 x 1023 atoms of Helium. If there is 1 mole of sodium chloride, then there is one mole of Na+ and one mole of Cl- ion in the solid. Also, if there is 1 mole of oxygen molecule, therefor there are 6.022 x 1023 molecule of oxygen atoms At Standard Temperature and Pressure 6.022 x 1023 particles of any gas will occupy 22.4L From this, the number of particles can be found when the volume at STP is given. For example, If 2.24L of oxygen gas at STP has given, then it can be easily conclude that the volume would contain one tenth of the mole of oxygen gas.