The solubility of solids in a solvent is very dependent on temperature. A solid tends to be more soluble in hot solvents than in cold. This is the reason the crystals precipitate out when the solution is cooled. There are no ideal solvents, however there are a few traits to look for when choosing the best solvent. Those that will dissolve the solute when the solution is hot but not cold.
The catalyst ensures that the reaction is fast enough for a dynamic equilibrium to be set up within the very short time that the gases are actually in the reactor. After having a look at the whole situation, we find the catalyst has a major role in product yielding. In the situation where catalyst is not present then we cannot
Kinetics Part I Lab: Uncertainty Report Qualitative: Systematic Errors: 1. The beakers that were used during the experiment might still have had some distilled water left from when the beaker was being washed out. This could have affected the content of the solution and the dilution, which could in turn have affected our results. 2. Sometimes our eye sight could fail us and we could have read the meniscus wrong, especially using a 10mL graduated cylinder.
Introduction: Calorimeter is a device which is used to determine the heat release or absorb in a chemical reaction. An exothermic reaction is one in which heat is given out into the surroundings. It results in a temperature increase in the solvent, container, and other immediate surroundings. The amount of heat evolved from the reaction can be expressed as qreaction = qsolution + qcalorimeter However, in this experiment, one assumption to be made as the calorimeter only absorb very little of heat, nearly zero. Thus, the equation simplifies to qreaction = qsolution Therefore, the temperature change caused by the addition of a given amount of heat will depend on the specific heat capacity, c, of the substance.
The correct amount of product that can be produced is the smaller of the two values since that reactant limits the production of product. In this lab, the theoretical yield of the equation is found by using the type of stoichiometry problem mentioned above, and the actual yield is how much product is actually formed in the lab. The percent yield, a comparison between theoretical and actual yield, is than found using the equation: Percent yield = (actual yield / theoretical yield) X 100% Data and Calculations: Balanced Equation: CaCl2(aq) + 2NaOH(aq) Ca(OH)2(s) +2NaCl(aq) Table 1: Results Volume of CaCl2
The success of this experiment was dependent largely on how well the apparatus was setup to ensure proper distillation of the solution. If the separate parts of the apparatus weren’t put together well is can case loss of solution through evaporation. The apparatus should also have no water in it since the presence of water can alter the intended results. The electrical attraction between water molecules causes dipole pulls which make t difficult to separate the molecules that move closer together. This raises the boiling point.
The determined ratio was 1:1.01, Mg:O. Even though the ratio does round to the accepted 1:1 relationship, error likely occurred, and was contributed through loss of product during combustion. While heating, it was difficult to detect smoke. This would lower the calculated mass of oxygen, and its ratio number, while increasing the ratio number for magnesium. Secondly, upon inspection of the contents in the crucible, not all of the contents appeared white, suggesting that possibly not all the magnesium reacted.
The ideal gas constant value is 0.082057 (L x atm)÷(mol x K). The purpose of the experiment was to be able to measure the temperature and the mass of each gas tested, to configure a constant and determine which gas is the most “ideal”. To configure “R” the equation will be, R= PV ÷ nT. Based on our knowledge of an ideal gas, we hypothesized Pentane will have the most ideal behavior to real gas and Cyclohexane to be the least ideal. We expect Pentane to be the most ideal because it’s boiling point is the furthest away from the boiling point of water.
The half reactions for this system are: Oxidation of 〖Fe〗^(2+): 〖Fe〗^(2+)→ 〖Fe〗^(3+)+1e^- Reduction of 〖MnO〗_4^-: 〖MnO〗_4^-+8H_3 O^++5e^-→ 〖Mn〗^(2+)+12H_2 O Which produces the following overall equation: 〖MnO〗_4^-+8H_3 O^++5〖Fe〗^(2+)→5〖Fe〗^(3+)+〖Mn〗^(2+)+12H_2 O Equilibrium is initially obtained at a very slow rate, therefore the titration is carried out in the presence of excess sulphuric acid (H_2 〖SO〗_4) at a high temperature; in order to drastically increase the rate at which equilibrium is attained. Potassium permanganate acts as its own satisfactory indicator since the reagent 〖MnO〗_4^- anion appears to be an intense purple colour while the product 〖Mn〗^(2+) cation has a colourless appearance. However, the end point must be read quickly as the permanganate end point gradually fades due to the 〖MnO〗_4^- reacting with the 〖Mn〗^(2+) that was formed during the titration. When performing the titration, one must be cautious as side reactions can occur and these side reactions must be prevented using appropriate chemical measures. If an insufficient amount of acid was
*Note: The volume of gas generated in the actual lab will contain water vapor in addition to hydrogen gas. In addition, there will be a vacuum that is created by the downward pressure due to the difference in water level in the graduated cylinder and in the water basin. This will cause a negligible increase in the volume compared to