For example, in part A the 1-chlorobutane eluted first because it has a lower boiling point of 78°C compare to the boiling point of 1-bromobutane of 101.4 to 102.9°C. This is why the retention times were lower for 1-chlorobutane with 1.61min compare to the retention time of 1-bromobutane with 1.82min. In addition, in part B 2-chloro-2-methylbutane that has a boiling point of 85 to 86°C, eluted first and then 2-bromo-2-methylbutane that has a boiling point of 108°C. This is also seen in the retention time of the compounds where 2-chloro-2-methylbutane had a time of 1.27 min and 2-bromo-2-methylbutane 1.44 min. The property that was most important in determining relative retention times was the boiling point and the volatility of 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.
For example, alkane’s boiling points increase as the chain of carbons increase. As seen in figure 2, between the first eight straight chained alkanes there is a smooth increase boiling point that coincides with the increase of number of carbons atom in the chain of each molecule.  Figure 2: Boiling Points of Alkanes Another property of an alkane is its flammability. When the molecular size of an alkane increases, the percentage of carbon in that alkane also increases. As a result, the number of carbons in a straight chain alkane increases, alkanes become less flammable.
For example, dopamine has a Km of 95 and a Vmax of 11 x 105. Both of these values are fairly high and would not make an effective substrate for catechol oxidase since the enzyme has a low affinity for the substrate. An example of a good alternative substrate would be 4-methylcatechol that has a low Km of 1.1 and a Vmax of 3.4 x 105. The enzyme now has a high affinity for the substrate and therefore requires less substrate to meet its Vmax. The substrate for the sweet potato catechol oxidase is N-phenylthiourea.
Record results to determine if NaHCO3 increases or decreases the pH of the water. 10. If NaHCO3 increases the pH of the water, it shows that it will increase the pH of the blood and if NaHCO3 decreases the pH of the water, it shows that it will decrease the pH of the blood. 11. Add 30ml of 0.1M HCl into a beaker containing distilled water.
In the sodium iodide test, the alkyl halide is added to sodium iodide in acetone. In this test, primary halides precipitate the fastest while secondary halides need to be heated in order for a reaction to occur. Comparison of the rates of precipitation of the obtained product to standard 1° and 2° bromide solutions will show whether the product is a primary or secondary
React ethanoic acid and 1-butanol under reflux with the presence of trace amount of concentrated sulfuric acid. In this step, the amount of 1-butanol and ethanoic acid used is the same so that a maximum 70% ester yield can be synthesised at the end of the experiment. After the reactants is accommodated in the reactant flask, the trace amount of concentrated sulfuric acid is added drops-by-drops into the reactant flask and the flask is swirled while adding the acid. Additional of concentrated sulfuric acid is to use as a catalyst to increase the rate of reaction by donating a proton to the oxygen atom in carboxylic acid to allow for the mechanism for esterification and thus, the satisfactory yield of ester can be achieved. Moreover, it acts as dehydrating agent, forcing the equilibrium to the products and lead to a greater yield of ester.
Fig. 6 (a) shows the effect of hydrogen peroxide increase on the MB removal at constant pH 3 and Fe3+ of 40 mg/L. The results show that the degradation rate of MB increases with an increase in initial H2O2 concentration from 100 to 400 mg/L, but in excess of about 400 mg/L; the H2O2 dose of 1000 mg/L, plot of the reaction rate curve is almost horizontal. This could be illustrated that the presence of H2O2 beyond the ratio with Fe3+ does not improve the MB degradation. According to Murry and Parson (2004) hydrogen
Although biodegradation of hydrocarbons can happen in a wide range of temperatures, biodegradation rate generally decreases with decreasing temperature (26). So in this study, the lowest rate of biodegradation was observed at lowest temperature. This is due to the fact that at low temperatures, the viscosity of the oil is increased while the volatilization of toxic short-chain alkanes is reduced, and its water solubility is decreased, thereby delaying the onset of biodegradation
The Rf value of the pure isolated caffeine was 0.28. This was 0.16 higher than the Rf value of the pure caffeine. Although the value of the isolated caffeine is nearly double that of the pure caffeine, it appears that caffeine was separated from Excedrin, but the sample was not pure. It still had other analgesics present. This is concluded from the location
To find the number of moles of each reactant added, volume in liters was multiplied by the molarity (concentration). 2. The second step is about finding the theoretical yield, which will help to determine the correct amount of Ca(OH)2 can be made in chemical reaction. However, before doing this, it’s necessary to find whether CaCl2 or NaOH is a limiting reagent. For each test, the limiting reagent is found by multiplying the number of moles of the reactant by 1 mole of Ca(OH)2 and dividing then by a number of moles of reactant from the reaction.
As seen in the trend of both buffer, once the pH is lower than 3, the slope of dv/dpH increase drastically, showing the decreasing effects of the buffer. On the other hand, in the trend of both buffer on the right side of graph shows when NaOH is added, the change in pH is more drastic once past about pH 5. Although buffer 1 and buffer 2 shows a similar trend, the plot of buffer 1 is above the plot of buffer 2. The reason for this is that buffer 1 is made by an acid and base with an almost equal concentration. This makes buffer 1 a greater buffer compared to buffer 2.