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Explain the consequences of thermal expansion
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According to Kissinger’s Equation, a plot of ln [(Tg)2/α] versus 1/Tg should be a straight line as shown in figure 4, and from its slope, the value of Eg can be calculated. And it is equal to 328.82 kJ/mol.
The crystallization process can be explained by the DSC results which obtained at different heating rates. The kinetic parameters (activation energy for crystallization, Ec, and Avrami exponent, n) for the crystallization peaks determined. Kissinger method used to determine the activation energy for crystallization (Ep) considering the heating rate (α) dependence of the peak crystallization temperature (Tp) as shown in figure 5.
The value of Ep can be determined from the slope of a plot of ln [(Tp)2/α] vs 1/Tp according to the following equation: [ 19, 20]
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1/Tp for the sample as shown in Figures 6 and 7. According to the slopes of the lines, the values of Ec1 and Ec2 can be calculated they are equal to 164.37 and 128.37 kJ/mol, respectively.
The fraction of crystallization (X) can be calculated from the DSC curves, according to the following ratio:
X =AT/A where AT is the area between Ti and T, while A is the total area of the exothermic peak between Ti and Tf, as shown in previous work [21].
Figure 8 shows the fraction of crystallization (x)
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This figure shows the presence of two exothermic peaks that represent the crystallization temperature.
The as the prepared sample was heat-treated at a temperature around this value and measured each time at DTA.
In addition, Figure 11 shows that the first exothermic peak disappeared when the sample heat treated at and higher than 530 OC and second exothermic peak appeared. And the two exothermic peaks disappeared when sample heat treated at 620 OC.
XRD
First, 100 mL of regular deionized water was measured using a 100 mL graduated cylinder. This water was then poured into the styrofoam cup that will be used to gather the hot water later. The water level was then marked using a pen on the inside of the cup. The water was then dumped out, and the cup was dried. Next, 100 mL of regular deionized water was measured using a 100 mL graduated cylinder, and the fish tank thermometer was placed in the water. Once the temperature was stabilizing in the graduated cylinder, the marked styrofoam cup was filled to the mark with hot water. Quickly, the temperature of the regular water was recorded immediately before it was poured into the styrofoam cup. The regular/hot water was mixed for a couple seconds, and the fish tank thermometer was then submerged into the water. After approximately 30 seconds, the temperature of the mixture leveled out, and was recorded. This was repeated three
This question refers to the example data given below. Using the rate law and the experimental values given below, calculate k.
The solvent should be easily removed from the purified product, not react with the target substances, and should only dissolve the target substance near it’s boiling point, but none at freezing. A successful recrystallization uses minimum amount of solvent, and cools the solution slowly, if done to fast, many impurities will be left in the crystals. Using the correct solvent, in this case ice water and ethyl acetate, the impurities in the compound can be dissolved to obtain just the pure compound. A mixed solvent was used to control the solubility of the product. The product is soluble in ethanol an insoluble in water. Adding water reduced solubility and saturates the solution and then the crystals
SE/TC is .413 and is not preferred because it is under 70% 7) The P/BV of 4.98
The thermometer’s original temperature before coming in contact with an outside object is represented by T. ∆T/∆t is the average temperature of the digital thermometer. represents the temperature of the heat flowing object. In this lab, the temperature of the air is represented by Tair=T. To= Thand is the temperature of the hand.
From figure 4 and formula, essentially explain the lack of thermal expansion activity during β phase is due to the fact that tilt angle δ has settled at zero which causes the removal of expansion components and discontinuity
The purpose of performing this lab was to find the specific heat capacity of an unknown metal.
-443.08 x (100.1 / 2.51) = -17670.2 J.mol. 1. H = -17.67 kJ.mol. 1.
The objective of this experiment was to identify a metal based on its specific heat using calorimetry. The unknown metals specific heat was measured in two different settings, room temperature water and cold water. Using two different temperatures of water would prove that the specific heat remained constant. The heated metal was placed into the two different water temperatures during two separate trials, and then the measurements were recorded. Through the measurements taken and plugged into the equation, two specific heats were found. Taking the two specific heats and averaging them, it was then that
n is the number of moles, R is the gas constant (8.31 J/Kmol), and ΔT is the change in temperature measured in Kelvins.
- Temperature was measured after and exact time i.e. 1 minute, 2 minutes, 3 minutes.
The anomalous temperature, 60°C, has been highlighted. red on the first graph. Here, the line can be seen clearly overtaking. the line at 70°C after 15 seconds, and the results at the 60°C are. anomalous as a result of the.
The Arrhenius equation ln k = ln A – (Ea / RT) can be shown