Comparative Evaluation Between Mathematical Models For Drug Release Of Curcumin Loaded Multifunctional Albumin Nanoparticles

By extrapolating from the curve, the concentration of the unknown solution is ≈ 1.9mg.mL. The rate of reaction increases linearly with an increase in the substrate concentration.

Rate= k [I-]1[H2O2]. (2.13675*10-5 ) = k [0.015] [0.015] then solve for k. For this trial, k=0.09497.

Time) because it had a correlation closest to 1. All three orders were graphed and a linear regression was used to see which graphed order was closest to 1. The order was determined by comparing the concentration and time to the mathematical predictions made using the integrated rate laws. Analyzing each graph and finding each correlation helped determine which graph was closest to 1. The more concentrated a solution is, the higher the absorbance of that solution. This is due to Beer’s Law. The law measures the absorbance of a solution by determining how much light passes through a solution. As the concentration of a solution increases, fewer wavelengths of light are able to pass through the concentrated solution. The absorbance at 60 seconds was 0.573 (Figure 1: Table1). To calculate the concentration (molarity), the Beer’s Law equation was used, Abs = slope(m)+b. Plugging in what is known into the Beer’s Law equation resulted in 0.573 = 3.172e+004 + 0, where the concentration is determined by M = 0.573-0/ 3.172e+004. So, the concentration at 60 seconds using the equation (M = 0.573-0 / 3.172e+004) was 1.824e-5 M. The 1st order graph resulted in k=0.006152 (Figure 1: Graph 1). Other groups also resulted in their decolorization of CV to be the 1st rate

3.6.3. 2, 4 – D (2, 4–Dichloro phenoxy acetic acid) stock solution (1mg/ml): 10.0mg of 2.4-D being weighed and dissolved completely in 1N NaOH to a final total volume

Is this a linear relationship? What happens to the initial reaction rate as substrate concentration increases?

The time for each trial will be measured. The time will start when the tablet is in the acid and it will end when the reaction is complete and the whole tablet has

The differential equations governing the instantaneous state of inventory level at any instance t are given by

Kr = Vmax per mg of enzyme x molecular weight x 10-3 mmol µmol-1 x 1 min per 60 seconds.

The slope of this graph, calculated using the average rise / run is also 1. As [S2O8-2] is held constant during these trials, the exponent n for [I-] equals one. The k constant can now be calculated.

formula Kw[Fex(C2O4)y]·zH2O. The variables x, y, and z were determined through the duration of the

Using two graduated cylinder, dispensed 2.00 mL of KI solution into a clean, dry 100 mL beaker and dispensed 3.00 mL of KI into a different 100 mL beaker. Moreover, dispensed 2.00 mL of Pb(NO3)2 into a clean, dry, small beaker using the digital Micropipette.

∆G˚= (-R)(T)(lnKsp) where R is a constant of 8.314 J/molK, temperatures values are plugged in for T, and the ln of Ksp is required. The average ∆G˚ found was -8325.4 J/mol. Using the Excel constructed graph and the slope intercept equation y= -2828.5x + 12.67 and the formula ∆G˚= ∆H˚-T∆S˚ the thermodynamic properties ∆H˚ and ∆S˚ can be calculated. ∆G˚ represents the spontaneity of a system, and because Potassium Nitrate is a relatively soluble salt, ∆G˚ should be a negative value. This experiment is a great practice to better understand all the thermodynamic valariables relative to the dissolving of KNO3 and how they all relate to one

Table 2 shows the concentrations of S2O8-2 and I- and the time of each reaction for each run. Rate of the reaction was calculated by dividing 1M with the time it took to complete the reaction. The constant k was calculated using the rate law (4), the x and y value was determined by the graphs below. Considering most of the k’s calculated by our laboratory class, the supposed real value of k must be 10/Ms for that reaction.

2) Calculate the a mean rate constant using orders of reactions and the rate equation allowing for the overall order or reaction to be found.

Table 3 : The time taken for potassium permanganate solution to decolourise in different concentration of glucose solution