Figure 1 shows the synthesized ionic liquid of CVD with studied acidic compounds (white crystalline materials) in 1:1 molar ratio after dissolving them in methanol and complete solvent evaporation after five days. CVD with CA, TA and SAC convert to a viscous yellow liquid form. This method was used to preparation of different ionic liquid form of drugs such as ketoconazole with TA and CA (24) and sulfasalazine and acyclovir with choline (28).
****Fig 1****
The DSC thermograms of CVD, CA, TA and their ionic liquid forms have been demonstrated in Figure 2. They shows that CVD and studied acidic compounds are crystalline and they have a distinct endothermic peaks corresponding to the melting of solid forms. The DSC curve of CA exhibits a broad
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According to the simulated NMR spectra of CA, TA and saccharin by ACD-ilab software (34), hydroxyl functional groups of them have a single peak at ~10 ppm which in is upfield to ~6 ppm in ionized from whenever the first carboxylic acids was converted to carboxylate (COO−). In SAC, disappearance of a single peak at 8.1 ppm of saccharin confirm ionization of 1,2-benzisothiazole ring of SAC. CVD has no peak in the studied regions (8 ppm and ~6 ppm for ionized form of SAC and studied carboxylic acid, respectively).
Similar spectra (Figure S1 in supplementary information) of CVD-CA and CVD-TA show the appearance of a peak at ~6 ppm. It established the ionization of the first carboxylic acid of CA and TA and salt formation. In CVD-SAC, disappearance of a single peak at 8.1 ppm of SAC in ionized form confirm ionization of 1,2-benzisothiazole
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The solubility of CVD-ILs with CA and TA is less than CVD. This observation could be related with pH of final solution. pHs of final solution for CVD-ILs in HCl 0.2 M are 2.5-3 for IL-CA and IL-TA. According to the pKa of CA and TA (Table 1), they are partially in non-ionized form and synthesized ionic liquid cannot increase the CVD’s solubility against of the ionic liquid’s solubility in acetate buffer solution. Nevertheless a high concentration of chloride ion in solution because of fully ionization of counter-ions, the solubility improved in compared with CVD. Moreover, based on previous study by Hamed et al. a change in CVD solubility could be related to ionic strength of solution. It is another important parameter in evaluating solubility of CVD and it decreases solubility (14). Ionic liquids can produce higher ionic strength in solution medium and it is a possible reason for decreasing of solubility in acidic medium whenever CVD is fully ionized.
Similar pattern is observed in solubility studies of CVD and corresponding IL forms in HCl 0.1 M and 0.01 M. However, solubility was considerably improved in compared with HCl 0.1 M, 0.2 M). It could be related to the low concentration of chloride ion. The maximum solubility was observed in CVD because in this pH, it is fully ionized and ionic strength in compared with CVD is less than CVD’s ionic
To begin, the solubility of the unknown compound in water was tested. If the compound is soluble in water, it can be inferred that it is either a polar covalent or ionic compound.
Goals The primary goal of this experiment was to identify an unknown compound by running various tests to determine the qualitative solubility, conductivity, and pH value of the compound. Tests were also performed for the presence of specific cations and anions in the compound. The second goal was to discover the reactivity of the unknown compound by reacting it with different types of substances. The third goal of this project was to calculate the quantitative solubility of the unknown compound in water.
These results a somewhat inaccurate due to the fact that when the solutions were actually freezing is difficult to tell. Some solutions froze slowly and showed a plateau, while others
As much was conducted throughout this lab, the projected completion of this lab displays that ultimately, the higher the temperature of the water, the faster the dissolving rate of the Alka-Seltzer is. In other words, the hotter the water temperature the quicker the tablet dissolves within the water in regards to the amount of time it took to dissolve. Furthermore, this experiment helps to explain that, if water is taken at a higher temperature and Alka-Seltzer is placed within the water, the Alka-Seltzer will take less time to dissolve because the higher temperatures cause the tablet to melt at a quicker rate. This compares to when Alka-Seltzer is placed in colder temperatures, where instead it takes more time to dissolve, because the lower
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)
The solubility rate of copper (II) chloride in methanol is 53g/100ml whilst the solubility rate for sodium chloride is 65g/ml. Although there solubility rate is fairly close the difference is enough that when little amounts of methanol is added only the copper (II) chloride dissolves. A factor that affects the solubilty of metals is their molecular mass. Copper (II) has a molecular mass of 63.546 whilst sodium has a molecular mass of 22.989769 meaning copper has higher solubility rate than sodium, this is because as the molecular mass of a metal increases it becomes difficult for molecules to hold onto their solute particles and when those particles break away they can easily dissolve into the solvent. Therefore because coppers molecular mass is greater than sodiums it’s solute particles breakaway with less resistant meaning copper dissolves better.
Aims of experiment • Determine the rate constants for hydrolysis of (CH3)3CCl in solvent mixtures of different composition (50/50 V/V isopropanol/water and 40/60 V/V isopropanol/water) • Examine the effect of solvent mixture composition on the rate of hydrolysis of (CH3)3CCl Introduction With t-butyl chloride, (CH3)3CCl, being a tertiary halogenoalkane, it is predicted that (CH3)3CCl reacts with water in a nucleophilic substitution reaction (SN1 mechanism), where Step 1 is the rate-determining step. The reaction proceeds in a manner as shown
(Schreiner 4) This elucidates how ammonia has few ions to conduct electricity. Moreover, “When table salt is dissolved in water, the solution conducts very well, because the solution contains ions. The ions come from table salt, whose chemical name is sodium chloride. Sodium chloride contains sodium ions, which have a positive charge, and chloride ions, which have a negative charge.
7.1.2 PREFORMULATION CHARACTERISTICS Table 23: Preformulation of powder blend of pantoprazole sodium core tablet Formulation code Angle of repose Bulk density (gm/cm2) Tapped density (gm/cm2) Hausners ratio %compressibility C1 25.7±0.03 0.34±0.02 0.42±0.03 1.24 19.04±0.01 C2 26.8±0.01 0.36±0.02 0.45±0.05 1.26
Olivia Lapchak Chemistry 113-004 23 February 2023 The Chemistry of Natural Waters Introduction When considering the amount of bottled water that is consumed every day, very little is contemplated when choosing what brand to buy. The contents in each of the water brands is vital information one should know before purchasing because it informs the consumer about many measurements including Calcium and Magnesium concentration, total dissolved solids, total hardness, and the water source for each brand. Each of these factors play a role in deciding whether a brand is healthy for the body, or not quite ideal.
Background Information In this lab KCl, NaCl, and a mixture of MgCl2 and NaCl are the independent variables that all lower the freezing point of water. Ice is used as the controlled variable because it is what the salts are lowering the freezing point of. Salt (Na) weakens intermolecular forces of water, thus lowering the the freezing point. This is why in colder climates where icy roads and walkways are a liability, salt is often scattered over areas that are slick with frozen water.
Various solids and liquids were immersed in solvents, ranging from water to TTE, in order to observe the solubility of the given solutes in each solvent. Each solute was placed in 3 different test tubes filled with water, ethanol and TTE. In part one of the lab, 2 small crystals of iodine, nonpolar solutes, were added to 5 ml of water, ethanol and TTE. Upon our observation iodine did not dissolve in all three solvents and instead produced a dark reddish-orange colour and magenta colour in ethanol and TTE respectively. The same procedure done to the 2 small iodine crystals was repeated with the 2 small ammonium chloride crystals, polar solutes.
Optimum quantities of calcium chloride, sodium citrate and calcium carbonate were used in the preparation of in-situ gel to maintain fluidity of the formulation before administration and resulted in gelation when the formulation was added to simulated gastric fluid. In-situ Gel Formulation of ETD was characterized to identify the best formulation. All the formulations were off white to pale yellow colored solution. They had pH in the range of 7.02-7.30. Viscosity of the formulation was determined using Brookfield Viscometer.
Strong acids and strong bases are strong electrolytes and are assumed to ionize completely in the presence of water. Weak acids however, only ionize to a limited extend in water. Any weak or strong acids when in contact with any weak or strong alkali will start to undergo neutralization regardless of their volume. When an indicator which is present in the acid-base mixture and have experienced colour change, it indicates that the mixture is in right proportions to neutralize each other and is also known as the equivalence point.
When a solid dissolves in a liquid, it then changes its physical state (from solid to liquid) by melting. Heat is needed to break the bonds holding the molecules in the solid together and at the same time; heat is given off during the formation of new bonds between solute and solvent. Results Table1. Results for titration of benzoic acid with NaOH at 20◦C V(NaOH) ml V(C_6