The longer the carbon chain of saturated fatty acid, the higher the melting point. Isomeric forms of fatty acids also affect the melting point. Fatty acid in cis form has lower melting point compared to fatty acid in trans form. Next, the polymorphism of the fat also can affect the melting point. β’ crystal form has high melting point compared to α form but lower than β form.
3.1 ABSTRACT The intermolecular interactions in the N-methylformamide with acetophenone, cyclic ketones (cyclopentanone and cyclohexanone) binary liquid systems are studied in combined experimental and computational methodology. The speed of sound (u), density (ρ) and viscosity ( ) values are measured for binary mixtures of N-methylformamide with ketones over the entire range of mole fraction at temperatures T= (303.15 to 318.15) K, at atmospheric pressure. From the experimental results, the values of excess molar volume ( ), excess isentropic compressibility (κsE), deviation in viscosity (η) and excess Gibbs free energy of activation of viscous flow ( ) are evaluated. The experimental results indicate the intermolecular association between the binary liquid
The difference between the rates of the reaction’s when 1 mole was inserted, and when the 3 mole was inserted is 118.49 seconds. This is the data the proves that my hypothesis is correct, this is because as the amount of mole increase, the amount of energry included in the reaction is more. for example, take the 1 mole of HCL, the 2 mole of HCL, and the 3 moles of HCL. The 1st mole of HCL has more particles than the 2nd mole of HCL, and thus when the 1st mole reacts with HCL, the rate of the reaction will be more because there will be more less energy in the reaction of HCL and Mg, compared to the 2nd mole of HCL and it reaction with Mg, (the 2 moles of HCL has more energy than the 1 mole), so hence, there will be more energy in the reaction . Both mole of HCL(the 2 moles of HCL and the 1 mole of HCL) have less energy , and thus when the 3 moles of HCL reacts with Mg, there is even more energy between the Mg and the HCL particles, so hence the rate of the reaction will be faster.
For example, If 2.24L of oxygen gas at STP has given, then it can be easily conclude that the volume would contain one tenth of the mole of oxygen gas. That is, total number of 6.022 x 1022 molecules of oxygen will be present. CONSEQUENCES OF AVOGADRO'S LAW There are a few important consequences of Avogadro's law The molar volume of all ideal gases at 1 atm pressure and 0°C is 22.4 L the volume increases. the amount of gas increases, only If temperature and pressure of a gas are constant when the volume decreases and amount of gas decreases, If pressure and temperature of a gas are constant. The fallowing graph shows the relationship between mass (n) and volume (v) as shown in Avogadro's Law.
A third peak with an integration of 3 was upfield of the benzene proton peaks. This peak corresponds to the protons of the methyl component of the ether substituent. Normally methyl protons would show up further upfield but the presence of the highly electronegative oxygen atom lead to increased deshielding of the protons. Carbon NMR showed six peaks the most deshielded carbon falling at a PPM of about 162. The most upfield of the carbons was at a PPM of 48 and belonged to the methyl carbon at the end of the ether substituent.
Initially, the conversion of benzyl alcohol in to benzaldehyde was chosen as a model reaction to optimize the reaction conditions. Effect of reaction time and mmol of H2O2 on progress of oxidation reaction was studied (Fig. 4. the experiment was performed with 20 mg catalyst, 10 ml acetonitrile and two different amount of H2O2 1 and 2 mmol for 1mmol benzyl alcohol at reflux condition (85 ºC ) and plotted with respect to the time. With increasing the mole ratio of BzOH : H2O2 from 1:1 to 1:2, the conversion of benzyl alcohol increased from 75% to 93%. The conversion also increased with increasing time of reaction and then remain constant at 180 min.
Given that 3.327 g of product was produced during the reaction, how many grams of water were released as water vapor? (moles = mass/molecular weight). From observation it appeared 1.673 grams of water were released as water vapor. 5. Write a balanced equation for the decomposition of copper carbonate hydroxide hydrate.
This is evident from the following figure. Figure 24 SEM images of (a) Sample with Tsat=200C, Tf=1100C and RD=84.1; (b) Sample with Tsat=00C, Tf=900C and RD=84.1. (Courtesy to ) Effect of foaming temperature on cell nucleation density and average cell size Variation in nucleation densities with temperature can be observed in the following figure Figure 25 Cell nucleation density as a function of foaming temperature for samples foamed at different saturation temperature. (Courtesy to ) The nucleation density increases with a decrease in foaming temperature. Cell nucleation density as high as 1014 cells/cm3 were obtained for saturation temperature of -100C for which CO2 concentration was 14.7% whereas for a saturation temperature of 600C the nucleation density was 109 cells/cm3 where CO2 concentration was 5%.
3.00 g of a sample was placed in a graduated cylinder with 10.0 mL of water. The volume of the water increased to 12.4 mL. What is the density of the sample? Challenge question: The nucleus of a carbon-12 atom is 2.7 x 10−3 pm and has a mass of 1.99 x 10−23 g. What is the density of the nucleus in g/cm3? Assume the nucleus is the shape of the sphere and use the formula v= 4 3 π r 3 v=43 π r3 First let’s find the volume.
There is a diverse range of chemical groups and reactions that give rise to the exquisite photochromic effect. For instance, cis-trans isomerizations or geometric isomerism, where the functional group of a molecule is rotated into a different position or orientation, retaining the same molecular formula. Then there are pericyclic reactions, following UV exposure, the molecule switches to form B by the rearrangement of bonds. The Spiropyrans, the most prevalent class of photochromes, are a good example. When exposed to high energy UV, the sp³-hybridized carbon-oxygen bond in a Spiropyran breaks and opens the ring such that the carbon becomes sp² hybridized.