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- Explain Pearson’s quantitative approaches to HSAB [hardness and electronegativity]Br Br > WHat Kind oF (caalon is this? E2IEi|SNI|SN2 1, MgBr 2. H20 HO heat CH3 CH2C AIC13More than 40% of the world's population has limited access to clean water. One proposed method for the production of clean water involves removing salt from seawater. Which technique would be used to separate the salt from the salt water mixture? a) chromatography b) fractional distillation c) filtration d) simple distillation
- Q.No.3 What is chromatography?. Why chromatography is considered as an important tool for analysis of organic compounds.14. Shown below are the mass spectra (without data tables) of four molecules. (The structure of each molecule is shown on the spectrum.) Mark the [M] peak for each, and indicate the molecular weight of the molecule. Relative Intensity Relative Intensity 100 80 60 40- 20 0-mm 10 100- 80 a. 20- 115-14-5501 0-mm 10 20 b. =0 20 30 30 40 40 50 50 60 m/z 60 m/z 70 70 80 80 90 100 90 100 Relative Intensity Relative Intensity 100- 80 40 20 0 100 80 40 10 20- 15-NU-5580 15 20 25 115--555 ont 10 20 30 30 35 40 45 50 55 m/z 40 50 m/z 60 60 70 65 70 75 80 Based on the five mass spectra you have seen, describe how one might identify the [M]* peak (molecular ion peak) on the mass spectrum of an unknown compound.The spectroscopic data in the table is generated with five solutions of known concentration. Concentration (M) 0.0179 0.0358 0.0716 0.143 0.286 Absorbance 0.1425 0.1896 0.5722 1.126 2.110
- Potentially Useful Information Spectrochemical Series: I < Br < SCN° < Cl° < NO3¯ < F < OH° < C2O4²-~ H2O < NCS < NH3 < en < PPH3 A student has three test tubes containing a metal (M) nitrate solution M(NO3)2 (aq) (where "M" represents a generic transition metal). The student adds aqueous ammonia (NH3) to one test tube, aqueous hydrochloric acid (HCI) to the second tube, and nothing more to the third tube, but forgets to label the tubes. After this, one test tube contains a red solution, one an orange solution, and one a yellow solution. Note: This metal, M, would follow the same rules that we used for Cu(II), Ni(II), and Co(II). The Jahn-Teller Effect does not apply. Complete each statement or answer each question below about the solutions in the three test tubes. The coordination complex causing the orange solution causes a larger splitting between the d orbitals of the metal then the coordination complex causing the... O red solution. O yellow solution. What color light does the…The spectroscopic data in the table is generated with five solutions of known concentration. Concentration (M) Absorbance 0.0170 0.0586 0.0340 0.117 0.0680 0.234 0.136 0.469 0.272 0.937 If the concentration is on the x-axis, what is the slope of the line formed by these points? slope = M-! * TOOLS x10Potentially Useful Information Spectrochemical Series: | < Br < SCN¯ < Cl° < NO3¯ < F° < OH° < C2O4²¯ ~ H2O < NCS° < NH3 < en < PPH3 く A student has three test tubes containing a metal (M) nitrate solution M(NO3)2 (aq) (where "M" represents a generic transition metal). The student adds aqueous ammonia (NH3) to one test tube, aqueous hydrochloric acid (HCI) to the second tube, and nothing more to the third tube, but forgets to label the tubes. After this, one test tube contains a red solution, one an orange solution, and one a yellow solution. Note: This metal, M, would follow the same rules that we used for Cu(II), Ni(II), and Co(II). The Jahn-Teller Effect does not apply. Complete each statement or answer each question below about the solutions in the three test tubes. The aqua complex causes less splitting of the d orbitals of the metal than the... ammonia complex. O chloro complex.
- The figure below shows 2 images obtained by different techniques. The sample is the same y corresponds to TiO2 only. Answer the following questions. With what other technique can you confirm that it is titanium oxide? explain whyWhat is in a fractionating column? Silica gel Collected fractions Glass or plastic beads Liquid nitrogenChemistry In a TLC experiment, three spots were made. In spot A was a crude sample of a mixture of three compounds ethyl 4- aminobenzoate (base), salicylic acid, and naphthalene as the neutral compound. In spot B, was just the neutral compound, naphthalene. In spot C was the pure compounds of spot A. When dipped with KMnO4, it is said that all spots should turn yellow, but while doing the experiment it was only the bottom compounds that turned yellow for me which I found out was the acid. Should all spots really turn yellow? Why would this happen?
