Label the functional groups and identify the correct compound based on the IR spectrum. It is hard to see that: a) has a double bond b) has a triple bond c) has a triple bond d) has both a double and triple bond e) has a triple bond % Transmittance 100 50 0 4000 b) 3000 OH d) 2000 1500 Wavenumber (cm-¹) H quan 1000 500

Label the functional groups and identify the correct compound based on the IR spectrum. It is hard to see that: a) has a double bond b) has a triple bond c) has a triple bond d) has both a double and triple bond e) has a triple bond % Transmittance 100 50 0 4000 b) 3000 OH d) 2000 1500 Wavenumber (cm-¹) H quan 1000 500

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Calculation of AH; of Magnesium Oxide Mg + 2 HCI MgCl2 * Hz2 AH, = kJ/mole MgCl2 +H2O > Mgo + 2 HCI -AH2 = kJ/mole He + 를 O2 + H2O AH3 = - 286 kJ/mole net: Mg + 을 O2- MgO 사Hnet = kJ/mole The literature value for the AHf of magnesium oxide is -602 kJ/mole.
Metallic Titanium and its alloys (especially those with aluminum and vanadium) combine the advantages of high strength and light weight and are therefore used widely in the aerospace industry for the bodies and engines of airplanes. The major natural source for titanium is the ore rutile, which contains titanium dioxide (TiO₂). a. An intermediate in the preparation of elemental titanium from TiO2 is a volatile chloride of titanium (boiling point = 136°C) that contains 25.24% titanium by mass. Determine the empirical formula of this compound. b. At 136°C and atmospheric pressure, the density of this gaseous metal chloride is 5.65 g/L. Determine the molecular formula of this compound. (Hint: determine its molecular mass first)
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produce carbon dioxide and water, as described in the following reaction: C.H»O, (s) + _ O: (g) → _ H;O (1) + CO; (g) are come
g In P Englis 9 Self Se G Gramn Tn Course G traduc * Balanc * Parap 4 Build a C How M G om/course.html?courseld%3D17244391&OpenVellumHMAC%=D9815e7584135fc006cabedb9aa0dda22#10001 The balanced reaction for the combustion of ethane is shown in the table. You allow 32 mol of ethane (CH6) to react with 4.1 mol oxygen (02). may be done using the Simulation, to indicate the amounts of all chemical species for the initial, change, and final conditions. Drag the appropriate amounts to their respective targets. • View Available Hint(s) Reset Help 2.3 mol 1.2 mol 0.0 mol 4.1 mol 2.0 mol 6.4 mol 3.2 mol 3.5 mol 11.2 mol 9.6 mol 2C,H.(g) + 7O2(g) → 4CO2(g) + 6H2O(g) Initial Change- Final Pearson Copyright © 2022 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy | Permissions | C
5. When 5.0 moles of potassium metal reacts with an excess of H₂0, 3.5 g of H2 are produced. 2 K (s) + 2 H20 (1) 2 KOH (aq) H₂ (g) a. Calculate the number of moles of water needed to react with all of the potassium metal. (Ans: 5 mol) + b. Calculate the theoretical yield of H2, in grams, formed when all 5.0 moles of potassium metal are consumed. (Ans: 5.04 g) c. Calculate the percent yield of the reaction, given 3.5 g (actual yield) are produced. (Ans: 69.4%)
Gravimetric analysis of Fe3O4 (MW = 232 g/mole) may be undertaken with the following reactions: Fe3O4 → Fe2O3 → Fe (OH)3. Weight of sample containing 8.00% Fe3O4 that must be taken to obtain a precipitate of Fe(OH)3 (MW = 107 g/mole) that weighs 150 mg is . a. 0.108 g b. 0.325 g c. 1.355 g d. 4.065 g Amount of Fe2O3 (MW = 160 g/mole) from which 150 mg of Fe(OH)3 (MW = 107 g/mole) may be obtained is . a. 0.112 g b. 0.224 g c. 0.448 g d. none of the other choices
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25. An indlcation of the average formula weight of a fat is its saponification number, expressed asthe mlligrams KOH requlred to hydrolyze (saponify) 1 g of the fat: CH,CO,R CH,OH CHCO,R + 3KOH → CHOH + 3RCO,K CH;CO,R CH,OH where R can be variable. A 1.10-g sample of butter is treated with 25.0 mL of 0.250 M KOH solution. After the saponification is complete, the unreacted KOH is back-titrated with O.250 M HCI, requiring 9.26 mL. What is the saponification number of the fat and what is its average formula weight (assuming the butter is all fat)?