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- What products are expected from the oxidation with hot KMnO4 (a) solution of (a) PGE1, and (b) PGE1α?Which is the most readily reduced? a.) Pb+2 + 2e- -> Pb Eo=-0.13V b.) Ag+ + e- -> Ag Eo=0.80V c.) Cl2 + 2e- -> 2Cl- Eo=1.36V d.) F2 + 2e- -> 2F- Eo=2.87VFor 20.7 the solutions for part c and d dont seem right. Shouldn't reacting with 1 eq of H2 just take away the C=C bond and not the C=O bond. (H2 reduces C=C selectively to form a ketone according to page 738. Also the both the C=C and C=O bond should be reduced if excess H2 is used
- Ammonia (NH3) is a weak base that under acidic conditions becomes protonated to the ammonium ion in thefollowing reaction:NH3 + H+ → NH4 +NH3 freely permeates biological membranes, including thoseof lysosomes. The lysosome is a subcellular organelle witha pH of about 4.5–5.0; the pH of cytoplasm is about 7.0.What is the effect on the pH of the fluid content of lysosomes when cells are exposed to ammonia? Note: Ammonium (NH4 +) does not diffuse freely across membranes.One of the following is reducing agent: a-S4O6^-2 b-more than one are true c-S2O5^-2 d-S2O3^-2Draw themajor product of this reaction. Ignore inorganic byproducts. PPC, CH2Cl2
- Sulfate reduction 5C2H4O2 (acetate) + 10H2O = 10CO2(g) + 40e- +40H+ E0’= (-0.28)*(-1) 4SO4- + 40e- + 40H+ = 4H2S(g) + 16H2O E0’= (-0.22) 5C2H4O2 + 4SO4 = 4H2S + 10CO2 + 6H2O full reaction= 0.06 How is 4SO4- + 40e- + 40H+ = 4H2S(g) + 16H2O E0’= (-0.22)? I don't get how it is -0.22 based off the redox tower.6 Please provide the correct answers. Thanks in advanceDraw the organic prouduct of the reaction with a) HBr b) Br2 c) H2, Pt/C d) O3; (CH3)2S
- The name for Reaction 1 is ______ Choices: A. Acidification B. Oxidation, C. Reduction, D. None of the choices The reagent/s for REACTION 1 is/are ______ Choices: A. CH3MgBr in ether, B. HCl, C. CH3NH2, D. none of the choicesMay u olease help me with this one and explain thanksI just was wondering why its not 2e- for the second reaction and just e-