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- when 2.0 of NaOHCO3 was heated the weight of the reside product after heating was 1.0g calculate the percent weight loss on heatingHow to get the wt. sample, g.For 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
- A Student added 1.5 ml of 3 M Methyl Magnesium Bromide to a conical vial containing Benzophenone & diethyl ether, after 30 minutes he added dilute sulfuric acid and isolated the product. Identify the Major product. a) b) c) d)reaction: NaHCO3(aq) + HCl(aq) ® NaCl(aq) + CO2(g) + H2O(l) How would your results be affected if the residue in the beaker wasn’t completely dry?Calculate the ΔH°rxn for the following reaction. SiO2(s) + 4HCl(g) → SiCl4(g) + 2H2O(g) ΔH°f [SiO2(s)] = –910.9 kJ/mol; ΔH°f [SiCl4(g)] = –657.0 kJ/mol; ΔH°f [HCl(g)] = –92.3 kJ/mol; ΔH°f [H2O(g)] = –241.8 kJ/mol Group of answer choices –139.5 kJ –137.4 kJ –104.4 kJ 104.4 kJ 139.5 kJ
- Compounds Mass 3-nitrophthalic acid used 200 mg = 0.2g 8% aqueous hydrazine used 0.4mL 3-nitrophthalhydrazide obtained 130 mg = 0.13 g sodium hydrosulfite dihydrate 0.6 g luminol obtained 70 mg = 0.07 g compute yield for nitrophthalhydrazide in the first step (assume nitrophthalic acid is limiting reagent) compute yield for luminol in the second step (using nitrophthalhydrazide as limiting reagent) compute yield for the overall reactionThe polyprotic acid, H3X, has acid dissociation constants Ka1 = 1.3×10-1, Ka2 = 6.5×10-4, and Ka3 = 3.3×10-5. What is the Kc value of the following reaction at 25°C?HX2−(aq)+OH−(aq)⇋X3−(aq)+H2O(l)HX2−(aq)+OH−(aq)⇋X3−(aq)+H2O(l)Report your answer to two significant figures using E notation.ALL INFORMATION NEEDED IS PROVIDED, I NEED FIRST QUESTIONS ANSWERED FIRST BEFORE OTHERS CAN BE ANSWEREDRate = -(Delta[RCl])/Delta t andln[RCl]t=-k*t+ln[RCl]0 < [R-Cl]-time equation Test tube A = 0.30 of 0.10M NaOH(aq), 6.70ml DI H2O & 1 drop of bromophenol bue Test Tube B = 3.00M of 0.10M R-CL in acetone. Mix together in 3 water baths Bath A Trial 1= 20C, 53.6s Trial 2= 20C, 49s for color to changeBath B Trial 1= 32C, 18s Trail 2= 32C, 20s Bath C Trial 1= 10C, 204s Trial B= 10C, 200s average out both trials for each bathcolor change happens after 10% of R-Cl has reacted so In terms of R-Cl this means that 10% of the R-Cl has reacted and ____ of the R-Cl is left Q1. at the time of the color change [R-Cl]t=____[R-Cl]o (Fill in the blank) Q4. Substitute the value of [R-Cl]t into appropriate equation listed above and solve for k in terms of [R-Cl] and t. Q5. Use average time for each temp and determine the value of k (with units) at each temperature Q6. Rearrange the Arrhenius…