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Q: Determine the mass in grams of C,H10 that are required to completely react to produce 8.70 mol of…
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Q: how do i do this
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- Which of the following is a spontaneous reaction.? a. Rxn with ΔH =- 10Kj/mol ΔS= -5J/mol T= 300K b. NaCl +H20 -> NaOH + HCl 25C c. H20(l) -> H2O(s) Temp: 25C d. Dissolution of 100g of solid sugar in 100 mL ice tea. Consider following reaction: HgO (s) -> Hg(l) + ½ O2 (g) Delta H = +90.7 kj/mol. What quantity of heat in kj/mol is required to produce one mole HgO? Write your answer without units. Given the following data 2ClF(g) + O2(g) --> Cl2O(g) + F2O (g) Delta H= 167.4 kJ I 2ClF3(g) + 2O2(g) --> Cl2O(g) + 3F2 O (g) Delta H= 341.4 kJ II 2 F2(g) + O2(g) ---> 2F2O (g) Delta H= -43.4 kJ III Calculate the delta H in kJ for below reaction: ClF(g) + F2(g) ---> ClF3(g)Calculate ΔG°rxn for the following reaction. The ΔG°f for each species is shown below the reaction. 4 ABO3 (g) + 5 B2A4 (l) → 7 B2 (g) + 12 A2O(l) ABO3 (g) B2A4 (l) B2 (g) A2O(l) ΔG°f (KJ/mol) -73.5 149.3 0 -237.1Reaction of interest : S2O82-(aq) + 3I- (aq)→ 2SO42-(aq) + I3-(aq). rate= k[S2O82-]1[I-]1.
- fischer esterification: 5.4mL of isopentyl alcohol, molar mass 88.15g/mol: 8.5mL of acetic acid, molar mass 60.05 g/mol: Experimental final product(Isopentyl acetate) I got: 3.10g. 1.Calculate for theretical yield. 2.Percentage yield 3. % recovery.The given are SlGMATROPlC REACTlONS. Show the arrøw pushing then find their products.Calculate Trial I, Trial II, Trial III Vol.Ag(NO3)Added.
- Examine the experimental data shown in the table below for the following reaction: (make sure to show your work for each part) A + 5B + 6C ---> 3D + 3E Experiment [A] (M) [B] (M) [C] (M) Rate (M/s) 1 0.35 0.35 0.35 8.0 x 10-4 2 0.70 0.35 0.35 3.2 x 10-3 3 0.70 0.70 0.35 6.4 x 10-3 4 0.70 0.35 0.70 3.2 x 10-3 A. What is the reaction order of A? B. What is the reaction order of B? C. What is the overall rate equation for this reaction? D. What is the value of the rate constant? E. What are the units of the rate constant?Reducing NO Emissions Adding NH3 to the stack gases at an electric power generating plant can reduceNOx emissions. This selective noncatalytic reduction (SNR) process depends on the reaction between NH3 (an odd-electron compound) and NO.$$4NH3(g)+6NO(g)5N2(g)+6H2O(g)The following kinetic data were collected at 1200 K. Experiment [NH3] (M) [NO] (M) Rate (M/s) 1 1.00x10-5 1.00x10-5 0.120 2 2.00x10-5 1.00x10-5 0.240 3 2.00x10-5 1.50x10-5 0.360 4 2.50x10-5 1.50x10-5 0.450 What is the rate-law expression for the reaction? Do not add multiplication symbols to your answer. $$Rate=If all other variables were kept constant, determine theeffect that the following errors would have on the calculatedpercent yield of the product. Would the yield be expected toincrease, decrease, or would there be no effect? Explainyour reasoning.– The product was insufficiently dried before weighing.– Some of the product was lost during the transfer fromthe Buchner funnel to the evaporating dish.– 7.5 mL of FeCl3 was added instead of 3.0 mL asoutlined in the procedure.– 4.587g of K2C2O4H2O was used instead of exactly4.000g .– The recrystallization step was skipped and theexperiment went straight to vacuum filtration.
- Run Reactants Temperature (Celsius) Initial Rate (kPa/s) 1 10 ml 3% H2O2 = 5ml 0.5M KI 23.0 0.0798 2 5ml 3% H2O2 = 5ml 0.5 M KI 23.0 0.0391 3 5ml 3% H2O2 + 10ml 0.5 M KI 23.0 0.0794 Run Initial Rate (mol/L*s) [H2O2] after mixing [I-] after mixing 1 3.24 x 10^-5 0.147 M 0.0417 M 2 1.59 x 10^-5 0.0733 M 0.417 M 3 3.23 x 10^-5 0.0733 M 0.0833 M Calculate the rate law for the reactionTable 2: Molarity of H2O2 and KI and Reaction Rate Trial H2O2 Concentration, M KI Concentration, M Reaction Rate(Reciprocal Slope) 1 0.29 M 0.40 M 14.08 2 0.29 M 0.20M 25 3 0.023 M 0.40 M 20 Rate constant value Trial 1 ______________ Trial 2 ______________ Trial 3 ______________ Average _____________Table 2: Molarity of H2O2 and KI and Reaction Rate Trial H2O2 Concentration, M KI Concentration, M Reaction Rate(Reciprocal Slope) 1 0.29 M 0.40 M 14.08 2 0.29 M 0.20M 25 3 0.023 M 0.40 M 20 Order with respect to H2O2: -0.138 Order with respect to KI: -0.828 Rate constant value Trial 1: 5.84 Trial 2: 5.55 Trial 3: 5.56 Average: 5.6 -Please help me with this part, please 1. Should the rate constant (k) be the same for all three trials in this experiment? Explain your answer. 2.Write the complete rate law for this reaction.