6.3 A plot of In K vs. 1/T gives a slope of -3186 K and an intercept of -18.3. Calculate the value of delta S in J/mol K. O a. 26.5 J/mol K O b. 192 J/mol K O c. 383 J/mol K O d. -2.20 J/mol K none of these O f. -38.4 J/mol K Og. -152 J/mol K Clear my choice 6.4 When a 0.3885 M HCI solution is used to titrate 6.7 ml of a borate solution, the initial volume on the burette is 0.82 ml and the final volume is 18.15 ml. What is the number of moles of HCl added? (This data will be repeated in the next question.) (Give 4 digits of the calculator answer) O a. 17.33 mol Ob. 6.733 mol O c. none of these d. 0.003366 mol Oe. e. 0.08426 mol Of. 0.006733 mol g. 9.730 mol Clear my choice MacBook Air

6.3 A plot of In K vs. 1/T gives a slope of -3186 K and an intercept of -18.3. Calculate the value of delta S in J/mol K. O a. 26.5 J/mol K O b. 192 J/mol K O c. 383 J/mol K O d. -2.20 J/mol K none of these O f. -38.4 J/mol K Og. -152 J/mol K Clear my choice 6.4 When a 0.3885 M HCI solution is used to titrate 6.7 ml of a borate solution, the initial volume on the burette is 0.82 ml and the final volume is 18.15 ml. What is the number of moles of HCl added? (This data will be repeated in the next question.) (Give 4 digits of the calculator answer) O a. 17.33 mol Ob. 6.733 mol O c. none of these d. 0.003366 mol Oe. e. 0.08426 mol Of. 0.006733 mol g. 9.730 mol Clear my choice MacBook Air

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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Ee.113.
pdf is needed, but algebra & For an instantaneous rate of change, is used to denote geometry can be used to approximate the value. The symbol an instantaneous change in a value, 1) What is true about the reactant concentration over time? A[Reactant] constant? At d[Reactant] 3) Circle the time at which the magnitude of the (instantaneous rate of dt change) is the lowest (i.e., during which time is the tangential slope least negative? This is the time when the [reactant] is changing least over time) 0.20 Concentration of reactant vs time 0.15 Rate = -Slope = 1.06x10-2 M/sec jou 0.10 0.05 0.00 15 10. Time, sec. 20 25 https://wnww.chem.purdue.edugdhelphowfosolveilKineticsCalculaingRales.html
For a certain process at 455 K, A G=-12.1 kJ and A H=-9.2 kJ. Therefore, A S for the Oa. 6.37 J/K mol process is Ob.0 J/K mol Oc. 3.98 J/K mol d. 8.2 J/K mol O e.-8.2 J/K mol
Calculate AH for the process in which 10.0 mL of water is converted from liquid at 5.0 °C to vapor at 75.0 °C? (Density of water is 0.99 g/mL and AvapH° = 44.0 Kj/mol H20) Lütfen birini seçin: O a. 27.1 kJ O b. 15.8 kJ O c. 35.2 kJ O d. 10.3 kJ O e. 33.5 kJ
Experiment 14 Advance Study Assignment: Heat Effects and Calorimetry 1. A metal sample weighing 147.90 g and at a temperature of 99.5°C was placed in 49.73 g of water in a calo- rimeter at 23.0°C. At equilibrium the temperature of the water and metal was 41.8°C. metal a. What was At for the water? (At = tanal - tinitial) T左= 99.5% 4t<(418を-23.6℃ /8.8% 18.8 °C H20 49.739 b. What was At for the metal? TE= 23,0°C At = (48%-79.5) = 57.72 -57.7 38こ How much heat flowed into the water? (Take the specific heat of the water to be 4.18 J/g°C.) °C C. AH=4,1E) = 1°C e latm 99.5 x 4.18 3907.98 3907.98) joules 319.77 d. Calculate the specific heat of the metal, using Equation 3. 9ニmedt 3407-48 J (14746) Cs (-57.7C) O4579 joules/g°C e. What is the approximate molar mass of the metal? (Use Eq. 4.) g/mol 2. When 4.98 g of NaOH was dissolved in 49.72 g of water in a calorimeter at 23.7°C, the temperature of the solution went up to 50.1°C. Is this dissolution reaction exothermic? Why? a. b. Calculate…
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Hess's Law II Calculate A,H for the reaction, from the following information: 2 Ni(s) + 1/4 Sg(s) + 3 O₂(g) ----> 2 NISO3(s) NISO3(s) ----> NiO(s) + SO₂(9) 1/8 Sg(s) + O₂(g) ----> SO₂(g) Ni(s) + 1/2O₂(g) ----> NiO(s) AH 166.00 kJ AH = -247.00 kJ A,H= -251.00 kJ
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