For numbers 16-17: Calculate the following values at T = 400 K for argon that behaves as a Berthelot gas.3a(OP)=b-a = 4.7x10-³ Pa m6 K/mol²b = 2.5067x10-5 m³/molRT216. Joule-Thomson coefficient (K/atm), Cp = 20.79 J/mol KA. -0.122B. -0.27817. Inversion temperature (K)A. 280.5B. 8.225C. -0.334C. 7.311D. -0.412D. 281.4

Answered: Image /qna-images/answer/7d58c08c-944e-4b47-aa9d-d5699519db9e.jpg

For numbers 16-17: Calculate the following values at T = 400 K for argon that behaves as a Berthelot gas. Н =b- 3a RT2 a = 4.7x10-³ Pa m6 K/mol² b = 2.5067x10-5 m³/mol OP T 16. Joule-Thomson coefficient (K/atm), C₂ = 20.79 J/mol. K A. -0.122 B. -0.278 17. Inversion temperature (K) A. 280.5 B. 8.225 C. -0.334 C. 7.311 D. -0.412 D. 281.4

For numbers 16-17: Calculate the following values at T = 400 K for argon that behaves as a Berthelot gas. Н =b- 3a RT2 a = 4.7x10-³ Pa m6 K/mol² b = 2.5067x10-5 m³/mol OP T 16. Joule-Thomson coefficient (K/atm), C₂ = 20.79 J/mol. K A. -0.122 B. -0.278 17. Inversion temperature (K) A. 280.5 B. 8.225 C. -0.334 C. 7.311 D. -0.412 D. 281.4

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...

See similar textbooks

Similar questions

der the following reaction: 2NO2(g) N₂O4 (9) Part A Part B Calculate AG at 298 K if the partial pressures of NO₂ and N₂O4 are 0.36 atm and 1.64 atm, respectively. Express the free energy in kilojoules to two decimal places. |ΨΕΙ ΑΣΦ AG= -6288.99 Submit Previous Answers Request Answer ? Review I Consta kJ X Incorrect; Try Again; 4 attempts remaining This value of free energy is more negative than AG for the reaction (i.e., more spontaneous), but the partial pressure of ta about four times greater than that of the reactant, so the free-energy change is expected to be more positive than AG (i.e spontaneous). Given that AG = AG + RTln Q, calculate the value of Q (it should be greater than 1 for the specified conditions), and free energy from Part A to find the free-energy change at the given partial pressures. Make certain values are expressed in before summing them.
construct a balanced reaction for the formation of Fe2O3(s) from its elements in their standard states at 298 K. For a stoichiometric coefficients of 1,
C(s) + 2H2(g) CH4(g) ΔH° = –75 kJ exo/endo - thermic Δν(gas) = __, (decrease)Vtot --> which way will equation shift? what will happen if there is an increase in: P(CH4) shift? __ ...two ways to to lose mass(C)? ___________ higher T shift? __
O Microsoft OneDrive e teaching and lea x OWLV2 | Online teaching and lea X genow.com/ilrn/takeAssignment/takeCovalentActivity.do?locator=assignment-take Use the References to access important values if needed for this question. Consider the following system at equilibrium where AH° = -87.9 kJ/mol, and K. = 83.3 , at 500 K. PCI3 (g) + Cl2 (g) =PCI3 (g) When 0.30 moles of PCI5 (g) are added to the equilibrium system at constant temperature: The value of K. The value of Q. K. The reaction must Orun in the forward direction to restablish equilibrium. Orun in the reverse direction to restablish equilibrium. O remain the same. It is already at equilibrium. The concentration of Cl, will Submit Answer Retry Entire Group 9 more group attempts remaining Chp
On a sheet of paper, calculate AH for the following (fictional) reaction, using the given information (applying Hess Law). Upload a picture of your answer, showing all work. Solve for AH of: A2B (I) CD3 (g) + E2F (g) Given: Reaction AH (kJ) ½ G2B(g) + B2(g) - ½ A2B (I) -50.0 G2B(g) + 2B2(g) - CD 3(g) + E2F(1) -200.0 E2F(I) – E2F(g) E2F(g) 75.0 Attach File Browse My Computer
Please all solve the problem
3. Heats of reaction can also be calculated using heats of formation and the relationship - AHrxn = ΣnAHƒ.prod — ΣmAHƒ.reac Substance HCl(aq) MgCl2(aq) MgO(s) H₂O(/) H.(kJ/mol) -167.2 -796.8 -601.2 -285.8 0 a. Standard State Elements Calculate the theoretical AH° values for the following reaction Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l) Mg(s) + O2(g) → MgO(s)
17
Please don't provide handwritten solution .....
a) b)
3) Calculate the reaction heat of formation of PbO from Pb and O2 at rxn temperatures of 500K and 800K seperately (using below reactions). Pb(s) + 1/2 O2(g) Pb(1) + 1/2 O2(g)- AH°298.PbO =-219400 J/mole Cp (PbO.s)-52.3 +8.66 x 10T- 8.2x10 T J/molK (298-1159 K) CP (Pb.s)-24.3 +8.7 x 10 T Cp (Pb.)= 32.5-3.1 x 10 T Cp(02g)-29.9 +4.18 x 10 T-1.7x10 T2 Pb(s)→ PbO(s) PbO(s) J/molK (298-600 K) J/molK (600-1300 K) J/molK (298-2500 K) Pb(1). AHmelting 4800 J/mole, Tmelting 600K
00 Turkcell 3G 03:22 @ %27D 3. When 120 mg of naphthalene, CH, (s), was burned in a bomb calorimeter the temperature rose by 3.05 K a) Calculate the calorimeter constant. b) By how much will the temperature rise when 10 mg of phenol, C,HOH(s), is bumed in the calorimeter under the same conditions? (AH n.- -5156.3 kJ/mol and Mw CH 128.17 g/mol, AH° CESON = -3054 kJ/mol, Mw CO94.12 g/mol) 4 At 25°C, the density of a 50 % by mass ethanol/water solution is 0.914 g cm. Given that the partial molar volume of water in this solution is 17.4 cm'mol'. What is the partial molar volume of the ethanol? (d 0.789 g/cm': d-1.00 g cm": Mw for water-18 g mot', ethanol-46.07g mot 5. The addition of 5.00 g of a compound to 250 g of naphthalene lowered the freezing point of naphthalene by 0.780 K. Calculate the molar mass of the compound. (K, 694 KAgmol for naphthalene)