AH (25°C kJ/mol) AG (25°C kJ/mol) (25°C J/mol-K) S° Substance NANO, (s) NANO, (s) -467.9 -367.0 116.5 -358.7 -284.6 103.8 Cl,0 (g) 80.3 97.9 266.2 Cl,oS (g) -212.5 -198.3 309.8 PbO (s) -217.32 -187.89 68.7 Cl,Cro, (1) Cl,Cro, (g) -579.5 -510.8 221.8 -538.1 -501.6 329.8 Compute the AH°, AG°, and AS° for the reaction given below at two temperatures: 298 K and 315 K. PbCl2(s) + 3NaNO2(s) ---> PbO(s) + NaNO3(s) + 2NO(g) + 2NaCl(s) At 298 K: ΔΗ' is. AG° is AS° K = At 315 K: ΔΗ' is AG° is AS° K = kJ kJ J/K %D kJ kJ _J/K

AH (25°C kJ/mol) AG (25°C kJ/mol) (25°C J/mol-K) S° Substance NANO, (s) NANO, (s) -467.9 -367.0 116.5 -358.7 -284.6 103.8 Cl,0 (g) 80.3 97.9 266.2 Cl,oS (g) -212.5 -198.3 309.8 PbO (s) -217.32 -187.89 68.7 Cl,Cro, (1) Cl,Cro, (g) -579.5 -510.8 221.8 -538.1 -501.6 329.8 Compute the AH°, AG°, and AS° for the reaction given below at two temperatures: 298 K and 315 K. PbCl2(s) + 3NaNO2(s) ---> PbO(s) + NaNO3(s) + 2NO(g) + 2NaCl(s) At 298 K: ΔΗ' is. AG° is AS° K = At 315 K: ΔΗ' is AG° is AS° K = kJ kJ J/K %D kJ kJ _J/K

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

In one box of nails, there are 75 iron nails weighing 0.280 lblb . The density of iron is 7.86 g/cm3g/cm3. The specific heat of iron is 0.452 J/g∘CJ/g∘C. The melting point of iron is 1535 ∘C∘C. A)How much heat, in joules, must be added to the nails in the box to raise their temperature from 16 ∘C∘C to 123 ∘C∘C? B)How much heat, in joules, is required to heat one nail from 31 ∘C∘C to its melting point?
You are given the following data: H₂(g) 2H (g) Br₂(g) → 2 Br (g) H₂(g) + Br₂(g) 1 kJ mol → → 2 HBr (g) Calculate AH° for the reaction. H(g) + Br(g) HBr (g) Be sure your answer has the correct number of significant digits. x10 ΔΗ° = 436.4 X ΔΗ° = 192.5 ΔΗ° = - 72.4 kJ mol kJ mol kJ mol
Substance Specific Heat Capacity (J/g °C) water (l)* (liquid) 4.184 water (s) (ice) 2.03 water (g) (steam) 2.0 aluminum (s) 0.89 iron (s) 0.45 mercury (l) 0.14 carbon (s) 0.71 silver (s) 0.24 gold (s) 0.13 *The symbols (s), (l), and (g) indicate the solid, liquid, and gaseous states, respectively. 18.5 J of heat is applied to 4.5 g samples of each of the following substances: water (s), water (g), mercury (l), and gold (s). Which substance's temperature will increase the most? water (g)water (s)gold (s)mercury (l) 18.5 J of heat is applied to 4.5 g samples of each of the following substances: water (l), aluminum (s), iron (s), and carbon (s). Which substance's temperature will increase the least? iron (s)aluminum (s)water (l)carbon (s)
2 NO(g) N2(g) + O2(g) H = -180.7 kJ 2 NO(g) + O2(g) 2 NO2(g) H = -113.1 kJ 2 N2O(g) 2 N2(g) + O2(g) H = -163.2 kJ Use Hess's law to calculate H for the reaction below.3 NO(g) N2O(g) + NO2(g)H = _____k J
If 9.20 kJ of energy are required to change the temperature of a sample of water from 10.0 oC to 96.0 oC , what is the mass of water in g ? (cs = 4.184 J/(g oC)
Find the total energy involved in converting 450.0g of water from -21°C to 35°C. Hvan Water = 40.7 KJ/mol Hrus Water = 6.01 KJ/mol Solid C =2.03 J/g°C water Liquid Cwater =4.184 J/g°C Gas Cwater 2.01 J/g°C O 235 KJ O 150 KJ O 85 KJ 255 KJ
A chemist carefully measures the amount of heat needed to raise the temperature of a 399.0 g sample of a pure substance from 4.1 °C to 19.7°C. The experiment shows that 865. J of heat are needed. What can the chemist report for the specific heat capacity of the substance? Round your answer to 3 significant digits. -1 -1 J.g •K x10 Submit Assignme Continue 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center | Accessibili I IMG-6111.jpg IMG-6110.jpg IMG-6109.jpg IMG-6108.jpg IMG-6107.jpg MacBook Air 80 DII DD F2 F3 F4 F5 F6 F7 F8 F9 F10 F11 @ # $ % & 2 3 4 5 6 7 8 9
Find the final temperature Suppose we mix 90.0 grams of hot water (90.0°C) with 10.0 grams of cold water (10.0°C). Let x = the final temperature. C=4.184 J/g.°C a. Set up an expression for the energy released (q) by the hot water (Aqhot motCAThot) b. Set up an expression for the energy absorbed (q) by the cold water (Aqcold=mcold CATcold) c. Calculate Tf
2Fe2O3(s) 4Fe(s) + 302(g). Use the following data: So Fe(s) = 27.31 J/K, S° O2(g) = 205.15 J/K, S° Fe2O3(s) = 89.96 J/K. Express your answer in joules per kelvin to two decimal places. - ΑΣΦ AS°= = 468.4 © = ? J/K
15.0 J of heat is added to 10.00 g of gold, which is initially at 25°C.The heat capacity of gold is 25.41 J/(mol • °C). What would be the final temperature of the gold?
Draw the Lewis Structure of CH3OCH3 Molecular Geometry? Ideal Bond Angles? Are there Polar Bonds Present (Indicate on Lewis do structure Is there an Overall Dipole? Octet rule violator and how?
An 80 g sample of gas is heated from 25 C to 225 C. During this process, 346 J of work was done by the system and its internal energy is 8475 J. What is the specific heat capacity of the gas? (ans: 0.55 J/gC)