When a 4.18 g sample of solid sodium hydroxide was dissolved in a calorimeter in 120.0 g of water, the temperature rose from 6.5 oC to 47.4 oC . Calculate ΔH (in kJ/mol NaOH ) for the following solution process: NaOH(s)→Na+(aq)+OH−(aq) ____kj/mol

3 If we measure the heat change (q) of everything in a system except the calorimeter, we can use the first law of thermodynamics (∆Euniverse = 0) to calculate the heat change of the calorimeter (qcal).  In this reaction, we combine 5.75 g of 67.4 ºC water with 6.67 g of 19 ºC water in a calorimeter.  Since some of the heat is absorbed by the calorimeter, the final temperature of the water is 25.4ºC.  Since the hot and cold water are combined, the final temp of the water is the final temp for the hot water and the cold water.  Calculate the heat change of the calorimeter in J. Remember that all of the heat changes must add up to zero so you can calculate the heat change of the hot water (qhotwater) and the heat change of the cold water (qcoldwater) using for each where Cs = 4.186 J/gºC.  The first law of thermodynamics tells us that all of these changes sum up to zero so   Enter your answer to one decimal place (tenths)

The ΔH°rxn for the following reaction: NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) is –55.7 kJ/mol. Given the following enthalpies of formation, determine ΔH°f of H2O(l). ΔH°f NaOH(aq) = −470.1 kJ/mol ΔH°f HCl(aq) =  −167.2 kJ/mol ΔH°f NaCl(aq) =  −407.2 kJ/mol

13. Energy is stored in the body in adenosine triphosphate, ATP, which is formed by reaction between adenosine diphosphate, ADP, and dihydrogen phosphate ions. ADP3-(aq) + H2PO42-(aq) -----> ATP4-(aq) + H2O (l) ΔH° = 38 kJ Is the reaction endothermic or exothermic?

Suppose you burned 0.300 g of C(s) in an excess of O₂ (g) in a constant-volume calorimeter to give CO₂ (g). C(s) + O₂(g) → CO₂ (g) The temperature of the calorimeter, which contained 776 g of water, increased from 24.90 °C to 27.28 °C. The heat capacity of the bomb is 896 J/K. Calculate AU per mole of carbon. (The specific heat capacity of liquid water is 4.184 J/g. K.) AU = = kJ/mol C

Calculate the ΔH°rxn of the reaction in a bomb calorimeter: NaOH(s) → NaOH(aq) Given the following: Volume of solution = 100mL (when solid added to water) Concentration = 1M Mass of NaOH = 1.0359g Change in temperature = 2.5°C Specific heat capacity of solution = 4.184 J/°C Specific heat capacity of bomb calorimeter = 30 J/°C Molecular mass of NaOH = 40g/mol

Calculate ASº for the combustion of propane. C3H8(g) + 502(g) → 3CO2(g) + 4H20(g) Substance C3H8(g) O2{g) CO2(g) H2O(g) S° / (J/K•mol) 269.9 205.138 213.74 188.825 -72.5 J/K 100.9 J/K O72.5 J/K O -100.9 J/K O 877 J/K

Constants Periodic Table Zinc metal reacts with hydrochloric acid according to the following balanced equation. Zn(s) + 2HCI(aq)→ZnCl2 (aq) + H2 (g) When 0.112 g of Zn(s) is combined with enough HCl to make 51.5 mL of solution in a coffee-cup calorimeter, all of the zinc reacts, raising the temperature of the solution from 22,1°C to 24.1 C Part A Find A H,xn for this reaction as written. (Use 1.0 g/mL for the density of the solution and 4. 18 J/g. C as the specific heat capacity.) 國 AHxn = kJ/mol Submit Request Answer Provide Feedback Next> P Pearson Copyright © 2021 Pearson Education Inc. All rights reserved. | Terms of Use Privacy Pollcy Permissions | Contact Us Vi 1:23

Among the equations below, indicate all those that are exothermic: i. HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)                    ∆H = -53 kJ ii. Fe3O4(s) + CO(g) + 38 kJ → 3FeO(s) + CO2(g) iii. CaO(s) + H2O(l) → Ca(OH)2(aq) + 65,2 kJ iv. H2O2(l) → H2(g) + O2(g)                                                 ∆H = 188 kJ v. NaOH(s) + HCl(l) → NaCl(s) + H2O(l)                              ∆H = -178 kJ

A chemist measures the enthalpy change AH during the following reaction: 2 Al(s) + Fe₂O3(s)→Al₂O3(s) + 2 Fe(s) Use this information to complete the table below. Round each of your answers to the nearest kJ/mol. reaction 3A1₂03 (s) + 6Fe(s) Al₂O3(s) + 2Fe(s) AH-852. kJ 6Al(s) + 3Fe₂O3(s) 2Al(s) + Fe₂O₂ (s) Al(s) + Fe₂O₂ (s) → — Al₂O₂ (s) + Fe(s) = ΔΗ kJ kJ ☐ kJ ☐ x10 X