Calculate the total heat for the following process for 101 g of acetic acidcooling from 135.0° C to 5.00°C using the given information:= 395 , SHiquid= 2.05g°CJAHvapJJΔΗ;us 195, ΔΗ, apSH steam1.34 Ig°CJSHsolid =J= 1.06MUST SHOWg°CALL THE STEPSCooling Curve135°CSTEAM118°c.VaporizationCondensationLIQUIDFreezing16.6°cMelting5.00°C

Given: Mass of acetic acid = 101 g. Initial temperature of acetic acid = 135.0 oC And the final…

195 , Calculate the total heat for the following process for 101 g of acetic acid cooling from 135.0° C to 5.00°C using the given information: AH fus = 395 , SHtiquid 1.06 C J ΔΗ;us195,ΔΗap 2.05- g°C» J J SH solid 1.34 g°C SH steam MUST SHOW g°C ALL THE STEPS Cooling Curve 135°C STEAM 118°c. Vaporization Condensation LIQUID 16.6c- Freezing Melting

195 , Calculate the total heat for the following process for 101 g of acetic acid cooling from 135.0° C to 5.00°C using the given information: AH fus = 395 , SHtiquid 1.06 C J ΔΗ;us195,ΔΗap 2.05- g°C» J J SH solid 1.34 g°C SH steam MUST SHOW g°C ALL THE STEPS Cooling Curve 135°C STEAM 118°c. Vaporization Condensation LIQUID 16.6c- Freezing Melting

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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lgs everything else). e system (the part you are studyml (reaut nsi ost systems we are interested in will have a constant pressure, mean ing that q (heat or thermal energy) would be equivalent to the change in enthalpy (AH) We defined environments in which heat goes into the system surroundings get hot as exothermic ΔΗ- and the surroundings get cold as EX rmic AH + while environments in which heat goes out of the system and the For each of the environments below: (1) identify your system and surroundings and (2) predict whether it would be endothermic or exothermic. 1. a. Wood burning b. Water freezing c. Sweat evaporating d. Chemical hand-warmer
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Use Hess’s law to calculate the enthalpy change for the reaction WO3(s) + 3H2(g) —> W(s) +3H2O(g) from the following data: 2W(s) + 3O2(g) —> 2 WO3(s) H=-1685.4 2H2(g) + O2(g) —> 2H2O(g) H=-477.84
Find the mass of water that vaporizes when 2.10 kg of mercury at 205 °C is added to 0.10 kg of water at 30°C. Assume the following: Cmercury (specific heat capacity of mercury) - 139 J/(kg.°C): Cwater (specific heat capacity of water) = 4186 J/(kg.°C); and L, (water) = 2.26x106 J/kg
Given the following reactions: N2 (g) + O2 (g) → 2NO (g) AH = +180.7 kJ 2N20 (g) → O2 (g) + 2N2 (g) AH = -163.2 kJ the enthalpy of reaction for 2N20 (g) → 2NOo (g) + N2 (g) is kJ. -343.9 343.9 -145.7 145.7 O 17.5
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How many grams of CaO must be added to 50.0 mL of water to make the temperature increase from 25.0 Â°C to 58.87 Â°C? The reaction is CaO(s) + H 2O(l) ↔ Ca(OH) 2(aq) -- ΔH = -83.7 kJAssume the heat capacity of the solution is the same as pure water (4.184 J/g*Â°C), the density of the solution is 1.00 g/mL and there is no loss of heat to the surroundings.
Given the standard enthalpy changes for the following two reactions: (1) N2(g) + 202(g)N204(g) AH° = 9.2 kJ (2) 2N20(g)2N2(g) + O2(g) AH° = -164.2 kJ what is the standard enthalpy change for the reaction: (3) 2N½0(g) + 302(g)- →2N2O4(g) AH° = ? kJ
A food product containing 82% moisture content is being frozen. Predict the specific heat of the product at -8 ° C when 82% of the air is frozen. The specific heat of the dry product is 2.5 kJ / (kg ° C). It is assumed that the specific heat of water at -10 ° C is the same as that of water at 0 ° C, and that the specific heat of ice follows the function Cp ice = 0.0062 T frozen + 2.0649. Cp of frozen product = answer J / kg ° C.
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Given the following enthalpy change values for the reactions below: A + B → 2C; AH = -485.1 kJ mol-¹ C + 2D + E +F; AH = 251.4 kJ mol-¹ ½E + G +H; AH = 471.1 kJ mol-¹ Calculate A,H for the following reaction (in kJ mol-¹) using Hess's Law: A + B + 4D + 4G → 4H + 2F Answer: