The question is in the image. Study the reactions for the formation of compounds from their elements.1. C (s) + O2 (g) CO2 (g)2. H2 (g) +02 (g) → H20 (1)3. 2C (s) + 3H2 (g) → C2H6 (g)AH = -394 kJAH = -242 kJΔΗ--84 kJThe combustion of C2H6 is shown by the following equation:C2H6 (g) + 02 (g)→ 2CO2 (g) + 3H20 (1)AH. = ?O -1,430 kJ285O 1,430 kJO-286 kJ

Given data 1. C(s) + O2(g) ---> CO2(g), DH = -394 KJ 2. H2(g) + 1/2O2(g) ---> H2O(l), DH =…

Answered: The question is in the image.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter4: Energy And Chemical Reactions

Section: Chapter Questions

Problem 97QRT

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Similar questions

A piece of unknown solid substance weighs 437.2 g, and requires 8460 J to increase its temperature from 19.3 °C to 68.9 °C. (a) What is the specific heat of the substance? (b) If it is one of the substances found in Table 5.1, what is its likely identity?
Combustion of table sugar produces CO2(g) and H2O( l). When 1.46 g table sugar is combusted in a constant-volume (bomb) calorimeter, 24.00 kJ of heat is liberated. a. Assuming that table sugar is pure sucrose, C12H22O11 (s), write the balanced equation for the combustion reaction. b. Calculate E in kJ/mol C12H22O11 for the combustion reaction of sucrose.
A 29.1-mL sample of 1.05 M KOH is mixed with 20.9 mL of 1.07 M HBr in a coffee-cup calorimeter (see Section 6.6 of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest whole-number coefficients, is 55.8 kJ. Both solutions are at 21.8C prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is 1.00 g/mL, and volumes are additive.
The combustion of 1.00 mol liquid methyl alcohol (CH3OH) in excess oxygen is exothermic, giving 727 kJ of heat. (a) Write the thermochemical equation for this reaction. (b) Calculate the enthalpy change that accompanies the burning 10.0 g methanol. (c) Compare this with the amount of heat produced by 10.0 g octane, C8H18, a component of gasoline (see Exercise 5.41).
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9.31 A metal radiator is made from 26.0 kg of iron. The specific heat of iron is 0.449Jg1C1 . How much heat must be supplied to the radiator to raise its temperature from 25.0 to 5 5.0°C?
A 21.3-mL sample of 0.977 M NaOH is mixed with 29.5 mL of 0.918 M HCl in a coffee-cup calorimeter (see Section 6.6 of your text for a description of a coffee-cup calorimeter). The enthalpy of the reaction, written with the lowest whole-number coefficients, is 55.8 kJ. Both solutions are at 19.6C prior to mixing and reacting. What is the final temperature of the reaction mixture? When solving this problem, assume that no heat is lost from the calorimeter to the surroundings, the density of all solutions is 1.00 g/mL, the specific heat of all solutions is the same as that of water, and volumes are additive.
A sample of natural gas is 80.0% CH4 and 20.0% C2H6 by mass. What is the heat from the combustion of 1.00 g of this mixture? Assume the products are CO2(g) and H2O(l).
Consider the following reaction in a vessel with a movable piston. R(g)+T(g)X(g)As the reaction takes place, the piston loses 1072 J of heat. The piston moves down and the surroundings do 549 J of work on the system. What is E?
The temperature of the cooling water as it leaves the hot engine of an automobile is 240 F. After it passes through the radiator it has a temperature of 175 F. Calculate the amount of heat transferred from the engine to the surroundings by one gallon of water with a specific heat of 4.184 J/g oC.
When a 0.740-g sample of trinitrotoluene (TNT), C7H5N2O6, is burned in a bomb calorimeter, the temperature increases from 23.4 C to 26.9 C. The heat capacity of the calorimeter is 534 J/C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample?
The second law of thermodynamics is sometimes paraphrased as: you can't break even. Explain. Because energy cannot be created out of nothing. Because some energy is lost in all energy transactions. Because some energy is gained in all energy transactons.

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