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![ww What we have done (6x A) + (6 x B).-(C)-(6xD) - [6-3935] [6* - 241.5 ] - [-2102-0) - [6 x 90.7 =f2361 1912) kJ - 1450.8 x](http://img.homeworklib.com/questions/0aac1790-e887-11ea-8c3e-ad0f5ef312e7.png?x-oss-process=image/resize,w_560)
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Imported From Sa... 4. Use Hess' Law to calculate the enthalpy change (AHY) for this reaction:...
I especially need help with understanding what the second question is asking. 4. Use Hess' Law...
I especially need help with understanding what the second
question is asking.
4. Use Hess' Law to calculate the enthalpy change (AHⓇt) for this reaction: (12 pts.) 6C(s) + 6H2(g) + 3029) → C6H12O6(S) + using the following equations: A. C(s) + O2(g) → CO2(g) B. H2(g) + 1/2O2(g) → H2O(g) C. C6H12O6(s) + 6020) — 6CO2(g) + 6H2O() D. H2O0) → H2O(g) AH'= -393.5 kJ AH'= -241.8kJ AH'= -2803.0 kJ AH°= +40.7 kJ Why is this change in enthalpy...
4. Use Hess' Law to calculate the enthalpy change (AHºr) for this reaction: (12 pts.) 6C(s)...
4. Use Hess' Law to calculate the enthalpy change (AHºr) for this reaction: (12 pts.) 6C(s) + 6H2(g) + 302(g) → C6H12O6(s) using the following equations: A. C(s) + O2(g) → CO2(g) B. H2(g) + 1/2O2(g) → H2O(g) C. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O)- D. H20m) → H2O(g) AH°= -393.5 kJ AH°= -241.8kJ AH°= -2803.0 kJ AH°= +40.7 kJ C 6111260) At = AHT (66 tn 4) AH3= DH, + A M₂ A4 = A Hz - H₂...
*ATTACHED THE PICTURE BELOW TO HAVE A BETTER UNDERTSTANDING* *Also please showwork on how tou solve...
*ATTACHED THE PICTURE BELOW TO HAVE A BETTER
UNDERTSTANDING*
*Also please showwork on how tou solve the problem to
umderstand it* TY
4. Use Hess' Law to calculate the enthalpy change (∆H°f) for
this reaction:
6C(s) + 6H2(g) + 3O2(g)C6H12O6(s)
using the following equations:
A. C(s) + O2(g)CO2(g)
B. H2(g) + 1/2O2(g)H2O(g)
C. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)
D. H2O(l)H2O(g)
∆H°= -393.5 kJ
∆H°= -241.8kJ
∆H°= -2803.0 kJ
∆H°= +40.7 kJ
Why is this change in enthalpy given...
3. A compound is analyzed and found to contain 65.4% C, 5.5% H, and 29.1%O by...
3. A compound is analyzed and found to contain 65.4% C, 5.5% H, and 29.1%O by mass. In a separate experiment, the compound is found to effuse through an opening at a rate of 3.0 x 10 mol/min. A sample of neon gas effused through the same opening at a rate of 7.0 x 10-5 mol/min. A. What is the empirical formula of the compound? (9 pts.) B. What is the molecular formula of the compound? (6 pts.) 4. Use...
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below:...
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below: Table 1. Change in Enthalpy for Reactions Reactions Change in Enthalpy (AH) AH = -393.5 kJ/mol (1) C() (2) Hz((g) + + O2(g) O2(g) → CO2(8) → H2O(l) AH2 = -285.8 kJ/mol (3) 2CH.(g) + 702(g) → 4 CO2(g)+ 6H2O(1) AH = -283.0 kJ/mol Calculate the enthalpy change for the reaction: 2 C(s) + 3H2(g) → CzH6(g) AH = kJ/mol
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below:...
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below: Table 1. Change in Enthalpy for Reactions Reactions Change in Enthalpy (AH) AH = -393.5 kJ/mol (1) C() (2) Hz((g) + + O2(g) O2(g) → CO2(8) → H2O(l) AH2 = -285.8 kJ/mol (3) 2CH.(g) + 702(g) → 4 CO2(g)+ 6H2O(1) AH = -283.0 kJ/mol Calculate the enthalpy change for the reaction: 2 C(s) + 3H2(g) → CzH6(g) AH = kJ/mol
Calculate the change in enthalpy (ΔH) for the heat of formation of ethane, C2H6, using Hess' Law and the following reactions:
Calculate the change in enthalpy (ΔH) for the heat of formation of ethane, C2H6, using Hess' Law and the following reactions:a) 2C(S) + 2O2(g) → 2CO2(g), ΔH = -188 kcalb) C2H6(g) + (7/2)O2(g) → 2CO2(g) + 3H2O(l), ΔH = -373 kcalc) H2(g) + (1/2)O2(g) → H2O(l), ΔH = -68.3 kcal
20. What is the standard enthalpy change for the following reaction? 3CH.(g) + 40.(g) + 3C02(g)...
20. What is the standard enthalpy change for the following reaction? 3CH.(g) + 40.(g) + 3C02(g) + 6H2O(g) AH (kJ/mol) CHig) - 74.87 O3(g) +142.7 CO2(g) --393.5 H2O(g) -241.8 a.-2285.1 kJ b.-2977.5 kJ c. +2977.5 kJ d. +2285.1 kJ e. -3426.5 kJ
20. What is the standard enthalpy change for the following reaction? 3CH.(g) + 40,(g) - 3C02(g) + 6H2O(g) AH® (kJ/m...
20. What is the standard enthalpy change for the following reaction? 3CH.(g) + 40,(g) - 3C02(g) + 6H2O(g) AH® (kJ/mol) CHig) - 74.87 03(g) +142.7 CO2(g) --393.5 H2O(g) -241.8 a. -2285.1 kJ b.-2977.5 kJ c. +2977.5 kJ d. +2285.1 kJ e. -3426.5 kJ
Use Hess’s Law to find the standard enthalpy change for the reaction CO2(g) → C(s) +...
Use Hess’s Law to find the standard enthalpy change for the reaction CO2(g) → C(s) + O2(g) using only the following information. Show all your work, including any equations you use to obtain your answer and showing clearly how you obtained that answer. (3 pts.) H2O(l) → H2(g) + 1/2O2(g) C2H6(g) → 2C(s)+ 3H2(g) 2CO2(g) + 3H2O(l) → C 2H6(g) + 7/2O2(g) ∆Ho (kJ) 643 kJ 190.6kJ 3511.1 kJ
I especially need help with understanding what the second
question is asking.
4. Use Hess' Law to calculate the enthalpy change (AHⓇt) for this reaction: (12 pts.) 6C(s) + 6H2(g) + 3029) → C6H12O6(S) + using the following equations: A. C(s) + O2(g) → CO2(g) B. H2(g) + 1/2O2(g) → H2O(g) C. C6H12O6(s) + 6020) — 6CO2(g) + 6H2O() D. H2O0) → H2O(g) AH'= -393.5 kJ AH'= -241.8kJ AH'= -2803.0 kJ AH°= +40.7 kJ Why is this change in enthalpy...
4. Use Hess' Law to calculate the enthalpy change (AHºr) for this reaction: (12 pts.) 6C(s) + 6H2(g) + 302(g) → C6H12O6(s) using the following equations: A. C(s) + O2(g) → CO2(g) B. H2(g) + 1/2O2(g) → H2O(g) C. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O)- D. H20m) → H2O(g) AH°= -393.5 kJ AH°= -241.8kJ AH°= -2803.0 kJ AH°= +40.7 kJ C 6111260) At = AHT (66 tn 4) AH3= DH, + A M₂ A4 = A Hz - H₂...
*ATTACHED THE PICTURE BELOW TO HAVE A BETTER
UNDERTSTANDING*
*Also please showwork on how tou solve the problem to
umderstand it* TY
4. Use Hess' Law to calculate the enthalpy change (∆H°f) for
this reaction:
6C(s) + 6H2(g) + 3O2(g)C6H12O6(s)
using the following equations:
A. C(s) + O2(g)CO2(g)
B. H2(g) + 1/2O2(g)H2O(g)
C. C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(l)
D. H2O(l)H2O(g)
∆H°= -393.5 kJ
∆H°= -241.8kJ
∆H°= -2803.0 kJ
∆H°= +40.7 kJ
Why is this change in enthalpy given...
3. A compound is analyzed and found to contain 65.4% C, 5.5% H, and 29.1%O by mass. In a separate experiment, the compound is found to effuse through an opening at a rate of 3.0 x 10 mol/min. A sample of neon gas effused through the same opening at a rate of 7.0 x 10-5 mol/min. A. What is the empirical formula of the compound? (9 pts.) B. What is the molecular formula of the compound? (6 pts.) 4. Use...
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below: Table 1. Change in Enthalpy for Reactions Reactions Change in Enthalpy (AH) AH = -393.5 kJ/mol (1) C() (2) Hz((g) + + O2(g) O2(g) → CO2(8) → H2O(l) AH2 = -285.8 kJ/mol (3) 2CH.(g) + 702(g) → 4 CO2(g)+ 6H2O(1) AH = -283.0 kJ/mol Calculate the enthalpy change for the reaction: 2 C(s) + 3H2(g) → CzH6(g) AH = kJ/mol
6. Hess's Law. Use the data in Table 1 to calculate AH for the reaction below: Table 1. Change in Enthalpy for Reactions Reactions Change in Enthalpy (AH) AH = -393.5 kJ/mol (1) C() (2) Hz((g) + + O2(g) O2(g) → CO2(8) → H2O(l) AH2 = -285.8 kJ/mol (3) 2CH.(g) + 702(g) → 4 CO2(g)+ 6H2O(1) AH = -283.0 kJ/mol Calculate the enthalpy change for the reaction: 2 C(s) + 3H2(g) → CzH6(g) AH = kJ/mol
20. What is the standard enthalpy change for the following reaction? 3CH.(g) + 40.(g) + 3C02(g) + 6H2O(g) AH (kJ/mol) CHig) - 74.87 O3(g) +142.7 CO2(g) --393.5 H2O(g) -241.8 a.-2285.1 kJ b.-2977.5 kJ c. +2977.5 kJ d. +2285.1 kJ e. -3426.5 kJ
20. What is the standard enthalpy change for the following reaction? 3CH.(g) + 40,(g) - 3C02(g) + 6H2O(g) AH® (kJ/mol) CHig) - 74.87 03(g) +142.7 CO2(g) --393.5 H2O(g) -241.8 a. -2285.1 kJ b.-2977.5 kJ c. +2977.5 kJ d. +2285.1 kJ e. -3426.5 kJ
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