If 6.500 moles of water react completely, according to the following equation, what is the total change in enthalpy?
6CO2+6H2O⟶C6H12O6+6O2
ΔHrxn=2,803 kJ
If 6.500 moles of water react completely, according to the following equation, what is the total...
If 9.500 moles of water react completely, according to the following equation, what is the total change in enthalpy? 6 CO2 + 6H2O +602 6 ΔΗΓκι → C6H20 = 2,803 kJ = Provide your answer below: kJ
Glucose (C6H12O6) is an important energy-rich compound, produced by photosynthesis according to the equation below. What mass of glucose can be produced from 2.50 g of CO2 and the necessary water? 6CO2(g) + 6H2O(l) → C6H12O6(l) + 6O2(g)
What mass of water is produced when 100L of oxygen react completely in the following equation at STP? 2 C2H6(E) + 702(e) — 4CO3(e) + 6H2O()
The respiration of 1 mole of glucose is represented by the following equation C6H12O6 + 6O2 --> 6CO2 + 6H2O + 686 Kcal Where is the energy that is released or absorbed (depends on the reaction) found?
*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...
According to the following reaction, if 10 moles of oxygen (O2) were to react completely, how many moles of water would be produced? Enter answer as a whole number. 2H2+O2→2H2O
How many moles of O2 are required to burn completely 63.5 g of C6H6, according to the following equation? 2C6H6 + 15O2 ---> 12CO2 + 6H2O
When 4 moles of HCKg) react with O2(g) to form H2O(g) and Cl2(g) according to the following equation, 114 kJ of energy are evolved. 4HCK(g) + O2(g) 2H2O(g) + 2C12(e) Reactants Products Enthalpy Enthalpy Products Reactants (A) (B) Which of the enthalpy diagrams above represents this reaction? Is this reaction endothermic or exothermic?
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...
32. Based on the following balanced equation, how many moles of O2 will be needed to react completely with of 4 moles of CH4?