4.838 g H2O will be formed from the combustion of 8.064 g C6H12O6 . Explanation: C6H12O6 + 6O2 → 6CO2 + 6H2O. Molar Masses of Glucose and ...
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Between 0.5 and 1.0 g of water. Explanation: C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(g). Moles of glucose, = 8.064⋅g 180.16⋅g ⋅mol−1 ...
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Click here to get an answer to your question ✍️ Heat of reaction for C6H12O6(s) + 6O2 (g) → 6CO2(g) + 6H2O(v) at constant pressure is - 651 kcal at ...
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With 37.5g of C6H12O6, b. The first step of glucose metabolism is the phosphorylation of glucose. In this reaction, glucose is reacted with ATP to ...
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6CO2(g) + 6H2O(l) -------> C6H12O6(s) + 6O2(g). –(– 2808 kJ). 6 x Rxn B. 6C(s) + 6O2(g) -----> 6CO2(g). 6(– 394 kJ). Reverse Rxn C x 3. 3O2(g) + 6H2(g) ...
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Part K. C6H12O6(s)+6O2(g)→6CO2(g)+6H2O(g). Express your answers as signed integers. Enter the initial and the final oxidation states separated by a comma.
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The heat of combustion of glucose: C6H12O6(s) + 6O2(g) = 6CO2(g) + 6H2O(g) is deltaH= -2970.4 kJ. a.) Is this an endothermic or exothermic process?
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C6H12O6(s) + 6O2(g) --> 6H2O(g) + 6CO2(g) How many liters of CO2 are produced when you start with 13.2 grams of glucose (C6H12O6) in excess oxygen at STP? 0.439 ...
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Answer to: Consider the combustion of glucose: C6H12O6(s) + 6O2(g) arrow 6CO2(g) + 6H2O(g); Delta H = -2800 kJ/mol. Which condition would result in the...
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C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ΔH = kJ/mol For every mole of glucose you burn, you are burning _____ calories! This process releases energy, ...
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How many grams of H2O will be produced when 8.064g of glucose is burned? The reaction can be written asC6H12O6(s)+6O2(g)→6CO2(g)+6H2O(g. Moles ...
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The chemical formula for glucose is C₆H₁₂O₆, not C₆H₁₂O. Balanced equation C₆H₁₂O₆ + 6O₂(g) → 6CO₂(g) + 6H₂O(g) Use stoichiometry to calculate the moles of ...
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... of glucose given the following reactions: C6H12O6(s) → 2C2H5OH(l) + 2CO2(g) ΔH = ? C6H12O6(s) + 6O2 → 6CO2(g) + 6H2O (l) ΔH = -2820kJ.
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chemistry. Consider the combustion of glucose: C6H12O6(s) + 6O2(g) 6CO2(g) + 6H2O(g) H = −3 kJ/mol. Which condition. I) Adding more carbon dioxide;
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3. Which one of the following will change the value of an equilibrium constant? a) changing temperature ... 6CO2 (g) + 6H2O (l) C6H12O6 (s) + 6O2 (g).
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