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In a butane lighter, 9.2g g of butane combines with 32.9g g of oxygen to form 27.8g g carbon dioxide and how many grams of water? -
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In a butane lighter, 9.5 of butane combines with 34.0 of oxygen to form 28.7 carbon dioxide and how many grams of water? -
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C4H10(g) + 13/2 O2> 4CO2(g)+ 5H2O(g) Hrxn -2658Kj What mass of butane in gms is necessary to produce 1.5 times ten ^3Kj of heat? What mass of CO2 is produced? -
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Consider the following reaction: 2Fe2O3 --> 4Fe + 3O2 ∆Hrxn° = +824.2 kJ The decomposition of 29.0 g of Fe2O3 results in a. the release of 150 kJ of heat b. the release of 12000 kJ of heat c. the absorption of 12000 kJ of heat -
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In a butane lighter, 9.2 g of butane combines with 32.9 g of oxygen to form 27.8 g carbon dioxide and how many grams of water? -
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what volume of oxygen gas is needed to completely combust 0.202 L of butane (C4H10) gas? -
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What will be the change in enthalpy when 100.0 g of butane, C4H10, is burned in oxygen as shown in the thermochemical equation below? 2 C4H10(l) + 13 O2(g) → 8 CO2(g) + 10 H2O(g) ΔH = −5271 kJ −2636 kJ −4534 kJ −9087 kJ
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The enthalpy of combustion of butane C4H10 is described by the reaction: C4H10(g) + (13/2) O2(g) -> 4CO2(g) + 5H2O(g) ΔH°rxn = –2613 kJ/mol Given the following enthalpies of formation: ΔH°f[CO2(g)] = -393.5 kJ/mol -
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Butane, C4H10, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is: 2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l) At 1.00 atm and 23 ∘C, what is the -
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how many moles of water vapour are formed when 10 liters of butane gas C4H10 is burned in oxygen at stp? -
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. Butane C4H10 undergoes combustion in the following reaction: 2 C4H10 + 13 O2 8 CO2 + 10 H2O In a reaction, 40.00 g of butane are reacted with 150.00 g of oxygen gas. A) What mass of carbon dioxide is produced?
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