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What volume of O2 is required to react with excess CS2 to produce 4.0L of CO2 Assume all gases at STP. Not sure what I am doing wrong, but the answer I am getting is not one of the answers. 4L/X mole=22.4 L/1 mole ===>.1786 mols
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The combustion of carbon disulfide in the presence of excess oxygen yields carbon dioxide and sulfur dioxide: CS2(g)+3O2(g)--->CO2(g)+2SO2(g) The combustion of 17.2 g of CS2 in the presence of excess oxygen yields __________ g of
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At a certain temperature , 0.960 mol of SO3 is placed in a 3.50 L container. 2SO3 ----> 2SO2 + O2 At equilibrium , 0.190 mol of O2 is present. Calculate kc. i know that it starts off by 2SO3 -----> 2SO2 + O2 I 0.960 0 0 C -2x +2x
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Which reaction shows that the enthalpy of formation of CS2 is Hf = 89.4 kJ/mol? A. C(s) + 2S(s) CS2(l) + 89.4 kJ B. C(s) + 2S(s) + 89.4 kJ CS2(l) C. C(s) + 2S(s) + 89.4 kJ CS2(l) + 89.4 kJ D. C(s) + 2S(s) CS2(l) I think the answer
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Carbon disulfide (CS2) reacts with excess chlorine (Cl2) to produce carbon tetrachloride (CCl4) and disulfur dichloride (S2Cl2). If 67.1 g of CS2 yields 45.2 g of CCl4, what is the percent yield? (Hint, you must first write the
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Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is s2(g)+C(s)-->CS2(g) Kc=9.40 at 900k How many grams of CS2(g) can be prepared by heating 14.5 moles of S2(g) with excess carbon in a 8.15 L
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Carbon disulfide is prepared by heating sulfur and charcoal. The chemical equation is S2 +C =CS2 Kc= 9.40 at 900k How many grams of CS2(g) can be prepared by heating 14.3 moles of S2(g) with excess carbon in a 8.55 L reaction
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