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Additional resources for Cracking the AP Chemistry Exam, 2017 Edition (College Test Preparation)
D The molecular weight of BaCl2 is 208 g/mol, with a purpose to determine what percentage moles of the hydrate we've. Moles = Moles of hydrate = mole The misplaced mass used to be because of water, so 1 mole of the hydrate should have contained nine grams of H2O. Moles = Moles of water = moles So if mole of hydrate includes mole of H2O, there needs to be 2 moles of H2O for each mole of hydrate, and the formulation for the hydrate needs to be BaCl2 • 2 H2O. 22. E Let’s say we now have a hundred grams of the compound. Moles = So moles of carbon = = 6. 7 moles and moles of hydrogen = = 20 moles based on our tough calculation, there are approximately thrice as many moles of hydrogen within the compound as there are moles of carbon, so the empirical formulation is CH3. The molar mass for the empirical formulation is 15 g/mol, so we have to double the moles of every aspect to get a compound with a molar mass of 30 g/mol. That makes the molecular formulation of the compound C2H6. 23. C Backsolving doesn’t paintings so good subsequently simply because there are diverse compounds that comprise oxygen at the correct part of the equation, which makes the method type of complicated. as a substitute, let’s simply try out plugging in values for the main advanced compound within the equation, CuFeS2. What if there’s 1 CuFeS2? That’s very unlikely simply because there are 2 Cu’s at the correct. What if there are 2 CuFeS2’s? Then definitely the right facet has 1 Cu2S to stability the Cu and a pair of FeO to stability the Fe. the fitting part also needs to have three SO2 to stability the S. Now there are eight O’s at the correct, so there needs to be four O2’s at the left, and the equation is balanced. 2 CuFeS2 + four O2 → 1 Cu2S + 2 FeO + three SO2 24. C Let’s say we've a hundred grams of the compound. Moles = So moles of carbon = = 6 moles and moles of hydrogen = = 25 moles in keeping with our tough calculation, there are approximately 4 occasions as many moles of hydrogen within the compound as there are moles of carbon, so the empirical formulation is CH4. 25. C Let’s say we now have a hundred grams of the compound. Moles = Moles of chlorine = = approximately 1 mole Moles of fluorine = = approximately three moles in accordance with our tough calculation, there are approximately thrice as many moles of fluorine within the compound as there are moles of chlorine, so the empirical formulation is ClF3. difficulties 1. (a) CaCO3 + 2 HCl → CaCl2 + H2O + CO2 (b) Use the perfect gasoline equation to discover the variety of moles of CO2 produced. be sure to convert to the correct devices. From the balanced equation, for each mole of CO2 produced, 1 mole of CaCO3 was once fed on. So zero. 035 moles of CaCO3 have been ate up. (c) we all know the variety of moles of CaCO3, so one can locate the mass. Grams = (moles)(MW) Grams of CaCO3 = (0. 035 mol)(100 g/mol) = three. 50 grams percentage via mass = (d) Mass of SiO2 = 10. zero g – three. five g = 6. five g Moles = Moles of SiO2 = = zero. eleven mol Molar ratio = = zero. 32 2. (a) all the hydrogen within the water and all the carbon within the carbon dioxide should have come from the hydrocarbon. Moles of H2O = = zero. a hundred and twenty moles each mole of water comprises 2 moles of hydrogen, so there are zero.