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Answer (2)
Answer:Loss of viable land, erosion, and other factors decrease the ability of farmers to use land. Other factors include inflation and government restrictions.
Answer: 1. How many moles of KClO₃ are needed to make 3.50 moles of KCl? - Assumption: The reaction is the decomposition of potassium chlorate: 2KClO₃ → 2KCl + 3O₂ - Solution: From the balanced equation, 2 moles of KClO₃ produce 2 moles of KCl. Therefore, the mole ratio is 1:1. To produce 3.50 moles of KCl, you need 3.50 moles of KClO₃. - Answer: 3.50 moles KClO₃ 2. How many moles of O₂ are produced when 1.26 moles of H₂O is reacted? - Assumption: The reaction is the electrolysis of water: 2H₂O → 2H₂ + O₂ - Solution: From the balanced equation, 2 moles of H₂O produce 1 mole of O₂. The mole ratio is 2:1. Therefore, 1.26 moles of H₂O will produce (1.26 moles H₂O) * (1 mole O₂ / 2 moles H₂O) = 0.63 moles of O₂. - Answer: 0.63 moles O₂ 3. How many grams of Fe₂O₃ are produced when 42.7 grams of Fe is reacted? - Assumption: The reaction is the rusting of iron: 4Fe + 3O₂ → 2Fe₂O₃ - Solution: - Find moles of Fe: Molar mass of Fe ≈ 55.85 g/mol. Moles of Fe = 42.7 g / 55.85 g/mol ≈ 0.764 moles - Mole ratio: 4 moles Fe produce 2 moles Fe₂O₃ (ratio is 2:1) - Moles of Fe₂O₃ produced: 0.764 moles Fe * (2 moles Fe₂O₃ / 4 moles Fe) ≈ 0.382 moles Fe₂O₃ - Grams of Fe₂O₃: Molar mass of Fe₂O₃ ≈ 159.69 g/mol. Grams of Fe₂O₃ = 0.382 moles * 159.69 g/mol ≈ 61.0 g - Answer: 61.0 grams Fe₂O₃ 4. What volume of a .35M AgNO₃ is required to completely react with 55 mL of a 0.24M NaCl solution? - Assumption: The reaction is a precipitation reaction: AgNO₃ + NaCl → AgCl + NaNO₃ - Solution: - Moles of NaCl: Moles = Molarity * Volume (in Liters) = 0.24 mol/L * 0.055 L ≈ 0.0132 moles - Mole ratio: 1 mole AgNO₃ reacts with 1 mole NaCl (1:1 ratio) - Moles of AgNO₃ needed: 0.0132 moles - Volume of AgNO₃: Volume (in Liters) = Moles / Molarity = 0.0132 moles / 0.35 mol/L ≈ 0.0377 L ≈ 37.7 mL - Answer: 37.7 mL AgNO₃
