This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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You're on a roll — Step 1: Calculate the concentration of HCl. 3) Step 1: Write the balanced chemical equation for the neutralization reaction. NaOH(aq) + HCl(aq) NaCl(aq) + H_2O(l) From the equation, the mole ratio of NaOH to HCl is 1:1. Step 2: Calculate the moles of NaOH used. Given: Volume of NaOH = 26 cm^3 = 0.026 dm^3 Concentration of NaOH = 0.1 M Moles of NaOH = Concentration × Volume Moles of NaOH = 0.1 mol/dm^3 × 0.026 dm^3 = 0.0026 mol Step 3: Determine the moles of HCl reacted. Since the mole ratio of NaOH to HCl is 1:1, Moles of HCl = Moles of NaOH = 0.0026 mol Step 4: Calculate the concentration of HCl. Given: Volume of HCl = 12.4 cm^3 = 0.0124 dm^3 Concentration of HCl = Moles of HClVolume of HCl Concentration of HCl = 0.0026 mol0.0124 dm^3 ≈ 0.209677 mol/dm^3 Rounding to three significant figures: Concentration of HCl ≈ 0.210 mol/dm^3 The concentration of HCl is 0.210 mol/dm^3. 1) Step 1: Write the balanced chemical equation for the combustion of carbon(II) oxide. 2CO(g) + O_2(g) 2CO_2(g) Step 2: Determine the limiting reactant using the given volumes and the mole ratio from the balanced equation. According to the equation, 2 volumes of CO react with 1 volume of O_2. Given: Volume of CO = 36 cm^3 Volume of O_2 = 72 cm^3 For 36 cm^3 of CO, the required volume of O_2 is: 36 cm^3 CO × 1 volume O_22 volumes CO = 18 cm^3 O_2 Since 18 cm^3 of O_2 is required and 72 cm^3 of O_2 is available, O_2 is in excess. CO is the limiting reactant. Step 3: Calculate the volume of CO_2 produced. From the equation, 2 volumes of CO produce 2 volumes of CO_2. Volume of CO_2 = 36 cm^3 CO × 2 volumes CO_22 volumes CO = 36 cm^3 CO_2 Step 4: Calculate the volume of excess O_2 remaining. Excess O_2 = Initial O_2 - O_2 reacted Excess O_2 = 72 cm^3 - 18 cm^3 = 54 cm^3 Step 5: Determine the total volume of the resulting gaseous mixture. The gaseous mixture consists of the product (CO_2) and the unreacted excess reactant (O_2). Total volume = Volume of CO_2 + Volume of excess O_2 Total volume = 36 cm^3 + 54 cm^3 = 90 cm^3 The total volume of the resulting gaseous mixture is 90 cm^3. 4) To prepare crystals of copper(II) chloride from copper metal: 1. Heat the copper metal in air to form copper(II) oxide. This is because copper does not react with dilute hydrochloric acid. 2Cu(s) + O_2(g) heat 2CuO(s) 2. Add the black copper(II) oxide to a beaker containing dilute hydrochloric acid. Stir the mixture gently. The copper(II) oxide will react with the acid to form a blue solution of copper(II) chloride. CuO(s) + 2HCl(aq) CuCl_2(aq) + H_2O(l) 3. Heat the solution gently to evaporate some of the water, concentrating the solution until it is saturated. This can be tested by dipping a clean glass rod into the solution and observing if crystals form on cooling. 4. Allow the concentrated solution to cool slowly at room temperature. Copper(II) chloride crystals will form. 5. Filter the crystals from the remaining solution (mother liquor) and wash them with a small amount of cold distilled water to remove impurities. 6. Dry the crystals between filter papers or in a desiccator. 5) An isotope refers to atoms of the same element that have the same number of protons (atomic number) but different numbers of neutrons, resulting in different mass numbers. They have identical chemical properties but differ in their physical properties. Got more? Send 'em.