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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Here are the solutions to the questions: 2. Design a model to demonstrate how to calculate the relative atomic mass of different elements. To calculate the relative atomic mass of an element, a model would involve the following steps: • Identify all naturally occurring isotopes of the element. • Determine the atomic mass (or mass number) of each isotope. • Find the natural abundance (percentage or fractional) of each isotope. • Calculate the weighted average of these isotopic masses. The formula for relative atomic mass is: Relative Atomic Mass = (Isotope Mass × Fractional Abundance) For example, if an element has two isotopes, Isotope A with mass M_A and abundance P_A, and Isotope B with mass M_B and abundance P_B, the relative atomic mass would be: Relative Atomic Mass = (M_A × P_A) + (M_B × P_B) This model demonstrates that the relative atomic mass is not a simple average but a weighted average based on how common each isotope is. 3. Copper has two isotopes, copper-63 and copper-65. The percentage abundance of copper-63 is 69% and that of copper-65 is 31%. Sketch the mass spectrum of copper. First, let's calculate the relative atomic mass of copper, as this is often implied when isotope abundances are given. Step 1: Convert percentage abundances to fractional abundances. Fractional abundance of copper-63 = 69\% = 0.69 Fractional abundance of copper-65 = 31\% = 0.31 Step 2: Calculate the relative atomic mass. Relative Atomic Mass = (63 × 0.69) + (65 × 0.31) Relative Atomic Mass = 43.47 + 20.15 Relative Atomic Mass = 63.62 Now, let's sketch the mass spectrum of copper. A mass spectrum shows the relative abundance of ions versus their mass-to-charge ratio (m/z). For single-charged ions, the m/z value corresponds to the isotopic mass. The mass spectrum would show two peaks: • A peak at m/z = 63 with a relative intensity proportional to 69%. • A peak at m/z = 65 with a relative intensity proportional to 31%. Here is a conceptual sketch of the mass spectrum: ` Relative Abundance (Intensity) ^ | * (69%) | | | | | | * (31%) | | | | | | |______|________|________> m/z 63 65 ` 4. In a sample containing 100 atoms of boron, 80 were boron-11 atoms while 20 were the boron-10 isotope. Determine the relative atomic mass of boron. Step 1: Determine the fractional abundance of each isotope. Total atoms = 100 Number of boron-10 atoms = 20 Number of boron-11 atoms = 80 Fractional abundance of boron-10 = (20)/(100) = 0.20 Fractional abundance of boron-11 = (80)/(100) = 0.80 Step 2: Calculate the relative atomic mass using the formula. Relative Atomic Mass = (Mass of Boron-10 × Fractional Abundance of Boron-10) + (Mass of Boron-11 × Fractional Abundance of Boron-11) Relative Atomic Mass = (10 × 0.20) + (11 × 0.80) Relative Atomic Mass = 2.00 + 8.80 Relative Atomic Mass = 10.80 The relative atomic mass of boron is 10.80.