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 chemistry questions: b) Balance the following chemical equation i) H_2(g) + O_2(g) → H_2O_(l) Step 1: Balance oxygen atoms. There are 2 oxygen atoms on the left and 1 on the right. Place a coefficient of 2 in front of H_2O. H_2(g) + O_2(g) → 2H_2O_(l) Step 2: Balance hydrogen atoms. Now there are 2 hydrogen atoms on the left and 2 × 2 = 4 on the right. Place a coefficient of 2 in front of H_2. 2H_2(g) + O_2(g) → 2H_2O_(l) Step 3: Verify all atoms are balanced. Reactants: H = 4, O = 2 Products: H = 4, O = 2 The equation is balanced. 2H_2(g) + O_2(g) → 2H_2O_(l) ii) KClO_3(s) → KCl_(s) + O_2(g) Step 1: Balance oxygen atoms. There are 3 oxygen atoms on the left and 2 on the right. The least common multiple of 3 and 2 is 6. Place a coefficient of 2 in front of KClO_3 and 3 in front of O_2. 2KClO_3(s) → KCl_(s) + 3O_2(g) Step 2: Balance potassium and chlorine atoms. Now there are 2 potassium and 2 chlorine atoms on the left, but only 1 of each on the right. Place a coefficient of 2 in front of KCl. 2KClO_3(s) → 2KCl_(s) + 3O_2(g) Step 3: Verify all atoms are balanced. Reactants: K = 2, Cl = 2, O = 6 Products: K = 2, Cl = 2, O = 6 The equation is balanced. 2KClO_3(s) → 2KCl_(s) + 3O_2(g) iii) K_2CO_3(s) + HCl_(aq) → KCl_(aq) + CO_2(g) + H_2O_(l) Step 1: Balance potassium atoms. There are 2 potassium atoms on the left and 1 on the right. Place a coefficient of 2 in front of KCl. K_2CO_3(s) + HCl_(aq) → 2KCl_(aq) + CO_2(g) + H_2O_(l) Step 2: Balance chlorine atoms. Now there is 1 chlorine atom on the left and 2 on the right. Place a coefficient of 2 in front of HCl. K_2CO_3(s) + 2HCl_(aq) → 2KCl_(aq) + CO_2(g) + H_2O_(l) Step 3: Verify all atoms are balanced. Reactants: K = 2, C = 1, O = 3, H = 2, Cl = 2 Products: K = 2, C = 1, O = 2 (from CO_2) + 1 (from H_2O) = 3, H = 2, Cl = 2 All atoms are balanced. K_2CO_3(s) + 2HCl_(aq) → 2KCl_(aq) + CO_2(g) + H_2O_(l) 10 (a) Construct the Diagram to show the electronic structure in each of the following compounds I. Oxygen gas (O_2) Step 1: Determine the total number of valence electrons. Each oxygen atom has 6 valence electrons. For O_2, there are 2 × 6 = 12 valence electrons. Step 2: Draw a single bond between the two oxygen atoms. This uses 2 electrons. O-O Step 3: Distribute the remaining 12 - 2 = 10 electrons as lone pairs to complete octets. If a single bond is used, each oxygen would need 6 more electrons (3 lone pairs) to complete its octet, totaling 12 lone pair electrons. Since only 10 are available, a single bond is not sufficient. Step 4: Form a double bond between the oxygen atoms. This uses 4 electrons. O=O Step 5: Distribute the remaining 12 - 4 = 8 electrons as lone pairs. Each oxygen atom needs 4 more electrons (2 lone pairs) to complete its octet (4 (from bond) + 4 (lone pairs) = 8). :O=O: Step 6: Verify octets for both atoms. Each oxygen atom has 4 bonding electrons (from the double bond) and 4 non-bonding electrons (from two lone pairs), totaling 8 electrons. Both octets are satisfied. The electronic structure (Lewis structure) for oxygen gas (O_2) is: :O=O: