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 — here are the solutions for Exercise 6.8: 1. The altitude of a regular square pyramid is 6 cm. If one edge of the base has length 4 cm then find its volume. Step 1: Identify the given values and the formula for the volume of a pyramid. Given: altitude (h) = 6 cm, base edge (s) = 4 cm. The volume of a pyramid is V = (1)/(3) × Base Area × h. Step 2: Calculate the area of the square base. Base Area (A_b) = s^2 = (4 cm)^2 = 16 cm^2. Step 3: Calculate the volume of the pyramid. V = (1)/(3) × 16 cm^2 × 6 cm V = (1)/(3) × 96 cm^3 V = 32 cm^3 The volume of the pyramid is 32 cm^3. 2. A circular cone has an altitude 12 cm and a base radius 10 cm. What is its volume? Step 1: Identify the given values and the formula for the volume of a cone. Given: altitude (h) = 12 cm, base radius (r) = 10 cm. The volume of a cone is V = (1)/(3) r^2 h. Step 2: Substitute the values into the formula and calculate the volume. V = (1)/(3) (10 cm)^2 (12 cm) V = (1)/(3) (100 cm^2) (12 cm) V = (1200)/(3) cm^3 V = 400 cm^3 Using ≈ 3.14159: V ≈ 400 × 3.14159 cm^3 V ≈ 1256.64 cm^3 The volume of the cone is 400 cm^3 or approximately 1256.64 cm^3. 3. A right circular cone has height 10 cm and circumference of the base is 12 cm. Find its volume. Step 1: Identify the given values and use the circumference to find the radius. Given: height (h) = 10 cm, circumference (C) = 12 cm. The formula for circumference is C = 2 r. 12 cm = 2 r Divide both sides by 2: r = 12 cm2 r = 6 cm Step 2: Calculate the volume of the cone using the found radius and given height. The volume of a cone is V = (1)/(3) r^2 h. V = (1)/(3) (6 cm)^2 (10 cm) V = (1)/(3) (36 cm^2) (10 cm) V = (360)/(3) cm^3 V = 120 cm^3 Using ≈ 3.14159: V ≈ 120 × 3.14159 cm^3 V ≈ 376.99 cm^3 The volume of the cone is 120 cm^3 or approximately 376.99 cm^3. 4. The lateral edge of a regular tetrahedron is 6 cm. Find its total surface area and its volume. Step 1: Identify the given value and formulas for a regular tetrahedron. Given: lateral edge (a) = 6 cm. A regular tetrahedron has 4 equilateral triangular faces. Area of one equilateral triangle with side a: A_face = sqrt(3)4a^2. Total Surface Area (TSA) = 4 × A_face = sqrt(3)a^2. Volume (V) of a regular tetrahedron with edge a: V = sqrt(2)12a^3. Step 2: Calculate the total surface area. TSA = sqrt(3) (6 cm)^2 TSA = sqrt(3) (36 cm^2) TSA = 36sqrt(3) cm^2 Using sqrt(3) ≈ 1.732: TSA ≈ 36 × 1.732 cm^2 TSA ≈ 62.352 cm^2 Step 3: Calculate the volume. V = sqrt(2)12 (6 cm)^3 V = sqrt(2)12 (216 cm^3) V = 18sqrt(2) cm^3 Using sqrt(2) ≈ 1.414: V ≈ 18 × 1.414 cm^3 V ≈ 25.452 cm^3 The total surface area is 36sqrt(3) cm^2 or approximately 62.35 cm^2 and the volume is 18sqrt(2) cm^3 or approximately 25.45 cm^3. 5. A right circular conical vessel of altitude 20 cm and base radius 10 cm is kept with its vertex downwards. If one liter of water is poured into it, how high above the vertex will the level of the water be? Use = 3.14. Step 1: Identify the given dimensions of the conical vessel and the volume of water. Vessel altitude (H) = 20 cm, vessel base radius (R) = 10 cm. Volume of water (V_w) = 1 liter = 1000 cm^3. Use = 3.14. Step 2: Establish the relationship between the radius (r_w) and height (h_w) of the water using similar triangles. Since the vessel is vertex downwards, the water forms a smaller cone similar to the vessel. The ratio of radius to height is constant: (r_w)/(h_w) = (R)/(H). (r_w)/(h_w) = 10 cm20 cm r_w = (1)/(2) h_w Step 3: Write the volume of water in terms of its height (h_w). The volume of the water cone is V_w = (1)/(3) r_w^2 h_w. Substitute r_w = (1)/(2) h_w: V_w = (1)/(3) ((1)/(2) h_w)^2 h_w V_w = (1)/(3) ((1)/(4) h_w^2) h_w V_w = (1)/(12) h_w^3 Step 4: Solve for h_w using the given volume of water. 1000 cm^3 = (1)/(12) (3.14) h_w^3 Multiply both sides by 12: 12000 cm^3 = 3.14 h_w^3 Divide by 3.14: h_w^3 = (12000)/(3.14) cm^3 h_w^3 ≈ 3821.656 cm^3 Take the cube root: h_w = [3]3821.656 cm^3 h_w ≈ 15.63 cm The level of the water will be approximately 15.63 cm. What's next?