Numerical Response (1 mark) 1. The minimum amount of heat energy required to completely evaporate 150 g of water expressed in scientific notation , is btimes 10^ast J. If the heat of fusion of water is 6.01kJ/mol and water's heat of vaporization is 40.65kJ/mol, what is the value of b correct to three significant digits? (Record your answer in the box on the right).

Question

Numerical Response (1 mark) 1. The minimum amount of heat energy required to completely evaporate 150 g of water expressed in scientific notation , is btimes 10^ast J. If the heat of fusion of water is 6.01kJ/mol and water's heat of vaporization is 40.65kJ/mol, what is the value of b correct to three significant digits? (Record your answer in the box on the right).

Answer 1

Here's how to solve this problem:1. **Calculate the moles of water:** * The molar mass of water (H₂O) is approximately 18.015 g/mol. * Moles of water = mass / molar mass = 150 g / 18.015 g/mol ≈ 8.326 mol2. **Calculate the energy required for vaporization:** * Energy = moles × heat of vaporization * Energy = 8.326 mol × 40.65 kJ/mol = 338.3 kJ3. **Convert to joules and scientific notation:** * Energy = 338.3 kJ × 1000 J/kJ = 338300 J * In scientific notation, this is 3.383 × 10⁵ J4. **Identify the value of b:** * The question asks for the value of *b* in the expression *b* × 10^* J. * Therefore, *b* = 3.38 (rounded to three significant figures).**Final Answer:** 3.38

Question

Solution

Answer