Answer:
<u>225.6 kJ</u>, <em>assuming the water is already at 100 °C</em>
Explanation:
The correct answer to this question will depend on the initial temperature of the water to which heat is added to produce steam. Energy is required to raise the water temperature to 100°C. At that point, an energy of vaporization is needed to convert liquid water at 100 °C to water vapor at 100°C. The heat of vaporization for water is 2256.4 kJ/kg. The energy required to bring 100g of water from a lower temperature to 100°C is calculated at 4.186 J/g°C. We don't know the starting temperature, so this step cannot be calculated.
<em><u>Assuming</u></em> that we are already at 100 °C, we can calculate the heat required for vaporization:
(100.0g)(1000.0g/1 kg)(2256.4 kJ/kg) = 225.6 kJ for 100 grams water.
The answer is 80 atomic mass
<span>The atomic weight of 13C should be pretty close to 13.0. (If you have the exact mass, use it in the problem.) So,
9.00 g / 13.0 g/mol = 0.692 moles
Therefore, the answer should be 0.692 moles are in 9.00 g of 13C.</span>
