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1. Given that 50 grams of ice is heated at -20.0 °C to steam at 135.0 °C. i. Show the graph of the changes from ice to steam ii. Calculate the energy needed to change the ice to steam Please use these values: Heat of fusion =
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The molar heat capacity of C6H6(ℓ) is 136 J/mol ·◦C and of C6H6(g) is 81.6 J/mol · ◦C. The molar heat of fusion for benzene is 9.92 kJ/mol and its molar heat of vaporization is 30.8 kJ/mol. The melting point of benzene is
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Conclusion question(s) from a lab we did to find the heat of fusion of ice: Does the value obtained for the molar heat of fusion depend on the volume of water used? Does it depend on the mass of ice melted? Does it depend on the
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The molar heat capacity of C6H6(ℓ) is 136J/mol · ◦C and of C6H6(g) is 81.6 J/mol · ◦C. The molar heat of fusion for benzene is 9.92 kJ/mol and its molar heat of vaporization is 30.8 kJ/mol. The melting point of benzene is
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Using the values for the heat of fusion, specific heat of water, or heat of vaporization, calculate the amount of heat energy in each of the following: * calories removed to condense 125g of steam at 100 degree C to cool the
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Use the heating curve of 50 g of water and the list of values to answer the question. specific heat of ice = 2.10 J/(g·°C) specific heat of water = 4.18 J/(g·°C) specific heat of water vapor = 2.07 J/(g·°C) latent heat of
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How many kJ of heat are required to convert 1.00 g of ice at -25°C to steam at 125°C? s(ice) = 2.108 J/g·°C s(liquid water) = 4.184 J/g·°C s(steam) = 1.996 J/g·°C heat of fusion = 6.01 kJ/mol heat of vaporization = 40.79
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A quantity of ice at 0.0 degrees C was added to 33.6 of water at 21.0 degree C to give water at 0.0 degrees C. How much ice was added? The heat of fusion of water is 6.01 kJ/mol and the specific heat is 4.18 J/(g * degrees C) q =
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How much energy is required to change a 58 g ice cube from ice at −1◦C to steam at 101◦C? The specific heat of ice is 2090 J/kg ·◦ C and of water 4186 J/kg ·◦ C. The latent heat of fusion of water is 3.33 × 105 J/kg,
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