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The heat required to vaporize 500 grams of ice at its freezing point is the sum of the heat required to raise the temperature of the ice to its melting point, the heat of fusion to melt the ice, the heat required to raise the temperature of water to its boiling point, and finally the heat of vaporization to vaporize the water. The specific heat capacity of ice, heat of fusion of ice, specific heat capacity of water, and heat of vaporization of water are all needed to perform the calculations.

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1y ago

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Does it take more heat to vaporize 12 grams of CH4 or 12 grams of Hg?

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Does sugar the time it takes water to freeze?

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How many grams of sugar must be dissolved in 300 grams of water in order to change the freezing point 2.5 degrees Celsius?

To decrease the freezing point of water by 2.5 degrees Celsius, you would need to dissolve approximately 37.5 grams of sugar in 300 grams of water. This is calculated based on the colligative property that states freezing point depression is directly proportional to the molality of the solute in the solution.


How many joules of heat are needed to completely vaporize 24.40 grams of water at its boiling point?

The heat of vaporization of water is 40.79 kJ/mol. First, determine the number of moles in 24.40 grams of water. Then, convert moles to joules using the molar heat of vaporization. This will give you the amount of heat needed to vaporize 24.40 grams of water.


What temperature is manganese at freezing?

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What is the molecular mass of 8.02 grams of solute in 861 grams of water lower the freezing point to -0.430 degrees C?

To calculate the molecular mass of the solute, we need to use the formula for freezing point depression: ΔTf = Kf * m. Given that the ΔTf is -0.430°C, the molal concentration (m) of the solute can be found by dividing the grams of solute by the grams of water. Substituting these values into the formula allows us to solve for Kf, which can eventually be used to determine the molecular mass of the solute.


Why The refrigerant should have low boiling point and low freezing point?

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8.02 grams of solute in 861 grams of water lower the freezing point to -0.430 c?

The change in freezing point is -0.430°C. The molal freezing point depression constant for water is 1.86°C kg/mol. Calculate the molality of the solution using the formula: ΔTf = Kf * m. 8.02 g of solute in 861 g of water is equivalent to 8.02 / 58.44 = 0.1373 mol of solute.


What is the change in the boiling point and freezing point of 500 grams of water if 45 grams of glucose C6H12O6 were dissolved in it?

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What is the antonym for boiling point?

Freezing point.


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The freezing cannot be stopped; only the freezing temperature is lowered adding salts.