To change 1.34 kg of ice at 0°C to water, you would need to add approximately 448 kJ of heat.
To calculate the amount of heat needed to convert ice at 0°C to water, we use the formula Q = m * Lf. Where Q is the Heat Transfer, m is the mass of the substance (ice in this case), and Lf is the heat of fusion for ice, which is 334 kJ/kg.
Plugging in the values we have: Q = 1.34 kg * 334 kJ/kg = 447.56 kJ.
Therefore, you would need to add approximately 448 kJ of heat to convert 1.34 kg of ice at 0°C to water at the same temperature.
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The heat required to convert 1.34 kg of ice at 0°C to water at 0°C is approximately 450.24 kJ.
When the substance freezes (changes from liquid to solid), the same amount of energy is released back into the surroundings. When a substance changes its phase (solid to liquid or liquid to gas) at a constant temperature, the heat required for this process is known as the heat of fusion.
For ice at its meltingpoint (0°C), the heat of fusion is approximately 336 kJ/kg. Thus for 1.34 kg of ice, we can calculate the heat required by using the formula below:
Heat = mass × heat of fusion
Heat = 1.34 kg × 336 kJ/kg
Heat = 450.24 kJ.
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The complete question is -
How much heat (in kJ) must be added to 1.34 kg of ice at 0°C to convert it to water at 0°C
Considering the definition of molar mass and molarity, the volume of each compound is:
The molar mass of substance is a property defined as its mass per unit quantity of substance, in other words, molar mass is the amount of mass that a substance contains in one mole.
Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.
The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:
Molarity is expressed in units.
You know that the molar mass of each compound is:
On the other side, you know that the mass of each compound is:
Then, considering the definition of molar mass, the amount of moles of each compound is:
The molarity of each compound is:
Considering the molarity, the volume of each compound is calculated as:
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molar mass:
molarity:
True or False
b. alpha
c. proton
d. positron
b. will be easily separated.
c. lose their own properties.