6. How many moles of water would require 92.048 kJ of heat to raise its temperature from 34.0 °C to 100.0 °C? (3 marks)
1 answer:
Taking into account the definition of calorimetry, 0.0185 moles of water are required.
<h3>Calorimetry</h3>Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
So, the equation that allows to calculate heat exchanges is:
Q = c× m× ΔT
where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.
<h3>Mass of water required</h3>In this case, you know:
- Heat= 92.048 kJ
- Mass of water = ?
- Initial temperature of water= 34 ºC
- Final temperature of water= 100 ºC
- Specific heat of water = 4.186
Replacing in the expression to calculate heat exchanges:
92.048 kJ = 4.186 × m× (100 °C -34 °C)
92.048 kJ = 4.186 × m× 66 °C
m= 92.048 kJ ÷ (4.186 × 66 °C)
<u><em>m= 0.333 grams</em></u>
<h3>Moles of water required</h3>Being the molar mass of water 18 , that is, the amount of mass that a substance contains in one mole, the moles of water required can be calculated as:
<u><em>amount of moles= 0.0185 moles</em></u>
Finally, 0.0185 moles of water are required.
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Answer:
Explanation:
1. Add the atomic mass of all the elements.
39+55.8+12+14= 120.8
2.Divide atomic mass of potassium by total atomic mass
2. Multiply by 100
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