3. How much energy is needed to raise 45 grams of water from 40°C to 115 °C?
1 answer:
Answer:
Q = 114349.5 J
Explanation:
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In this case, since this a problem in which we need to calculate the total heat of the described process, it turns out convenient to calculate it in three steps; the first one, associated to the heating of the liquid water from 40 °C to 100 °C, next the vaporization of liquid water to steam at constant 100 °C and finally the heating of steam from 100 °C to 115 °C. In such a way, we calculate each heat as shown below:
Thus, the total energy turns out to be:
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Answer:
Mole fraction of = 0.58
Mole fraction of = 0.42
Explanation:
Let the mass of and
= x g
Molar mass of = 33.035 g/mol
The formula for the calculation of moles is shown below:
Thus,
Molar mass of = 46.07 g/mol
Thus,
So, according to definition of mole fraction:
Mole fraction of = 1 - 0.58 = 0.42
Answer:
The number of moles of KClO₃ reacted was 0,15 mol
Explanation:
For the reaction:
2KClO₃(s) → 2KCl(s) + 2O₂(g)
The only gas product is O₂.
Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:
749mmHg - 23,8mmHg = 725,2mmHg
Using gas law:
PV/RT = n
Where:
P is pressure (725,2mmHg ≡ <em>0,9542atm</em>)
V is volume (<em>5,76L</em>)
R is gas constant (<em>0,082 atmL/molK</em>)
And T is temperature (25°C ≡ <em>298,15K</em>)
Replacing, number of moles of O₂ are <em>0,2248 moles</em>
As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:
0,2248 mol O₂× = <em>0,15 mol of KClO₃</em>
I hope it helps!