<h2>Answer:</h2>
He is right that the energy of vaporization of 47 g of water s 106222 j.
<h3>Explanation:</h3>
Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.
In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.
It means for vaporizing 18 g, 40.65 kJ energy is needed.
So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ
Hence the student is right about the energy of vaporization of 47 g of water.
Thermoplastic and thermosetting
thermoplastic:- they are easily molded and extruded into films, fibres and packaging.For eg. PVC
thermosetting:-they are hard and durable and can be used for aircraft parts,tires and auto parts .For eg. phenolic resins.
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The question is missing the data sets.
This is the complete question:
A single penny has a mass of 2.5 g. Abbie and James
each measure the mass of a penny multiple times. Which statement about
these data sets is true?
O Abbie's measurements are both more accurate
and more precise than James'.
O Abbie's measurements are more accurate,
but less precise, than James'.
O Abbie's measurements are more precise,
but less accurate, than James'.
O Abbie’s measurements are both less
accurate and less precise than James'.
Penny masses (g)
Abbie’s data
2.5, 2.4, 2.3, 2.4, 2.5, 2.6, 2.6
James’ data
2.4, 3.0, 3.3, 2.2, 2.9, 3.8, 2.9
Answer: first option, Abbie's measurements are both more accurate
and more precise than James'.
Explanation:
1) To answer this question, you first must understand the difference between precision and accuracy.
<span>Accuracy is how close the data are to the true or accepted value.
</span>
<span>Precision is how close are the data among them, this is the reproducibility of the values.</span>
Then, you can measure the accuracy by comparing the means (averages) with the actual mass of a penny 2.5 g.
And you measure the precision by comparing a measure of spread, as it can be the standard deviation.
2) These are the calculations:
Abbie’s data
Average: ∑ of the values / number of values
Average = [2.5 + 2.4 + 2.3 + 2.4 + 2.5 + 2.6 + 2.6 ] / 7 = 2.47 ≈ 2.5
Standard deviation: √ [ ∑ (x - mean)² / (n - 1) ] = 0.11
James’ data
Average = [2.4 + 3.0 + 3.3 + 2.2 + 2.9 + 3.8 + 2.9] / 7 = 2.56 ≈ 2.6
Standard deviation = 0.53
3) Conclusions:
1) The average of Abbie's data are closer to the accepted value 2.5g, so they are more accurate.
2) The standard deviation of Abbie's data is smaller than that of Jame's data, so the Abbie's data are more precise.
