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castortr0y [4]
3 years ago
15

The chemical equation below shows the decomposition of ammonium nitrate (NH4NO3)NH4NO3 N20 2H20A chemist who is performing this

reaction starts with 160.1 g of NH4NO3. The molar mass of NH4NO3 is 80.03 g/mol; the molarmass of water (H20) is 18.01 g/mol. What mass, in grams, of H20 is produced?O 9.01O 18.01O 36.03O 72.06

Chemistry
1 answer:
Tcecarenko [31]3 years ago
4 0
<span>The answer is 72.06. The relative formula mass of N2O is 44 (14 + 14 + 16). The relative formula mass of 2H2O is 36 (2(1 + 1 + 16)). The total relative formula mass of the products is 44 + 36 = 80. The ratio of the relative mass of the products to the original mass of the reactants is 2:1. As the relative mass of water produced is 36, doubling this (to make the ratio) gives 72.06.</span>
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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.
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