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antiseptic1488 [7]
2 years ago
14

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? Abbie's measurements are both more accurate and more precise than James'. Abbie's measurements are more accurate, but less precise than James'. Abbie's measurements are more precise, but less accurate than James'. Abbie’s measurements are both less accurate and less precise than James'.
Chemistry
2 answers:
const2013 [10]2 years ago
7 0
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.
podryga [215]2 years ago
7 0
A. Abbie's measurements are both more accurate and more precise then James'
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The question is incomplete, here is the complete question:

Problem page in each of the molecules drawn below one chemical bond is colored red. Decide whether this bond is likely to be polar or not. if the bond is likely to be polar, write down the chemical symbol for the atom which will have more negative charge.

The image is attached below.

<u>Answer:</u>

<u>For carbon dioxide molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.

<u>For water molecule:</u> The bond is considered as polar and the elecvtronegative atom is oxygen.

<u>Explanation:</u>

There are two types of covalent bonds:

  • <u>Polar covalent bond:</u> This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond. <u>For Example:</u> CO_2,NO_2 etc..
  • <u>Non-polar covalent bond:</u> This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. <u>For Example:</u> N_2,O_2 etc..

<u>In carbon dioxide molecule:</u>

The given bond is present between C and O atom.

Electronegativity value of C = 2.5

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.5) = 1

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

<u>In water molecule:</u>

The given bond is present between H and O atom.

Electronegativity value of H = 2.1

Electronegativity value of O = 3.5

Electronegativity difference = (3.5 - 2.1) = 1.4

As, electronegativity difference is present. So, the bond is considered as polar and the elecvtronegative atom is oxygen.

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