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Kazeer [188]
3 years ago
9

Two groups were tasked to measure the mass of the metal objects given by their teacher. Rachel’s group has the following data fo

r 3 trials: 97 g, 98 g, and 99 g. Meanwhile, Ashley’s group has 99.5 g, 100.1 g, and 100.5 g. If the accepted value for the metal’s mass is 100 g, how will you describe the data of each group in terms of precision and accuracy?
Chemistry
1 answer:
Studentka2010 [4]3 years ago
5 0

The measurement of Rachel’s group is precise but not accurate while the measurement of Ashley’s group is accurate but not precise.

Precision has to do with how close together the values obtained from a scientific measurement is. If we take a look at the values obtained by  Rachel’s group, we will notice that the values are exactly 1.00 g apart. This means that the values are precise.

However, these values a far from the actual value which is 100.00 g therefore the measurement of Rachel’s group is precise but not accurate.

On the other hand, the values obtained by Ashley’s group are; 99.5 g, 100.1 g, and 100.5 g. These values are very close to the actual value which is 100.00 g  hence they are accurate.

The values obtained by Ashley’s group do not have consistent intervals therefore, they are not precise.

Learn more; brainly.com/question/15664210

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Answer:

Explanation:

water/wind/ice how does this agent continue to weathering lake annette

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Answer:co2

Explanation:

7 0
2 years ago
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Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

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