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3 years ago

What is the uncertainty for each of the following numbers? a) 65.74 b) 60.009 c) 66_ _d) 65.750

Chemistry

1 answer:

alex41 [277]3 years ago

5 0

Answer:

a) 65.74 ± 0.01

b) 60.009 ± 0.001

c) 66 ± 1

d) 65.750 ± 0.001

Explanation:

When reporting results, scientist usually specify a range of values that they expect this "true value" to fall within. The most common way to show this range of values is: measurement = best estimate ± uncertainty

Uncertainties are almost always quoted to one significant digit (example: ±0.05 s).

Wrong: 34.3 cm ± 4.3 cm

Correct: 34 cm ± 4 cm

Always round the experimental measurement or result to the same decimal place as the uncertainty.

Wrong: 1.237 s ± 0.1 s

Correct: 1.2 s ± 0.1 s

Therefore for to answer the question:

a) 65.74 ± 0.01

b) 60.009 ± 0.001

c) 66 ± 1

d) 65.750 ± 0.001

I hope you find this information useful and interesting! Good luck!

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Answer : The percent purity of $Fe_2O_3$ in the original sample is 87.94 %

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The given balanced chemical reaction is:

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$\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}$

Molar mass of Fe = 55.8 g/mole

$\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole$

Now we have to calculate the moles of $Fe_2O_3$

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of $Fe_2O_3$

So, $3.15\times 10^4mole$ of Fe produced from $\frac{3.15\times 10^4}{2}=15750$ mole of $Fe_2O_3$

Now we have to calculate the mass of $Fe_2O_3$

$\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3$

Molar mass of $Fe_2O_3$ = 159.69 g/mole

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Now we have to calculate the percent purity of $Fe_2O_3$ in the original sample.

Mass of original sample = $2.86\times 10^3kg$

$\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100$

$\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%$

Therefore, the percent purity of $Fe_2O_3$ in the original sample is 87.94 %

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