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romanna [79]
3 years ago
15

While doing a lab a student found the density of a piece of pure aluminum to be 2.85 g/cm3 the accepted value for the density of

aluminum is 2.70 g/cm3. What was the percent error
Chemistry
1 answer:
____ [38]3 years ago
3 0

Answer:

THE PERCENT ERROR IS 5.55 %

Explanation:

To calculate the percent error, we use the formula:

Percent error = Found value - accepted value / accepted value * 100

Found value = 2.85 g/cm3

Accepted value = 2.70 g/cm3

Solving for the percent error, we have:

Percent error = 2.85 g/cm3 - 2.70 g/cm3 / 2.70 g/cm3 * 100

Percent error = 0.15 / 2.70 * 100

Percent error = 0.05555 * 100

Percent error = 5.55 %

In conclusion, the percent error is 5.55 %

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GalinKa [24]

<u>Answer:</u> The moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

  • <u>For lead (II) nitrate:</u>

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L   (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

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Molarity of NaI solution = 0.00157 M

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Putting values in equation 1, we get:

0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol

For the given chemical reaction:

Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, 3.14\times 10^{-5} moles of NaI will react with = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 3.14\times 10^{-5} moles of NaI will produce = \frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is 1.57\times 10^{-5} moles

4 0
3 years ago
Which element has the ground state electron configuration 1s22s22p63s23p3?
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Answer:

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The ΔHºfs can be found in appropiate reference or texts.

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