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3 years ago

What mass of lead (density 11.4 g/cm3) would have an identical volume to 20.1 g of mercury (density 13.6 g/cm3)?

2 answers:

8 0

In order to determine the mass of lead, Pb, that would have an identical volume as 20.1 g of mercury, Hg, we first get the equivalent volume of 20.1 g of mercury. This can be done by using the given density of mercury which is equal to 13.6 g/cm^3. This is shown in the following equation:

20.1 g Hg/ 13.6g/cm^3 = 1.4779 cm^3 of Hg

Consequently, we use this obtained volume which is the target volume for lead. Again, we make use of the density of the substance, in this case, lead has a density of 11.4 g/cm^3. The resulting equation is then:

11.4 g/cm^3 x 1.4779 cm^3 = 16.8485 g of Pb

Thus, the amount of lead that has an identical volume as 20.1 g of mercury is 16.8485 g Pb.

inysia [295]3 years ago

3 0

Answer: The mass of lead is 16.872 g

Explanation:

To calculate mass of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$ ......(1)

For Mercury:

Density of mercury = $13.6g/cm^3$

Mass of mercury = 20.1 g

Putting values in equation 1, we get:

$13.6g/cm^3=\frac{20.1g}{\text{Volume of mercury}}\\\\\text{Volume of mercury}=1.48cm^3$

The volume of mercury is $1.48cm^3$ which is equal to the lead sample.

For lead:

We are given:

Density of lead = $11.4g/cm^3$

Volume of lead = $1.48cm^3$

Putting values in equation 1, we get:

$11.4g/cm^3=\frac{\text{Mass of lead}}{1.48cm^3}\\\\\text{Mass of lead}=16.872g$

Hence, the mass of lead is 16.872 g

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Explanation:

Steps followed to practice laboratory safety during the experiment are as follows.

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Careful observations were made so that correct calculations about the experiment can be carried out.

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in a simulation mercury removal from industrial wastewater, 0.020 L of 0.10 M sodium sulfide reacts with 0.050 L of 0.010 M merc

Margarita [4]

Answer: 0.1161 grams of mercury(II) sulfide) form.

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}$ .....(1)

a) Molarity of $Na_2S$ solution = 0.10 M

Volume of solution = 0.020 L

Putting values in equation 1, we get:

$0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol$

$\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol$

b) Molarity of $Hg(NO_3)_2$ solution = 0.010 M

Volume of solution = 0.050 L

Putting values in equation 1, we get:

$0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol$

$Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3$

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ reacts with 1 mole of $Na_2S$

Thus 0.0005 moles of $HgNO_3$ reacts with= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg(NO_3)_2$

Thus $Hg(NO_3)_2$ is the limiting reagent and $Na_2S$ is the excess reagent.

According to stoichiometry :

1 mole of $Hg(NO_3)_2$ forms= 1 mole of $Hg_2S$

Thus 0.0005 moles of $Hg(NO_3)_2$ forms= $\frac{1}{1}\times 0.0005=0.0005$ moles of $Hg_2S$

mass of $H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g$

Thus 0.1161 grams of mercury(II) sulfide) form.

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Volgvan

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puteri [66]

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Explanation:

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