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Hunter-Best [27]
2 years ago
7

Ancient Egyptians used a variety of lead compounds as white pigments in their cosmetics, including PbS, PbCO3, PbCl(OH), and Pb2

Cl2CO3. Which of these compounds contains the highest percentage of lead (by mass)
Chemistry
1 answer:
ser-zykov [4K]2 years ago
6 0

Answer : The compound contains the highest percentage of lead (by mass) is, PbS.

Explanation :

To calculate the percentage of lead in sample, we use the equation:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100

<u>For PbS :</u>

Mass of PbS = 239.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{239.3g}\times 100=86.58\%

The percentage of lead in the PbS is 86.58 %.

<u>For PbCO_3 :</u>

Mass of PbCO_3 = 267.2 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{267.2g}\times 100=77.55\%

The percentage of lead in the PbCO_3 is 77.55 %.

<u>For PbCl(OH) :</u>

Mass of PbCl(OH) = 259.7 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{259.7g}\times 100=79.78\%

The percentage of lead in the PbCl(OH) is 79.78 %.

<u>For Pb_2Cl_2CO_3 :</u>

Mass of Pb_2Cl_2CO3_ = 545.3 g

Mass of Pb = 207.2 g

Putting values in above equation, we get:

\%\text{ of lead}=\frac{\text{Mass of lead}}{\text{Mass of sample}}\times 100=\frac{207.2g}{545.3g}\times 100=37.99\%

The percentage of lead in the Pb_2Cl_2CO3_ is 37.99 %.

Hence, from this we conclude that, the compound PbS contains the highest percentage of lead (by mass).

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Cyclohexane has a freezing point of 6.50 ∘C and a Kf of 20.0 ∘C/m. What is the freezing point of a solution made by dissolving 0
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Answer:

The freezing point will be 2.9^{0}C

Explanation:

The depression in freezing point is a colligative property.

It is related to molality as:

Depressioninfreezing point=K_{f}Xmolality

Where

Kf= 20\frac{^{0}C}{m}

the molality is calculated as:

molality=\frac{moles_{solute}}{mass_{solvent}}

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massofcyclohexane=25g=0.025Kg

molarity=\frac{0.0045}{0.025}=0.18m

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What is the oxidation state of selenium in SeO3?​
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Answer: The oxidation state of selenium in SeO3 is +6

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SeO3 is the chemical formula for selenium trioxide.

- The oxidation state of SeO3 = 0 (since it is stable and with no charge)

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- the oxidation state of selenium in SeO3 = Z (let unknown value be Z)

Hence, SeO3 = 0

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We know both the elements involved in bonding are non-metals and the primary type of bonds involved in non-metals are covalent bonds. Covalent bonds are formed when two atoms share one or more electrons; thus we know that whatever the number of electrons shared, it has to be equal for both. We can eliminate choices A and B.
Next, we understand that it is easier for one atom to be shared among the two, rather than the 7. First, because Hydrogen needs only 1 electron to be stable and would require energy to compensate the remaining 6. Second, electrons are attracted towards the nucleus so it is inefficient to try and share 7 electrons when 1 provides the same amount of stability to Fluorine.
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2 years ago
Read 3 more answers
A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight
zhannawk [14.2K]

Answer: The molecular formula will be C_8H_8O_2

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 70.6 g

Mass of H = 5.9 g

Mass of O = 23.5 g

Step 1 : convert given masses into moles.

Moles of C =\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{70.6g}{12g/mole}=5.9moles

Moles of H =\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{5.9g}{1g/mole}=5.9moles

Moles of O =\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{23.5g}{16g/mole}=1.5moles

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = \frac{5.9}{1.5}=4

For H = \frac{5.9}{1.5}=4

For O =\frac{1.5}{1.5}=1

The ratio of C : H: O= 4: 4:1

Hence the empirical formula is C_4H_4O

The empirical weight of C_4H_4O = 4(12)+4(1)+1(16)= 68g.

The molecular weight = 136 g/mole

Now we have to calculate the molecular formula.

n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{136}{68}=2

The molecular formula will be=2\times C_4H_4O=C_8H_8O_2

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