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lutik1710 [3]
3 years ago
15

Even though lead is toxic, lead compounds were used in ancient times as white pigments in cosmetics. What is the percentage of l

ead by mass in lead(IV) carbonate, Pb(CO3)2
Chemistry
1 answer:
Gekata [30.6K]3 years ago
4 0

<u>Answer:</u> The mass percent of lead in lead (IV) carbonate is 63.32 %

<u>Explanation:</u>

The given chemical formula of lead (IV) carbonate is Pb(CO_3)_2

To calculate the mass percentage of lead in lead (IV) carbonate, we use the equation:

\text{Mass percent of lead}=\frac{\text{Mass of lead}}{\text{Mass of lead (IV) carbonate}}\times 100

Mass of lead = (1 × 207.2) = 207.2 g

Mass of lead (IV) carbonate = [(1 × 207.2) + (2 × 12) + (6 × 16)] = 327.2 g

Putting values in above equation, we get:

\text{Mass percent of lead}=\frac{207.2g}{327.2g}\times 0100=63.32\%

Hence, the mass percent of lead in lead (IV) carbonate is 63.32 %

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SO
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Question 7.

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\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}

Question 8. I

We can use the Ideal Gas Law to solve this question.

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\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}

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3 years ago
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