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Vanyuwa [196]
3 years ago
14

A lead sphere has a mass of 1.20 x 104 g, and its volume is 1.05 x 103 cm^3. Calculate the density of lead.

Chemistry
1 answer:
Hatshy [7]3 years ago
4 0

Answer: The density of lead will be 11.42g/cm^3

Explanation:

Density is defined as the mass contained per unit volume.

Density=\frac{mass}{Volume}

Given : Mass of lead=1.20\times 10^4g

Volume of lead =1.05\times 10^3cm^3

Putting in the values we get:

Density=\frac{1.20\times 10^4}{1.05\times 10^3}=11.42g/cm^3

Thus density of lead will be 11.42g/cm^3

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Using the mmoles of (35)-2,2,-dibromo-3,4-dimethylpentane calculated earlier and the molecular weight of the product (962 g/mol)
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The yield of the product in gram is \mathsf{{w_P}=0.26 \ gram}

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i.e

Theoretical yield = \dfrac{n_P}{n_R}\times 100\%

where;

n_P = \dfrac{w_P}{m_P}    and n_R = \dfrac{w_R}{m_R}

Theoretical yield = \dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

Given that the theoretical yield = 100%

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100\% =\dfrac{(\dfrac{w_P}{m_P})}{(\dfrac{w_R}{m_R})} \times 100\%

\dfrac{w_P}{m_P}=\dfrac{w_R}{m_R}

{w_P}=\dfrac{w_R \times m_P}{m_R}

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w_P = derived weight of the product

m_P =the molecular mass of the derived product

m_R = the molecular mass of the reagent

w_R = weight in a gram of the reagent

{w_P}=\dfrac{w_R \times m_P}{m_R}

{w_P}=\dfrac{0.697 \times 96.2}{257.997}

\mathsf{{w_P}=0.26 \ gram}

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