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3 years ago

What is the density of an 84.7 g sample of an unknown substance if the sample occupies 49.6cm cubed

Chemistry

1 answer:

Ahat [919]3 years ago

7 0

Answer:

1.70 g.cm⁻³

Solution:

Data Given;

Mass = 84.7 g

Volume = 49.6 cm³

Density = ?

Formula Used;

Density = Mass ÷ Volume

Putting values,

Density = 84.7 g ÷ 49.6 cm³

Density = 1.70 g.cm⁻³

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The tabulated data were collected for this reaction:

erastovalidia [21]

Answer:

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Explanation:

Equation of Reaction:

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If $A + B \rightarrow C + D$ , the rate of backward reaction is given by:

$Rate = k [A]^{a} [B]^{b}\\k = \frac{Rate}{ [A]^{a} [B]^{b}}\\k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$

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Using the information provided in lines 1 and 2 of the table:

$0.014 / [0.05]^a [0.05]^b = 00.029/ [0.100]^a [0.05]^b\\0.014 / [0.05]^a [0.05]^b = 00.029/ [2*0.05]^a [0.05]^b\\0.014 / = 0.029/ 2^a\\2^a = 2.07\\a = 1$

Using the information provided in lines 3 and 4 of the table and insering the value of a:

$0.041 / [0.100]^a [0.100]^b = 0.115 / [0.200]^a [0.200]^b\\0.041 / [0.100]^a [0.100]^b = 0.115 / [2 * 0.100]^a [2 * 0.100]^b\\$

$0.041 = 0.115 / [2 ]^a [2]^b\\ \[[2 ]^a [2]^b = 0.115/0.041\\ \[[2 ]^a [2]^b = 2.80\\\[[2 ]^1 [2]^b = 2.80\\\[[2]^b = 1.40\\b = \frac{ln 1.4}{ln 2} \\b = 0.5$

The rate law is: $Rate = k [CH_3 Cl] [Cl_2]^{0.5}$

The rate constant $k = \frac{Rate}{ [CH_3 Cl]^{a} [Cl_2]^{b}}$ then becomes:

$k = 0.014 / ( [0.050] [0.050]^(0.5) )\\k = 1.25$

b) Overall order of reaction = a + b

Overall order of reaction = 1 + 0.5

Overall order of reaction = 1.5

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