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Veronika [31]
2 years ago
11

Phosphorus-32 has a half-life of 14.0 days. Starting with 6.00 g of 32P, how many grams will remain after 42.0 days ?

Chemistry
2 answers:
Sladkaya [172]2 years ago
5 0
It it 23k bc they only have 2367 factors
aleksandrvk [35]2 years ago
4 0
6g...........3g..........1,5g........0,75g
0d..........14d.........28d.........42d

After 42 days 0,75g of Phoshorus-32 will remain.
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Answer:

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b) Overall order of reaction = 1.5

Explanation:

Equation of Reaction:

CH_{3} Cl (g) + 3 Cl_2 (g) \rightarrow CCl_4 (g) + 3 HCl (g)

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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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Hey there!

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