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Elan Coil [88]
3 years ago
5

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the rea

ding of 400. counts has diminished to 100. counts after 90.3 minutes , what is the half-life of this substance? Express your answer with the appropriate units. View Available Hint(s) t1/2 t 1 / 2 t_1/2 = nothing
Chemistry
1 answer:
taurus [48]3 years ago
8 0

Answer:

t₁/₂ = 45.1 min

Explanation:

The  radioactive decay equation is given by

N/N₀ = e^-kt  where N = counts after time t

                                 N₀ = counts initially

                                 k = decay constant

                                 t=  time elapsed

The question is what´s the half-life of this substance, and we can solve it once we have k from the expression above since

                                 t₁/₂ = 0.693/k

which is derived from  that equation, but for the case    N/N₀ is 0.5

Lets calculate k and  t₁/₂ :

N/N₀ = e^-kt      (taking ln in the two sides of the equation)

ln (N/N₀) =  ln e^-kt   = -kt   ⇒ k = -ln(N/N₀)/t

k = -ln(100/400)/90.3 min = 0.01535 min⁻¹

 t₁/₂ = 0.693/k  = 0.693/0.01535 min⁻¹  = 45.1 min

We can check this answer since the time in the question is the double of this  half-life and the data shows the material has decayed by a fourth: two half-lives.

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A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

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