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3 years ago

How can you simulate the radioactive half-life of an element?

Chemistry

1 answer:

ch4aika [34]3 years ago

6 0

Answer:

TRIAL 1:

For “Event 0”, put 100 pennies in a large plastic or cardboard container.

For “Event 1”, shake the container 10 times. This represents a radioactive decay event.

Open the lid. Remove all the pennies that have turned up tails. Record the number removed.

Record the number of radioactive pennies remaining.

For “Event 2”, replace the lid and repeat steps 2 to 4.

Repeat for Events 3, 4, 5 … until no pennies remain in the container.

TRIAL 2:

Repeat Trial 1, starting anew with 100 pennies.

Calculate for each event the average number of radioactive pennies that remain after shaking.

Plot the average number of radioactive pennies after shaking vs. the Event Number. Start with Event 0, when all the pennies are radioactive. Estimate the half-life — the number of events required for half of the pennies to decay.

Explanation:

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The decomposition reaction of carbon disulfide to carbon monosulfide and sulfur is first order with k = 2.80 ✕ ✕ 10−7 sec-1 at 1

madam [21]

Answer: a) 1.97 grams of carbon disulfide will remain after 37.0 days.

b) 2.85 grams of carbon monosulfide will be formed after 37.0 days.

Explanation: The decomposition of carbon disulfide is given as:

$CS_2(g)\rightarrow CS(g)+S(g)$

at t=0 4.83g 0 0

at t=37 days 4.83 - x x x

here,

x = amount of $CS_2$ utilised in the reaction

This reaction follows first order kinetics so the rate law equation is:

$k=\frac{2.303}{t}log\frac{A_o}{A}$

where, k = rate constant

t = time

$A_o$ = Initial mass of reactant

A = Final mass of reactant

a) For this, the value of

$k=2.80\times10^{-7}sec^{-1}$

t = 370 days = 3196800 sec

$A_o$ = 4.83

A = 4.83-x

Putting values in the above equation, we get

$2.8\times 10^{-7}sec^{-1}=\frac{2.303}{3196800sec}log\left(\frac{4.83}{4.83-x}\right)$

x = 2.85g

Amount of $CS_2$ remained after 37 days = 4.83 - x

= 1.97g

b) Amount of carbon monosulfide formed will be equal to "x" only which we have calculated in the previous part.

Amount of carbon monosulfide formed = 2.85g

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What is the name of the ionic compound rbcl?

Kamila [148]

Rubidium Chloride is the answer i believe

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