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ale4655 [162]
4 years ago
8

Find the age t of a sample, if the total mass of carbon in the sample is mc, the activity of the sample is a, the current ratio

of the mass of 14 6c to the total mass of carbon in the atmosphere is r, and the decay constant of 14 6c is λ. assume that, at any time, 14 6c is a negligible fraction of the total mass of carbon and that the measured activity of the sample is purely due to 14 6c. also assume that the ratio of mass of 14 6c to total carbon mass in the atmosphere (the source of the carbon in the sample) is the same at present and on the day when the number of 14 6c atoms in the sample was set. express your answer in terms of the mass ma of a 14 6c atom, mc, a, r, and λ.
Chemistry
1 answer:
Paha777 [63]4 years ago
6 0
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
 The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) = 
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).

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\text{Moles of} HgS=\frac{20.0 g}{233g/mol}=0.085moles

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4HgS(s)+4CaO(s)\rightarrow 4Hg(l)+3CaS(s))+CaSO_4(s)  

According to stoichiometry :  

4 moles of HgS produce =  4 moles of Hg

Thus 0.085 moles of HgS will require=\frac{4}{4}\times 0.085=0.085moles  of Hg

Mass of Hg=moles\times {\text {Molar mass}}=0.085moles\times 200.6g/mol=17.2g

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What is a natural resource that is plentiful, but strong enough to be used as currency?
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The two characteristics — “plentiful” and “currency” — are not really related.

Money is a symbolic tool to expedite trade. In earliest days, people might carry bushels of wheat to exchange for jars of oil. But that is cumbersome, and the search costs are too high. It’s hard to find another party who both wants what you have, AND who also has what you want!.

Instead, we all agree on trading with “money.” The farmer gives the store milk and gets money. Later you give the store money and get milk.

SO… The most important characteristic of currency is that it is hard to fake. That is to prevent cheating. Gold and silver were scarce so made good coins. Then printed currency was made hard to counterfeit. Trade with money is based on confidence in the money.

In an entirely different direction are the natural resources we take for granted - like air and water.

If you explain your question further in the comments, glad to respond.

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Real value equals nature which sustains all life.

This is indisputable & real value, nature & life are inseparable.

Nature is a singular compound billions years old recreative self-sustainable real value system of living organisms. Humans are a mere millions years old failing species of mammal.

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Fake “value“ or wealth is created by failed economics which creates fake income inequality that is used to manipulate the masses access to real value. Fake value is “shadow wealth” used by capitalists to control and manipulate access to natures real value.

Wealth is a fake value lien placed by the wealthy (capitalists) on natures real life sustaining value and is facilitated by economics and capitalism in particular.

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