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Fiesta28 [93]
2 years ago
9

2. What is the chance that a carbon-14 atom will decay in six thousand years?

Chemistry
1 answer:
tatyana61 [14]2 years ago
3 0

Answer:

C-14 when formed naturally decays by beta emission to N-14. Therefore, there's a 100% probability of naturally decaying from its birth past 6000 yrs and on through 11,200 yrs and on to an infinity of timed half-lives.

Explanation:

<em>The question is somewhat ambiguous in its wording but the following note will focus on defining the 1st order decay trend of C-14 and the half-life values that can be determined from the 1st order decay equation. </em>

Radioactive decay is described by 1st order kinetics and follows an exponential trend shown graphically below. Such a trend is represented by the expression A = A₀·exp(-k·t) where A = final activity, A₀ = initial activity, k = rate constant and t = time of decay.

Also in describing radioactive decay process is 'half-life'. It is defined as the time needed for the original amount of material to decrease to 1/2 of its original amount. The 'half-life' equation is derived from the above 1st order decay equation and is given by the expression t(1/2) = 0.693/k. This equation allows one to determine the time needed for 'one' half-life. The same equation can be used to determine the 'second' half-life, or the time it takes for the original material to decay to 1/4th of the original amount and so on.

In consideration of the decay of C-14 which has a published 1st half-life of 5,600 years, then a second half-life equals 2(5600 yrs) or 11,200 yrs and a 3rd half-life would be 3(5600 yrs) or 16,800 yrs and so on.

So, since C-14 naturally decays by beta emission which is high energy electron emissions (β⁻) to give N-14. Such will continue to decay past 5,600 years on to a 2nd half-life for as many half-lives one wished to consider.

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Hydrogen sulfide,H2S, is a very toxic gas with a smell of rotten eggs. using the following: H2S+3/2 O2=SO2+H2O H2+1/2O2=H2O S+O2
Serga [27]

Answer:

ΔH = -20kJ

Explanation:

The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:

H₂(g) + S(g) → H₂S(g)

Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:

<em>(1) </em>H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

The sum of -(1) + (2) + (3) gives:

<em>-(1) </em>SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ

<em>(2) </em>H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ

<em>(3) </em>S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ

<em>-(1) + (2) + (3): </em><em>H₂(g) + S(g) → H₂S(g) </em>

<em>ΔH =</em> +519kJ - 242kJ - 297kJ = <em>-20 kJ</em>

<em />

I hope it helps!

5 0
3 years ago
Type the correct answer in the box. Express your answer to three significant figures.
VladimirAG [237]

<u>Given:</u>

Mass of calcium nitrate (Ca(NO3)2) = 96.1 g

<u>To determine:</u>

Theoretical yield of calcium phosphate, Ca3(PO4)2

<u>Explanation:</u>

Balanced Chemical reaction-

3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2

Based on the reaction stoichiometry:

3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2

Now,

Given mass of Ca(NO3)2 =  96.1 g

Molar mass of Ca(NO3)2 =  164 g/mol

# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles

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Molar mass of Ca3(PO4)2 = 310 g/mol

Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g

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