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Svetradugi [14.3K]
3 years ago
14

The barium isotope 133ba has a half-life of 10.5 years. a sample begins with 1.1×1010 133ba atoms. how many are left after (a) 5

years, (b) 30 years, and (c) 190 years?
Chemistry
2 answers:
enot [183]3 years ago
7 0

Answer: Atoms of _{56}^{133}\textrm{Ba} left in

a) N=78.608\times 10^8atoms

b) N=14.650\times 10^8atoms

c) N=31.35\times 10^3atoms

Explanation: The given reaction is a type of radioactive decay and all the radioactive decay follows first order reactions. Hence, to calculate the rate constant, we use the formula:

k=\frac{0.693}{t_{1/2}}

t_{1/2}=10.5years

k=\frac{0.693}{10.3years}\\k=0.0672years^{-1}

To calculate how much amount of sample is left, we use the rate law expression for first order kinetics, which is:

N=N_oe^{-kt}    ....(1)

where,

k = rate constant  = 0.0672years^{-1}

t = time taken for decay process

N_o = initial amount of the reactant  = 1.1\times 10^{10}g

N = amount left after decay process

  • For a)

t = 5 years

Putting values in equation 1, we get:

N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 5years)}\\N=78.608\times 10^8atoms

  • For b)

t = 30 years

Putting values in equation 1, we get:

N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 30years)}\\N=14.650\times 10^8atoms

  • For c)

t = 190 years

Putting values in equation 1, we get:

N=(1.1\times 10^{10})\times e^{(-0.0672years^{-1}\times 190years)}\\N=31.35\times 10^3atoms

Sladkaya [172]3 years ago
4 0
<span>a) 7.9x10^9 b) 1.5x10^9 c) 3.9x10^4 To determine what percentage of an isotope remains after a given length of time, you can use the formula p = 2^(-x) where p = percentage remaining x = number of half lives expired. The number of half lives expired is simply x = t/h where x = number of half lives expired t = time spent h = length of half life. So the overall formula becomes p = 2^(-t/h) And since we're starting with 1.1x10^10 atoms, we can simply multiply that by the percentage. So, the answers rounding to 2 significant figures are: a) 1.1x10^10 * 2^(-5/10.5) = 1.1x10^10 * 0.718873349 = 7.9x10^9 b) 1.1x10^10 * 2^(-30/10.5) = 1.1x10^10 * 0.138011189 = 1.5x10^9 c) 1.1x10^10 * 2^(-190/10.5) = 1.1x10^10 * 3.57101x10^-6 = 3.9x10^4</span>
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Brums [2.3K]

Answer : The standard enthalpy of formation of ethylene is, 52.4 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

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The formation reaction of C_2H_4 will be,

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The intermediate balanced chemical reaction will be,

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Now we will reverse the reaction 1, multiply reaction 2 and 3 by 2 then adding all the equation, we get :

(1) 2CO_2(g)+2H_2O(l)\rightarrow C_2H_4(g)+3O_2(g)     \Delta H_1=+1411kJ

(2) 2C(s)+2O_2(g)\rightarrow 2CO_2(g)    \Delta H_2=2\times (-393.5kJ)=-787kJ

(3) 2H_2(g)+2O_2(g)\rightarrow 2H_2O(l)    \Delta H_3=2\times (-285.8kJ)=-571.6kJ

The expression for enthalpy of formation of C_2H_4 will be,

\Delta H_{formation}=\Delta H_1+\Delta H_2+\Delta H_3

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Therefore, the standard enthalpy of formation of ethylene is, 52.4 kJ

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