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netineya [11]
4 years ago
10

A nuclear waste site. cesium-137 is a particularly dangerous by-product of nuclear reactors. it has a half-life of 30 years. it

can be readily absorbed into the food chain and is one of the materials that would be stored in the proposed waste site at yucca mountain (see the article opening this section). suppose we place 3000 grams of cesium-137 in a nuclear waste site.how much cesium-137 will be present after 60 years, or two half-lives?
Chemistry
1 answer:
hram777 [196]4 years ago
4 0
The mass decay rate is of the form
m(t) = m_{0} e^{-kt}
where
m₀ = 3000 g,the initial mass
k = the decay constant
t = time, years.

Because the half-life is 30 years, therefore
e^{-30k} =  \frac{1}{2} \\\ -30k = ln(0.5) \\ k =  \frac{ln(0.5)}{-30} =0.0231

After 60 years, the mass remaining is
m = 3000 e^{-0.0231*60} = 750 \, g

Answer: 750 g

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Equal molar amounts of H2 and I2 are placed into an evacuated container and heated to 445°C, where the following equilibrium is
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<u>Answer:</u> The initial concentration of H_2\text{ and }I_2 are 0.0192 M and 0.0192 M respectively.

<u>Explanation:</u>

We are given:

Equilibrium concentration of HI = 0.030 M

Moles of hydrogen gas = Moles of iodine gas (concentration will also be the same)

For the given chemical equation:

                  H_2+I_2\rightleftharpoons 2HI

<u>Initial:</u>            x     x        -

<u>At eqllm:</u>     x-c    x-c     2c

Calculating the value of 'c'

2c=0.030\\\\c=\frac{0.030}{2}=0.015M

The expression of K_{eq} for above reaction follows:

K_{eq}=\frac{[HI]^2}{[H_2]\times [I_2]}

We are given:

K_{eq}=50.2

[H_2]=(x-c)=(x-0.015)

[I_2]=(x-c)=(x-0.015)

Putting values in above equation, we get:

50.2=\frac{(0.030)^2}{(x-0.015)\times (x-0.015)}\\\\x=0.0108,0.0192

Neglecting the value of x = 0.0108 M, because the initial concentration cannot be less than the equilibrium concentration.

x = 0.0192 M

Hence, the initial concentration of H_2\text{ and }I_2 are 0.0192 M and 0.0192 M respectively.

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evablogger [386]

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Explanation:

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8 0
3 years ago
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A chemist fills a reaction vessel with 7.92 atm nitrogen (N2) gas, 2.02 atm hydrogen (H2) gas, and 2.11 atm ammonia (NH3) gas at
djverab [1.8K]

<u>Answer:</u> The Gibbs free energy of the given reaction is -40 kJ

<u>Explanation:</u>

The given chemical equation follows:

N_2(g)+3H_2(g)\rightarrow 2NH_3(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(NH_3(g))})]-[(1\times \Delta G^o_f_{(N_2(g))})+(3\times \Delta G^o_f_{(H_2(g))})]

We are given:

\Delta G^o_f_{(NH_3(l))}=-16.45kJ/mol\\\Delta G^o_f_{(H_2(g))}=0kJ/mol\\\Delta G^o_f_{(N_2(g))}=0kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(2\times (-16.45))]-[(1\times (0))+(3\times (0))]\\\\\Delta G^o_{rxn}=-32.9kJ/mol

The equation used to Gibbs free energy of the reaction follows:

\Delta G=\Delta G^o+RT\ln Q_{eq}

where,

\Delta G = free energy of the reaction

\Delta G^o = standard Gibbs free energy = -32.9 kJ/mol = -32900 J/mol  (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314J/K mol

T = Temperature = 25^oC=[273+25]K=298K

Q_{eq} = Ratio of concentration of products and reactants at any time = \frac{(p_{NH_3})^2}{(p_{H_2})^3\times p_{N_2}}

p_{NH_3}=2.11atm

p_{N_2}=7.92atm

p_{H_2}=2.02atm

Putting values in above equation, we get:

\Delta G=-32900J/mol+(8.314J/K.mol\times 298K\times \ln (\frac{(2.11)^2}{(2.02)^3\times 7.92}))\\\\\Delta G=-39553.04J/mol=--39.55kJ=-40kJ

Hence, the Gibbs free energy of the given reaction is -40 kJ

7 0
3 years ago
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