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1 year ago

What is the yield of uranium from 2.50 kg U3O8?

Chemistry

1 answer:

mr_godi [17]1 year ago

8 0

Answer: Thus the yield of uranium from 2.50 kg $U_3O_8$ is 2.12 kg

Explanation:

According to avogadro's law, 1 mole of every substance weighs equal to molecular mass and contains avogadro's number $(6.023\times 10^{23})$ of particles.

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

moles of $U_3O_8=\frac{2.50\times 1000g}{842g/mol}=2.97mol$ (1kg=1000g)

As 1 mole of $U_3O_8$ contains = 3 moles of U

2.97 mole of $U_3O_8$ contains = $\frac{3}{1}\times 2.97=8.91moles$ moles of U

Mass of Uranium= $moles\times {\text {Molar mass}}=8.91mol\times 238g/mol=2120g=2.12kg$

( 1kg=1000g)

Thus the yield of uranium from 2.50 kg $U_3O_8$ is 2.12 kg

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1 year ago

Consider a general reaction A ( aq ) enzyme ⇌ B ( aq ) A(aq)⇌enzymeB(aq) The Δ G ° ′ ΔG°′ of the reaction is − 5.980 kJ ⋅ mol −

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Answer : The value of $K_{eq}$ is, 11.2

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Explanation :

The relation between the equilibrium constant and standard Gibbs free energy is:

$\Delta G^o=-RT\times \ln K_{eq}$

where,

$\Delta G^o$ = standard Gibbs free energy = -5.980 kJ/mol = -5980 J/mol

R = gas constant = 8.314 J/K.mol

T = temperature = $25^oC=273+25=298K$

$K_{eq}$ = equilibrium constant = ?

Now put all the given values in the above formula, we get:

$\Delta G^o=-RT\times \ln K_{eq}$

$-5980J/mol=-(8.314J/K.mol)\times (298K)\times \ln K_{eq}$

$K_{eq}=11.2$

Thus, the value of $K_{eq}$ is, 11.2

Now we have to calculate the $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

The given reaction is:

$A(aq)\rightleftharpoons B(aq)$

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$

$\Delta G_{rxn}=\Delta G^o+RT\ln \frac{[B]}{[A]}$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -30.5 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = $37.0^oC=273+37.0=310K$

Q = reaction quotient

[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})$

$\Delta G_{rxn}=-9035.75J/mol=-9.04kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -9.04 kJ/mol

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2 years ago

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zmey [24]

solution:

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Volume of water is V= 10 ml

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Total amount of caffeine that can be unextracted is given by

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