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m_a_m_a [10]
3 years ago
8

After 11.5 days, 12.5% of a sample of radon-222 that originally weighed 42g remains. What is the half-life of this isotope?

Chemistry
1 answer:
Bond [772]3 years ago
5 0
Half-life is defined as the amount of time it takes a given quantity to decrease to half of its initial value.  The equation to describe the decay is
Nt=N0(1/2) ^{t/t(1/2)}  where N0 is the initial quantity, Nt is the remaining quantity after time t, t1/2 is the half-time.  So work out the equation, t1/2 = t (-ln2)/ln(Nt/N0) = 11.5*(-ln2)/ln(12.5/100) = 3.83 days
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Write a balanced equation for its combustion reaction.
laila [671]
<h3>Answer:</h3>

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

<h3>Explanation:</h3>

The balanced chemical equation for the combustion of the hydrocarbon in question is;

C₅H₁₂O(l)+15/2O₂(g)→5CO₂(g)+6H₂O(l)

  • A balanced chemical equation is one in which the number of atoms of each element is equal on both sides of the equation.
  • In this case;
  • Reactant side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
  • Product side has; 5 carbon atoms, 12 hydrogen atoms and 16 Oxygen atoms
  • An equation is balanced by putting appropriate coefficients on reactants and products involved in the reaction.
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3 years ago
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zmey [24]
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3 years ago
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Se hace reaccionar 4,00 g de aluminio y 42,00 g de bromo, según la reacción: Al(s)+Br2(l)⟶AlBr3(s) Calcular las moles de AlBr3(s
NeX [460]

Answer:

0.145 moles de AlBr3.

Explanation:

¡Hola!

En este caso, al considerar la reacción química dada:

Al(s)+Br2(l)⟶AlBr3(s)

Es claro que primero debemos balancearla como se muestra a continuación:

2Al(s)+3Br2(l)⟶2AlBr3(s)

Así, calculamos las moles del producto AlBr3 por medio de las masas de ambos reactivos, con el fin de decidir el resultado correcto:

n_{AlBr_3}^{por\ Al}=4.00gAl*\frac{1molAl}{27gAl} *\frac{2molAlBr_3}{2molAl}=0.145mol AlBr_3\\\\n_{AlBr_3}^{por\ Br_2}=42.00gr*\frac{1molr}{160g Br_2} *\frac{2molAlBr_3}{3molBr_2}=0.175mol AlBr_3

Así, inferimos que el valor correcto es 0.145 moles de AlBr3, dado que viene del reactivo límite que es el aluminio.

¡Saludos!

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