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kvv77 [185]
3 years ago
11

How many half-lives must elapse until 84 % of a radioactive sample of atoms has decayed? -g?

Chemistry
1 answer:
Ivahew [28]3 years ago
4 0
Half-life is the time required for decay of 50% of radio-active nuclei.

Thus, when radio-active material crosses 1st half-life, 100/2 = 50% radio-active material is left and remaining 50% is elapsed.

When, when radio-active material crosses 2nd half-life, 50/2 = 25% radio-active material is left and remaining 75% is elapsed.

When radio-active material crosses 3rd half-life, 25/2 = 12.5% radio-active material is left and remaining 87.5% is elapsed.

Thus, 2 <span>half-lives must elapse until 84 % of a radioactive sample of atoms has decayed.</span>
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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!

3 0
2 years ago
Plz can anyone help me I'm a bit stuck in this one ​
trasher [3.6K]

Answer:

just see if i am not wrong

learning balancing in chemistry it take time

hope i am correct

8 0
2 years ago
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