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CaHeK987 [17]
3 years ago
13

The half-life of substance-X is 5 minutes. If you start with 20g of substance-X, how much substance-X is leftover after 5 minute

s? How about 10 minutes?
Chemistry
1 answer:
timurjin [86]3 years ago
4 0
<span>At 20g with half-life of 5 minutes there would be 10g leftover in 5 minutes. In 10 mins, there would be 5g leftover. Half-life is actually what it says, it refers to time it takes for half of the active elements, etc to break down at which time the potency of the "product" is half as strong. This term is used mainly with radioactive items or medicines.</span>
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+4

Explanation:

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Supongo que quiere preparar una solución al 10% de sulfato de magnesio en peso. El frasco del producto químico que se tiene indi
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Answer:

Se requerirán 14.57 gramos de MgSO₄·7H₂O, que se disolverían en 105.43 gramos de agua.

Explanation:

Si tenemos 120 gramos de una solución al 10% de sulfato de magnesio en peso, habrán en la solución (120*10/100) 12 gramos de sulfato de magnesio (MgSO₄).

Sin embargo, el reactivo que está disponible es heptahidratado (MgSO₄·7H₂O), por lo que hay que calcular <em>cuántos gramos de sulfato de magnesio heptahidratado contendrán 12 gramos de MgSO₄</em>.

<u>Calculamos las moles de 12 gramos de MgSO₄</u>, usando su masa molecular:

  • 12 g MgSO₄ ÷ 120.305 g/mol = 0.0997 mol MgSO₄.

<u>Después calculamos la masa de MgSO₄·7H₂O que contendrá 0.0997 mol MgSO₄</u>, usando la masa molecular de MgSO₄·7H₂O:

  • 0.0997 mol * 246.305 g/mol = 14.57 g MgSO₄·7H₂O

Para saber la cantidad de agua en la que se disolverá el reactivo, restamos la masa de soluto de la masa total de la solución:

  • 120 g - 14.57 g = 105.43 g

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3 years ago
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5 0
2 years ago
Rxn
givi [52]

Answer: The enthalpy of formation of SO_3 is  -396 kJ/mol

Explanation:

Calculating the enthalpy of formation of SO_3

The chemical equation for the combustion of propane follows:

2SO_2(g)+O_2(g)\rightarrow 2SO_3(g)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta H^o_f_{(SO_2(g))})+(1\times \Delta H^o_f_{(O_2(g))})]

We are given:

\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(SO_2(g))}=-297kJ/mol\\\Delta H^o_{rxn}=-198kJ

Putting values in above equation, we get:

-198=[(2\times \Delta H^o_f_{(SO_3(g))})]-[(2\times \Delta -297)+(1\times (0))]\\\\\Delta H^o_f_{(SO_3(g))}=-396kJ/mol

The enthalpy of formation of SO_3 is -396 kJ/mol

4 0
3 years ago
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