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3 years ago

The half-life of I-37 is 8.07 days. If 25 grams are left after 40.35 days, how many grams were in the original sample?

Chemistry

1 answer:

andreev551 [17]3 years ago

3 0

Answer:

800 g

Explanation:

We can express the decay of I-37 using the formula:

Final Mass = Initial Mass * $0.5^{\frac{Time}{Half-Life} }$

We input the data:

25 g = Initial Mass * $0.5^{(\frac{40.35}{8.07})}$

And solve for Initial Mass:

25 g = Initial Mass * $0.5^{5}$

25 g = Initial Mass * 0.03125

Initial Mass = 800 g

Meaning that out 800 grams of I-37, only 25 will remain after 40.35 days.

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3 years ago

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Answer:

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a difference in water temperature is the answer

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6 0

3 years ago

La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácid

Tanya [424]

La aspirina se prepara haciendo reaccionar ácido salicílico con exceso de anhídrido etanoico. En un experimento, 50.05 g de ácido salicílico se convirtieron en 55.45 g de aspirina. ¿Cuál fue el porcentaje de rendimiento?

In English:

Aspirin is prepared by reacting salicylic acid with excess ethanoic anhydride. In one experiment, 50.05 g of salicylic acid was converted to 55.45 g of aspirin. What was the yield percentage?

Answer:

el rendimiento porcentual para la cantidad dada de ácido salicílico es 84.99 %

In English:

the percent yield for the given amount of salicylic acid is 84.99%

Explanation:

La ecuación química equilibrada para la reacción se puede escribir como:

C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + HC₂H₃O₂

Para la reacción mostrada arriba; El reactivo limitante de la reacción es el ácido salicílico. Ahora; calcular el porcentaje de rendimiento; se espera que primero determinemos el rendimiento teórico de la reacción.

Entonces; la fórmula para calcular el porcentaje de rendimiento: $\mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }$

El rendimiento teórico se determina de la siguiente manera:

50.05 g * 1 mol / 138.21 g / mol de C₇H₆O₃ * 1 mol de C₉H₈O₄ / 1 mol de C₇H₆O₃ * 180.157 g / mol de C₉H₈O₄ = 65.24 g de C₉H₈O₄

Porcentaje de rendimiento $\mathbf {= \frac{55.45 }{65.24 } *100 }$

Porcentaje de rendimiento = 84.99%

Por lo tanto, el porcentaje de rendimiento para la cantidad dada de ácido salicílico es 84.99%

In English:

The balanced chemical eqaution for the reaction can be written as:

C₇H₆O₃ + C₄H₆O₃ → C₉H₈O₄ + HC₂H₃O₂

For the reaction shown above; The limiting reactant from the reaction is salicylic acid. Now; to calculate the percentage yield ; we are expected to first determine the theoretical yield of the reaction.

So; the formula for calculating the percentage yield $\mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }$

The theoretical yield is determined as follows:

50.05 g * 1 mol/ 138.21 g/mol of C₇H₆O₃ * 1 mol of C₉H₈O₄/ 1 mol of C₇H₆O₃ * 180.157 g/mol of C₉H₈O₄ = 65.24 g of C₉H₈O₄ is produced

Percentage yield $\mathbf {= \frac{55.45 }{65.24 } *100 }$

Percentage yield = 84.99%

Thus, the percent yield for the given amount of salicylic acid is 84.99%

7 0

3 years ago

Rusting of iron is a very common chemical reaction. It results in one form from Fe reacting with oxygen gas to produce iron (III

Vlada [557]

Answer: The given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$

Explanation:

We are given:

Moles of iron = 12.0 moles

The chemical equation for the rusting of iron follows:

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For oxygen gas:

By Stoichiometry of the reaction:

4 moles of iron reacts with 3 moles of oxygen gas

So, 12.0 moles of iron will react with = $\frac{3}{4}\times 12.0=9.0mol$ of oxygen gas

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By Stoichiometry of the reaction:

4 moles of iron produces 2 moles of iron (III) oxide

So, 12.0 moles of iron will produce = $\frac{2}{4}\times 12.0=6.0mol$ of iron (III) oxide

Hence, the given amount of iron reacts with 9.0 moles of $O_2$ and produce 6.0 moles of $Fe_2O_3$

5 0

3 years ago