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Pie
3 years ago
12

Why do the nuclides that are used as radiotracers have short half-lives?

Chemistry
1 answer:
Flauer [41]3 years ago
4 0
Well Some radioisotopes used in nuclear medicine have short half-lives, which means they decay quickly and are suitable for diagnostic purposes; others with longer half-lives take more time to decay, which makes them suitable for therapeutic purposes
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Where on the periodic table are the transition metals located​
Allisa [31]

Answer:

The lanthanides and the actinides at the bottom of the table are sometimes known as the inner transition metals because they have atomic numbers that fall between the first and second elements in the last two rows of the transition metals

6 0
3 years ago
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Which statement is true for a solution when its concentration of hydroxide ions increases? It becomes more acidic. It becomes le
makvit [3.9K]

Answer:

Its pH value increases.

Explanation:

pH is the measure of alkalinity or acidity of a compound.

pH = - log [H+]

and pH + pOH = 14

where pOH is the measure of basicity of a solution, given by -log[OH-]

As a solution gets more basic that is higher [OH-], the pH increases, and on the other hand, as the pH of a solution decreases by one pH unit, the concentration of H+ increases by ten times.

5 0
3 years ago
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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?

<em>In English:</em>

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 %

<em>In English:</em>

<em>the percent yield for the given amount of salicylic acid is </em><em>84.99%</em>

<em></em>

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%

<em>In English:</em>

<em>The balanced chemical eqaution for the reaction can be written as:</em>

<em>C₇H₆O₃ + C₄H₆O₃    →    C₉H₈O₄ + HC₂H₃O₂</em>

<em>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. </em>

<em>So; the formula for calculating the percentage yield </em>\mathbf {= \frac{actual \ yield }{theoretical \ yield } *100 }<em>  </em>

<em />

<em>The theoretical yield is determined as follows:</em>

<em>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</em>

<em />

<em>Percentage yield </em>\mathbf {= \frac{55.45 }{65.24 } *100 }<em />

<em>Percentage yield = 84.99%</em>

<em />

<em>Thus, the percent yield for the given amount of salicylic acid is </em><em>84.99%</em>

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3 years ago
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"Compounds A and B react to give a single product, C. Write the rate law for each of the following cases and determine the units
olasank [31]

Answer:

Part a: <em>Units of k is </em>M^{-2}s^{-1}<em> where reaction is first order in A and second order in B</em>

Part b: <em>Units of k is </em>M^{-1}s^{-1}<em> where reaction is first order in A and second order overall.</em>

Part c: <em>Units of k is </em>M^{-1}s^{-1}<em> where reaction is independent of the concentration of A and second order overall.</em>

Part d: <em>Units of k is </em>M^{-3}s^{-1}<em> where reaction reaction is second order in both A and B.</em>

Explanation:

As the reaction is given as

A+B \rightarrow C

where as the rate is given as

r=k[A]^x[B]^y

where x is the order wrt A and y is the order wrt B.

Part a:

x=1 and y=2 now the reaction rate equation is given as

r=k[A]^1[B]^2

Now the units are given as

r=k[A]^1[B]^2\\M/s =k[M]^1[M]^2\\M/s =k[M]^{1+2}\\M/s =k[M]^{3}\\M^{1-3}/s =k\\M^{-2}s^{-1} =k

The units of k is M^{-2}s^{-1}

Part b:

x=1 and o=2

x+y=o

1+y=2

y=2-1

y=1

Now the reaction rate equation is given as

r=k[A]^1[B]^1

Now the units are given as

r=k[A]^1[B]^1\\M/s =k[M]^1[M]^1\\M/s =k[M]^{1+1}\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k

The units of k is M^{-1}s^{-1}

Part c:

x=0 and o=2

x+y=o

0+y=2

y=2

y=2

Now the reaction rate equation is given as

r=k[A]^0[B]^2

Now the units are given as

r=k[B]^2\\M/s =k[M]^2\\M/s =k[M]^{2}\\M^{1-2}/s =k\\M^{-1}s^{-1} =k

The units of k is M^{-1}s^{-1}

Part d:

x=2 and y=2

Now the reaction rate equation is given as

r=k[A]^2[B]^2

Now the units are given as

r=k[A]^2[B]^2\\M/s =k[M]^2[M]^2\\M/s =k[M]^{2+2}\\M/s =k[M]^{4}\\M^{1-4}/s =k\\M^{-3}s^{-1} =k

The units of k is M^{-3}s^{-1}

7 0
3 years ago
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