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Nookie1986 [14]
4 years ago
15

When the nuclide iron-59 undergoes beta decay: The name of the product nuclide is . The symbol for the product nuclide is .

Chemistry
1 answer:
Travka [436]4 years ago
5 0

Answer:

Explanation:

⁵⁹₂₆Fe --------- ⁰₋₁e + ⁵⁹₂₇Co

Co- 59 is known as Cobalt

its symbol is ⁵⁹₂₇Co

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i have an unknown volume of gas at a pressure of 0.50 atm a temperature of 325 k. If i raise the pressure to 1.2 atm, decrease t
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Assuming that the number of mols are constant for both conditions:
\frac{P_1V_1}{T_1} =  \frac{P_2V_2}{T_2}
Now you plug in the given values. V_1 is the unknown. 
\frac{0.50 atm*V_1}{325K} = \frac{1.2 atm* 48 L}{230K}

Separate V_1
V_1= \frac{1.2 atm* 48 L * 325K }{230K*0.50 atm }
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Explain why elements and compounds are pure substances
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2 years ago
Se mezclan 20 gramos de agua (1cal/g°C) a 40 °C con 15 gramos de alcohol (0,58cal/g°C) a 30 °C. ¿Cuál ha sido la temperatura de
madreJ [45]

Answer:

T_{EQ}=37\°C

Explanation:

¡Hola!

En este caso, para los problemas de equilibrio térmico, consideramos que la energía liberada por la sustancia que inicialmente está caliente (agua), es absorbida por la sustancia que inicialmente está fría (alcohol); thus, we can write:

Q_{agua}=-Q_{alcohol}

La cual puede ser escrita en términos de masa, calor specifico y temperaturas:

m_{agua}C_{agua}(T_{EQ}-T_{agua})=-m_{alcohol}C_{alcohol}(T_{EQ}-T_{alcohol})

De este modo, al resolver para la temperature de equilibrio térmico, obtenemos la siguiente expresión:

T_{EQ}=\frac{m_{agua}C_{agua}T_{agua}+m_{alcohol}C_{alcohol}T_{alcohol}}{m_{agua}C_{agua}+m_{alcohol}C_{alcohol}}

Así, al reemplazar los valores en esta, obtenemos:

T_{EQ}=\frac{(20g)(1cal/g\°C)(40\°C)+(15g)(0.58cal/g\°C)(30\°C)}{(20g)*(1cal/g\°C)+(15g)(0.58cal/g\°C)} \\\\T_{EQ}=37\°C

¡Saludos!

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3 years ago
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