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Reil [10]
3 years ago
15

Sodium-24 has a half life how much sodium-24 will remain in an 18 g sample after 75 hours

Chemistry
1 answer:
frutty [35]3 years ago
3 0

Answer:

\boxed{\text{0.56 g}}

Explanation:

The half-life of Na-24 (15 h) is the time it takes for half of it to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.  

We can construct a table as follows:  

  No. of                 Fraction          Mass

<u>half-lives   t/da   Remaining   Remaining/g </u>

      0             0           1                      18

       1            15            ½                    9.0

      2           30            ¼                    4.5

      3           45            ⅛                    2.2

      4           60            ⅟₁₆                   1.1

      5           75            ⅟₃₂                   0.56

      6           90           ⅟₆₄                   0.28

We see that  \boxed{\textbf{0.56 g}} remain after five half-lives (75 h).

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Para formar bronce, se mezclan 150g de cobre a 1100°C y 35g de estaño a 560°C. Determine la temperatura final del sistema.
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Answer:

La temperatura final del sistema es 1029,346 °C.

Explanation:

Asumamos que el sistema conformado por el cobre y el estaño no tiene interacciones con sus alrededores. Por la Primera Ley de la Termodinámica, el cobre cede calor al estaño con tal de alcanzar el equilibrio térmico. El cobre se encuentra inicialmente en su punto de fusión, mientras que el estaño está por encima de ese punto, de modo que la transferencia de calor es esencialmente sensible:

m_{Cu}\cdot c_{Cu}\cdot (T-T_{Cu}) = m_{Sn}\cdot c_{Sn}\cdot (T_{Sn}-T)

(m_{Cu}\cdot c_{Cu} + m_{Sn}\cdot c_{Sn})\cdot T = m_{Sn}\cdot c_{Sn}\cdot T_{Sn} + m_{Cu}\cdot c_{Cu}\cdot T_{Cu}

T = \frac{m_{Sn}\cdot c_{Sn}\cdot T_{Sn}+m_{Cu}\cdot c_{Cu}\cdot T_{Cu}}{m_{Cu}\cdot c_{Cu}+m_{Sn}\cdot c_{Sn}} (1)

Donde:

m_{Sn} - Masa del estaño, en gramos.

m_{Cu} - Masa del cobre, en gramos.

c_{Sn} - Calor específico del estaño, en calorías por gramo-grados Celsius.

c_{Cu} - Calor específico del cobre, en calorías por gramo-grados Celsius.

T_{Sn} - Temperatura inicial del estaño, en grados Celsius.

T_{Cu} - Temperatura inicial del cobre, en grados Celsius.

Si sabemos que m_{Cu} = 150\,g, m_{Sn} = 35\,g, c_{Cu} = 0,093\,\frac{cal}{g\cdot ^{\circ}C}, c_{Sn} = 0,060\,\frac{cal}{g\cdot ^{\circ}C}, T_{Sn} = 560\,^{\circ}C y T_{Cu} = 1100\,^{\circ}C, entonces la temperatura final del sistema es:

T = \frac{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (560\,^{\circ}C)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)\cdot (1100\,^{\circ}C)}{(35\,g)\cdot \left(0,060\,\frac{cal}{g\cdot ^{\circ}C} \right)+(150\,g)\cdot \left(0,093\,\frac{cal}{g\cdot ^{\circ}C} \right)}

T = 1029,346\,^{\circ}C

La temperatura final del sistema es 1029,346 °C.

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