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3 years ago

The half-life of cobalt-60 is 5.3 years. after __________ years, 1/4 of the original amount of cobalt-60 will remain.

Chemistry

2 answers:

aleksandr82 [10.1K]3 years ago

7 0

Answer:

10.6

Explanation:APEX

cestrela7 [59]3 years ago

4 0

The number of years required for 1/4 cobalt-60 to remain after decay is calculated as follows

after one half life 1/2 of the original mass isotope remains

after another half life 1/4 mass of original mass remains

therefore if one half life is 5.3 years then the years required

= 2 x 5.3years = 10.6 years

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A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

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3 years ago

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emmainna [20.7K]

Molar mass of butane:-

4(12)+10=58u

Moles of Butane:-

100/58=1.7mol

#16

2mols of Butane produce 8mol CO_2
1mol of butane produces 4mol CO_2

Moles of CO_2

4(1.7)=6.8mol

Mass of CO_2:-

44(6.8)=299.2g

#17

2mols of butane need 13mol O_2
1mol of butane needs 6.5mol O_2

Moles of O_2

6.5(1.7)=11.05mol

Mass of O_2

11.05(32)=353.6g

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2 years ago

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bulgar [2K]

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Explanation:

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