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3 years ago

How does an element's atomic mass compare to the mass number of its most abundant isotope?

2 answers:

8 0

Atomic mass is an average mass of an element whereas the mass number is a mass of a single atom of an element.

Explanation:

Atomic mass

It is the average weight of the element's atom.
it depends upon the atomic mass of isotopes and their abundances.
It is calculated by considering all the isotopes of an element and its abundance.
It has decimal as it deals with the average weight of an element

Mass number

It is the weight of nucleons of the atom.
it does not depend upon the atomic mass of isotopes and their abundances.
It is calculated separately for each atom of an element.
It is a simplified number that is no decimal is present.

So, from this, we can conclude that atomic mass is an average mass of an element whereas the mass number is a mass of a single atom of an element.

Learn more about atomic mass and mass number here:

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dybincka [34]3 years ago

3 0

Answer:

It's very similar to it

Explanation:

Because the atomic mass in this situation is an average so it represents all isotopes in this situation due to their abundance

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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

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Answer:

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

Explanation:

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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}$}$

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