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zaharov [31]
3 years ago
8

Explain why materials with metallic lattice structures can be used to make wires and connections that conduct electricity in ele

ctrical circuits.
Chemistry
1 answer:
kogti [31]3 years ago
7 0
<span>All metals have similar properties BUT, there can be wide variations in melting point, boiling point, density, electrical conductivity and physical strength.<span>To explain the physical properties of metals like iron or sodium we need a more sophisticated picture than a simple particle model of atoms all lined up in close packed rows and layers, though this picture is correctly described as another example of a giant lattice held together by metallic bonding.</span><span>A giant metallic lattice – the <span>crystal lattice of metals consists of ions (NOT atoms) </span>surrounded by a 'sea of electrons' that form the giant lattice (2D diagram above right).</span><span>The outer electrons (–) from the original metal atoms are free to move around between the positive metal ions formed (+).</span><span>These 'free' or 'delocalised' electrons from the outer shell of the metal atoms are the 'electronic glue' holding the particles together.</span><span>There is a strong electrical force of attraction between these <span>free electrons </span>(mobile electrons or 'sea' of delocalised electrons)<span> (–)</span> and the 'immobile' positive metal ions (+) that form the giant lattice and this is the metallic bond. The attractive force acts in all directions.</span><span>Metallic bonding is not directional like covalent bonding, it is like ionic bonding in the sense that the force of attraction between the positive metal ions and the mobile electrons acts in every direction about the fixed (immobile) metal ions of the metal crystal lattice, but in ionic lattices none of the ions are mobile. a big difference between a metal bond and an ionic bond.</span><span>Metals can become weakened when repeatedly stressed and strained.<span><span>This can lead to faults developing in the metal structure called 'metal fatigue' or 'stress fractures'.</span><span>If the metal fatigue is significant it can lead to the collapse of a metal structure.</span></span></span></span>
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(For example, the relative atomic mass of \rm H is 1.008 means that the mass of one mole of \rm H\! atoms would be approximately 1.008\! grams on average.)

The question counted the number of \rm H_2SO_4 molecules without using any unit. Avogadro's Number N_{\rm A} helps convert the unit of that count to moles.

Each mole of \rm H_2SO_4 molecules includes exactly (1\; {\rm mol} \times N_\text{A}) \approx 6.02\times 10^{23} of these \rm H_2SO_4 \! molecules.

3.27 \times 10^{23} \rm H_2SO_4 molecules would correspond to \displaystyle n = \frac{N}{N_{\rm A}} \approx \frac{3.27 \times 10^{23}}{6.02 \times 10^{23}\; \rm mol^{-1}} \approx 0.541389\; \rm mol of such molecules.

(Keep more significant figures than required during intermediary steps.)

The formula mass of \rm H_2SO_4 gives the mass of each mole of \rm H_2SO_4\! molecules. The value of the formula mass could be calculated using the relative atomic mass of each element:

\begin{aligned}& M({\rm H_2SO_4}) \\ &= (2 \times 1.008 + 32.06 + 4 \times 15.999)\; \rm g \cdot mol^{-1} \\ &= 98.702\; \rm g \cdot mol^{-1}\end{aligned}.

Calculate the mass of approximately 0.541389\; \rm mol of \rm H_2SO_4:

\begin{aligned}m &= n \cdot M \\ &\approx 0.541389\; \rm mol \times 98.702\; \rm g \cdot mol^{-1}\\ &\approx 53.3\; \rm g\end{aligned}.

(Rounded to three significant figures.)

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