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Metals - Uses of metals and alloys

Grade 12IGCSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Alloys are mixtures of a metal with one or more other elements (usually other metals or carbon). For example, Steel is an alloy of Iron (FeFe) and Carbon (CC).

In a pure metal, atoms are of the same size and arranged in regular layers. These layers can slide over each other easily, making pure metals relatively soft and malleable.

In an alloy, the introduction of different-sized atoms (e.g., CC atoms in FeFe) disrupts the regular lattice arrangement. This prevents the layers from sliding over each other easily, making alloys harder and stronger than pure metals.

Aluminum (AlAl) is widely used in the manufacture of aircraft because of its low density and its ability to resist corrosion due to a thin, tough layer of aluminum oxide (Al2O3Al_2O_3) on its surface.

Copper (CuCu) is used for electrical wiring because it has very high electrical conductivity and is ductile (can be drawn into wires). It is also used in cooking utensils due to its high thermal conductivity.

Stainless steel is an alloy of iron (FeFe) containing chromium (CrCr) and nickel (NiNi). It is used for cutlery and chemical plants because the chromium forms a passive oxide layer that prevents rusting.

Zinc (ZnZn) is used for galvanizing iron (FeFe). This provides sacrificial protection because ZnZn is more reactive than FeFe and will oxidize preferentially according to the half-equation: ZnightarrowZn2++2eZn ightarrow Zn^{2+} + 2e^-.

📐Formulae

Density(ρ)=mass(m)volume(V)\text{Density} (\rho) = \frac{\text{mass} (m)}{\text{volume} (V)}

4Al(s)+3O2(g)2Al2O3(s)4Al(s) + 3O_2(g) \rightarrow 2Al_2O_3(s) (Formation of protective oxide layer)

FeFe2++2eFe \rightarrow Fe^{2+} + 2e^- (Oxidation of Iron during rusting)

ZnZn2++2eZn \rightarrow Zn^{2+} + 2e^- (Sacrificial protection by Zinc)

💡Examples

Problem 1:

Explain why brass, an alloy of copper (CuCu) and zinc (ZnZn), is harder than pure copper (CuCu).

Solution:

Pure CuCu consists of atoms of identical size arranged in a regular lattice. When a force is applied, these layers slide over each other easily. In brass, the ZnZn atoms have a different atomic radius than CuCu atoms. These larger/smaller atoms disrupt the regular layers, making it more difficult for the layers to slide, resulting in a harder material.

Explanation:

The hardness of an alloy is a direct result of the structural disruption caused by atoms of different sizes in the metallic lattice.

Problem 2:

Calculate the mass of an aluminum (AlAl) component used in an engine if its volume is 500extcm3500 ext{ cm}^3 and the density of the aluminum alloy is 2.7extg/cm32.7 ext{ g/cm}^3.

Solution:

m=ρ×Vm = \rho \times V m=2.7extg/cm3×500extcm3=1350extgm = 2.7 ext{ g/cm}^3 \times 500 ext{ cm}^3 = 1350 ext{ g}

Explanation:

Using the density formula ρ=mV\rho = \frac{m}{V}, we rearrange for mass to find the physical weight of the alloy part.

Uses of metals and alloys - Revision Notes & Key Formulas | IGCSE Grade 12 Chemistry