5083 Aluminum Armor Plate

Hot Products

Mill Finish Plain Aluminum with High Quality and Competitive Price

Aluminum Plain Sheets for Marine Manufactured in China

Aluminum Circle Aluminium Round Plate DC&CC

Plain aluminium sheet with competitive price

ASTM 6061T6 Aluminium Sheet / 6061 Aluminum Plate

Stucco Pattern Embossed Aluminum Checkered Anti-Slip Coated Sheet and Coil

Aluminum Thick Plate for shoes mold 6061

7075 T651/T6 Aluminium thick plates 8-150mm

Aluminum Circle Aluminum Round Aluminium

2023 New production High Quality 3003 Alloy Aluminium Circle and Aluminum Disc

FAQ

Yes, aluminum sheets can be used as heat shields. Aluminum has high thermal conductivity, which means it can effectively absorb and distribute heat, making it an excellent material for heat protection. It is commonly used in various applications such as automotive, aerospace, and industrial settings to shield sensitive components from high temperatures.

could aluminum sheet be attracted by magnet?

no, it couldn't.

Maybe you didn't know but in the year 2025 the earth's population will be up to 12 billion people, which is double what we have now. And i am doing a project on Aluminum, and I have to know if there is enough AL for this many people in the year 2025. So will there be enough AL in the world to support us all?

Aluminium is the third most abundant element on earth after oxygen and silicon. I am quite sure that we will not run out of aluminium in the future. Oxygen - or at least pure good quality oxygen - is probably a more worrying scenario.

Yes, aluminum sheets can be used as a substitute for steel in certain applications. Aluminum has a high strength-to-weight ratio, is corrosion-resistant, and has excellent thermal conductivity. This makes it suitable for various industries such as automotive, aerospace, and construction, where weight reduction, durability, and resistance to environmental factors are important. However, the choice between aluminum and steel ultimately depends on the specific requirements and constraints of the application.

Some common sizes of aluminum sheet include 4x8 feet (48x96 inches), 4x10 feet (48x120 inches), and 5x10 feet (60x120 inches).

A sample of aluminum and a sample of steel have superficially identical microstructures. Which would you expect to be stronger, and why?

If you define stronger to be a higher yield strength, then it's about dislocation movement. Dislocations form as a material begins to deform. As more deformation occurs, dislocations begin to move. Since steel is iron with impurities (namely carbon), these impurities impede dislocation movement. Aluminum, in pure form, does not have anything to impede dislocation movement except for the grain boundaries of the microstructure, which the steel also has if the microstructures are identical. If you define stronger as in ultimate tensile strength, then it becomes an issue of bond strength. Basically how tightly iron atoms bond to one another versus aluminum atoms. There are some quantum mechanical considerations at this point, but I won't get into those details.

Aluminum sheet can be used for insulation, but it is not as effective as other insulation materials, such as fiberglass or foam. Aluminum has a low thermal conductivity, meaning it is a poor conductor of heat. However, it does not have good insulating properties on its own because it does not provide much resistance to heat flow. In order to use aluminum sheet for insulation, it is commonly combined with other insulating materials to enhance its effectiveness. For example, aluminum foil is often used as a radiant barrier in buildings to reflect heat and reduce the amount of heat transferred through walls and roofs. Overall, while aluminum sheet can contribute to insulation, it is usually used in conjunction with other materials to provide better thermal insulation.

Calculate the mass in grams of Iodine (I2) that will react completely with 20.4 g of aluminum to form aluminum oxide (AlL3)

Aluminium oxide? Shouldn't it be Aluminium iodide? 1. Form a balanced equation 3I2 + 4Al - 2Al2I3 2. Number of moles of Al 20.4/(Mr of Al) = 20.4/27 = 0.7555mol 3. Number of moles of I2 0.7555/4 x 3 = 0.5666mol 4. Mass of I2 required = 0.5666 x Mr of I2 = 0.5666 x (127x2) Ans: 143.9164g