Aluminum Circle Aluminium Round Aluminum Plate

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5 m.t.
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100 m.t./month

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Product Description:

1.Specification


ALUMINIUM CIRCLE


ALLOY:AA1***(AA1050,AA1070,AA1100etc)


      AA3***(AA3003 etc)


TEMPER:O,H14


THICKNESS:0.5mm-4mm


DIAGONAL:100mm-1200mm


STANDARD:GB/T 3880-2006


Special specification is available on customer's requirement


2.Description


Aluminium foil has a shiny side and a matte side. The shiny side is produced when the aluminium is rolled during the final pass. It is difficult to produce rollers with a gap fine enough to cope with the foil gauge, therefore, for the final pass, two sheets are rolled at the same time, doubling the thickness of the gauge at entry to the rollers. When the sheets are later separated, the inside surface is dull, and the outside surface is shiny. This difference in the finish has led to the perception that favouring a side has an effect when cooking. While many believe that the different properties keep heat out when wrapped with the shiny finish facing out, and keep heat in with the shiny finish facing inwards, the actual difference is imperceptible without instrumentation.


Thin sheets of aluminium are not very effective at attenuating low-frequency magnetic fields. The shielding effectiveness is dependent upon the skin depth. A field traveling through one skin depth will lose about 63 percent of its energy (it is attenuated to 1/e = 1/2.718... of its original energy). Thin shields also have internal reflections that reduce the shielding effectiveness. For effective shielding from a magnetic field, the shield should be several skin depths thick. Aluminium foil is about 1 mil (25 µm); a thickness of 10 mils (250 µm) (ten times thicker) offers less than 1 dB of shielding at 1 kHz, about 8 dB at 10 kHz, and about 25 dB at 100 kHz.


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We can supply the best quality and best service!


4.Pictures

Aluminum Circle Aluminium Round Aluminum Plate

Aluminum Circle Aluminium Round Aluminum Plate

Aluminum Circle Aluminium Round Aluminum Plate


5.FAQ


1)Can we supply samples?

Yes, we can!


2)What can you do if you have other quastions?


You can contact us freely at any time!


3)How about our payment term?

LC and TT and other.


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Q:Can you make aluminum oxide by electrolysis?
it is called hard anodization a lot of machine shops can do it - I think there is a certain soup to put it in The Al would be the anode
Q:What happens to the aluminum atoms in this reaction?
This reaction occurs because Al is above Cu in the activity series.Al is oxidized and the copperis reduced. 2 Al^0(s) + 3Cu2+(aq) -- 2 Al3+(aq) + 3 Cu^0(s)
Q:If aluminum, with an atomic weight of 27...?
aluminum oxide is Al2O3 , or 2 part Al to 3 part O. 27 grams Al2 *2 / 54 grams Al available = 16 grams of O * 3 / X grams of O solve for X: X = 48 grams of oxygen needed next guy is right....the formula for aluminum oxide is Al2O3..sorry...I readjusted the formula above to show the change
Q:Calculate the mass in grams of Iodine (I2) that will react completely with 20.4 g of aluminum?
actually the guy before me did it right but made a mistake at one point 0.07 mole is the # of moles of Al which needs to be changed to moles of product, and then back to find mole of iodide. once thats found then you can use the MW to change it to grams.
Q:Analysis of a Magnesium-Aluminum Alloy?
P_H2 = 0.950 atm (Dalton's Law of partial pressures) n=Pv/RT = (0.950 atm)(0.311 L) / (0.08206 (L*atm)/(mol*K))(302.15 K) n_H2 = 0.011915983 mol Balanced equations: Al + 3HCl -- 3/2H2 + AlCl3 Mg + 2HCl -- H2 + MgCl2 By these equations, we know that every mole of Al will give us 1.5 moles of H2, and every mole of Mg will give 1 mole of H2. We can therefore set up an equation for the mass of Al like this: *Let a = the mass of MAGNESIUM* Al = 0.250 g - a With this equation in mind, we can setup two equations solving for 'n' of each element by dividing by its molar mass and multiplying by the molar ratio: n_Mg = a / 24.30 (1:1 ratio, so we don't have to multiply) -- number of moles of H2 produced by the reaction of Mg (now written as n_H') = a / 24.30 n_Al = (0.250 g - a) / (26.98 g/mol) Because of the molar ratio shown above, we must multiply n_Al by 1.5 in order to get n_H2 produced by the reaction of aluminum, hereafter known as n_H2 Since we know the number of moles produced by the sum of the reactions, we can add these equations together and solve for n_H2. (**note that your value will be different because you have a different volume**) Set up the equation like this: n_H2' + n_H2 = n_H2 = 0.011915983 mol Sub in your individual equations for n_H2' and n_H2: (a/24.3) + 1.5[(0.250-a)/26.98] = 0.011915983 mol Rearrange and solve for a (mass of MAGNESIUM): (26.98a + 9.1125 - 36.45a) / (24.3)(26.98) = 0.011915983 0.011915983 = 9.47a a = 0.137298281 g Once you have your 'a' value, divide it by the total mass (0.250 g) and multiply by 100%. This gives you the percentage of Mg. (0.137298281 g / 0.250 g) * 100% = 54.9193 % Since you want ALUMINUM, you must subtract the percentage of Mg from 100. 100 - 54.9193 = 45.08% So, the mass percentage of aluminum is 45.08%. I hope this is helpful!
Q:how to distinguish the physical property of copper sheet from the one of aluminum sheet?
physical property: copper sheet: golden, heavy, aluminum sheet: white, light.
Q:Doesn't the acidity of soda drinks dissolve aluminium cans?
Because of the aluminum chemical properties. Aluminum Oxide (Al2O3) is an extremely hard and compact material, so when the metal is exposed, it gets a very thin coat of Al2O3 that has a 9 hardness (only diamond is harder), that protects the rest of the metal from the atmosphere. Also, you should question if the phosphoric acid actually attacks aluminum, that is because that acid is found in Pepsi and coke..., BUT, as far as i know cans have a little plastic coat inside..., so the liquid does not get in touch with the aluminum.
Q:Aluminum Foil's 'Dull' Side: Myth or Not?
Straight from Reynolds: Which side of Reynolds Wrap® Aluminum Foil should I use, the shiny or the dull side? Actually, it makes no difference which side of the aluminum foil you use—both sides do the same fine job of cooking, freezing and storing food. The difference in appearance between dull and shiny is due to the foil manufacturing process. In the final rolling step, two layers of foil are passed through the rolling mill at the same time. The side coming in contact with the mill's highly polished steel rollers becomes shiny. The other side, not coming in contact with the heavy rollers, comes out with a dull or matte finish. The exception is when using Reynolds Wrap® Release® Non-Stick Aluminum Foil. The non-stick coating is applied during manufacturing to the dull side of the foil. Always place the non-stick (dull) side toward the food.
Q:What masses of iron(III) oxide and aluminum must be used to produce 10.0 g iron?
i'm uncertain regardless of the incontrovertible fact that it must be relaxing, are you making thermite? i might guess which you would be able to artwork this out from the atomic weights of each, and taking the valencies under consideration - 2 aluminium atoms for 3 oxygen atoms. 2 iron atoms for 2 oxygen (a million:a million).
Q:Would Lava Melt an Aluminum Can?
Temperature of lava - 700 degrees Celsius. Melting point of aluminum - 660 degrees Celsius. I'll let you figure out if the can would melt...

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