Extruded Aluminum Panel For Roofing Building Application

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Item specifice:

Grade: 1000 Series,3000 Series,4000 Series,5000 Series,6000 Series,7000 Series,2000 Series Surface Treatment: Coated,Embossed,Anodized,Polished,Mill Finish,Color Coated,Oxidized,Enameled Wire,Brushed,Printed,Composited,Holographic Impression,Sand Blasted,Powder Coating Shape: Angle,Square,T-Profile,Round,Flat,Rectangular,Oval,Hexagonal
Temper: T3-T8,O-H112,T351-T651,T351-T851,Soft,Half Hard,Hard Application: Liner & Wad,Decorations,Door & Window,Heat Sink,Transportation Tools,Glass Wall,Food,Kitchen Use,Pharmaceutical,Seal & Closure,Insulation Material,Label & Tag

Product Description:

Extruded Aluminum Panel For Roofing Building Application


Alloy NO.








1050 1060 

1100 3003 










5052 5083

5454 5754 

















Stick blue film→plastic film→waterproof paper→1~2 tons on a export 

standard pallet(corner protection) 


decoration:ceilings,walls,furniture,cabinets,elevators,signs,name plate,

transportation, cookware, household appliances:refrigerators,microwave

ovens,  machinery, mold making,aerospace and military aspects, auto,

PP cap , construction etc


Chemical Composite: GB/T 3190-2008, JIS, EN, ASTM

Mechanical Property: GB/T 3880.2-2012, JIS, EN, ASTM



Kind attention 

Specifications can be customized as the customer’s requirements


Extruded Aluminum Panel For Roofing Building Application


Extruded Aluminum Panel For Roofing Building Application


Q: Can you provide free samples?

A: Yes, free samples will be sent to you on freight at destination.


Q: Can I get your latest catalogue?

A:  Yes, it will be sent to you in no time.


Q: What are your payment terms?

A: We accept L/C, D/A, D/P, T/T, West Union, etc


Q: Can you provide free samples?
A: Yes, free samples will be sent to you on freight at destination.


Q: Can I get your latest catalogue?

A:  Yes, it will be sent to you in no time.

Q: What is the MOQ?
A:  5 tons.

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Q:Aluminum Mig Welding?
You MUST use 100% Ar for MIG welding aluminum. Molten Al is highly reactive. Al powder is the active ingredient in thermite, for example. It will react with CO2, stripping away the oxygen and creating a nasty mess. Also, you really need a MIG welder with twice as much current output. along the lines of 200-300 amps. 130 amps will not perform well even with a spool gun, and running the weak, prone-to-jamming .030 Al wire. You could try to do it that way, but I would not advise. (There is no way in hell you'd be able to cram wimpy .030 Al wire through a 6 foot traditional welding lead. It'll kink and jam after 5 seconds.) MIG welding aluminum REQUIRES using the high-current, high feed speed, spray transfer mode. Here are some typical parameters: 030 wire, 500 IPM, 22 volts. 120-130 amps. 035 wire, 500 IPM, 23 volts, 150 amps. 045 wire, 400 IPM, 24 volts, 200 amps. EDIT, the guy above me has the right idea. They use electricity to force the reaction between aluminum oxide (dissolved in cryolite) and carbon, producing molten Al and CO2 gas. Normally the reaction would tend to go the other way. Aluminum has a higher affinity for oxygen.
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:aluminum elements question?
Aluminum is element 13, so it has 13 protons. This also means that it normally has 13 electrons (provided it is not in an ionic state). Normally it has 14 neutrons, but may have less or more in various isotopes.
Q:Calculate Number of Aluminium ions ?
The formula for aluminium oxide is Al2O3 Its molecular weight is 2*27+3*16= 54+48 =102 Now, 102g of Al2O3 contains 2*6.023*10^23 number of Al ions so 1g contains (2*6.023*10^23)/102 number of Al ions so 0.051g Al2O3 contain (2*6.023*10^23)/102 *0.051 number of Al ions which is equal to 6.023*10^23*10^-3 = 6.023*10^20 number of Al ions(Answer)
Q:Aluminum standard 30*25*2500 to 1000 1 how to calculate the number of materials needed?. 2 how to calculate the cutting loss?
How many density units should be used in international units per kilogram of cubic meters, so that the unit of mass is kilogram? Weight = quality *9.81. The unit of cattle per kilogram is 5.4 kg of pure aluminum, and the weight is about 53 cows
Q:Will copper and an aluminum solution react? *BEST ANSWER POINTS!?
No. Copper metal will not reduce aluminum ions to aluminum metal. We can observe that copper is below aluminum in the activity series, which indicates that the Cu + Al3+ reaction is nonspontaneous. The other way round will work, sort-of. Aluminum metal will reduce copper ions. But it takes freshly prepared aluminum metal. Aluminum metal passivates, that is, it reacts with oxygen to form an Al2O3 layer on the surface of the metal. The presence of chloride ion will help provide a clean Al surface.
Q:How many grams of oxygen are needed to react with 3.00 moles of aluminum?
Well, aluminum oxide is Al2O3, right? So for every 2 aluminum atoms there are 3 oxygen atoms, So, if you have three moles of aluminum, you'll need 4 and a half moles of oxygen to react completely with the aluminum. One mole of oxygen atoms weighs 16 grams, so 4.5 moles weighs 72 grams, right? Now it could be that your teacher told you three point zero zero moles of aluminum to get you to use two significant figures, so if you've studied significant figures you may want to look up a more accurate number for the atomic weight of aluminum, and carry the calculation through to three significant figures. Otherwise, 72 is a fine answer.
Q:How to distinguish aluminum sheet from magnesium sheet?
put them into NaOH, the one generating bubble is aluminum sheet, because aluminum sheet can react with it, generating hydrogen. Or light them in the air, the burning one is magnesium sheet.
Q:When you mix sodium hydroxide with water then add aluminum hydrogen gas is given off, is this from the water?
Aluminium does not normally react with water because of a surface coating of the highly unreactive Al2O3 that forms on exposure to atmospheric oxygen. In the presence of strong base, however, this compound dissolves due to complexation by hydroxide, similarly to how silver chloride dissolves in ammonia. Al2O3(s) + 2OH-(aq) + 3H2O(l) ----- 2[Al(OH)4]-(aq) Once this has occured, aluminium metal, a very strong reducing agent, is exposed to water. 2Al(s) + 6H+(aq) ----- 2Al3+(aq) + 3H2(g) Even though the solution is alkaline, there is still a low concentration of H+ formed by the autoprotolytic dissociation of water. The Al3+ formed then reacts with hydroxide to produce more [Al(OH)4]-. Yes, the hydrogen gas is from the water.
Q:student combines 43.1 mL of a 0.279 M aluminum chloride solution with 14.6 mL of a 0.464 M aluminum sulfate (a?
You have to be able to determine the number of moles of Al+3 that each solution contributes to the final solution: 1. aluminum chloride: AlCl3 Multiply the volume (in L) by the molarity to determine the number of moles of each compound dissolved in solution. 0.0431 L x 0.279 M = 0.0120 mol AlCl3 in solution When the AlCl3 in placed in the water, it dissociates to form the following ions: AlCl3 -- Al+3 + 3Cl- So for every one mole of AlCl3 placed in the water, one mole of aluminum ions will dissociate. Therefore, since we have 0.0120 mole of AlCl3 in the solution, that means that the aluminum chloride will contribute 0.0120 Al+3 ions to the final solution. 2. Aluminum sulfate: Al2(SO4)3 Multiply the volume (in L) by the molarity to determine the number of moles of each compound dissolved in solution. 0.0146 L x 0.464 M = 0.00677 mol Al2(SO4)3 in solution When the Al2(SO4)3 in placed in the water, it dissociates to form the following ions: Al2(SO4)3 -- 2Al+3 + 3(SO4)-2 So for every one mole of Al2(SO4)3 placed in the water, two mole of aluminum ions will dissociate. Use the mole ratios of the dissociation reaction to determine the number of moles of Al+3 ions that the aluminum sulfate contributes to the solution. 0.00677 mol Al2(SO4)3 x (2 mol Al+3 ions / 1 mol Al2(SO4)3) = 0.01354 mol Al+3 ions Therefore, the total number of Al+3 ions in solution is the sum: 0.01354 + 0.0120 = 0.02554 mol Al+3 ions The last piece of information needed to determine the concentration of the final solution is the volume of the final solution. Since the two volumes were mixed, the volume of the final solution will be the sum of the two solutions. 0.0431 L + 0.0146 L = 0.0577 L Therefore, to calculate molarity: Molarity = moles of solute / liters of solution Molarity = 0.02554 mol / 0.0577 L = 0.443 M Al+3

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