Item specifice:
Grade: 
1000 Series,8011,5000 Series,6000 Series,3000 Series 
Surface Treatment: 
Mill Finish 
Shape: 
Square,Round,Flat 
Temper: 
OH112,Half Hard 
Application: 
Door & Window,Food 
Technique: 
D/C, C/C 
Thickness: 
0.235 
Width: 
62.5mm 
Packaging: 
Wooden pallet 
Product Description:
Aluminium Can Pulling Stock And Body Material
Alloy:aa3104，aa5182，aa5052
Temper:h19，h48
Thickness:0.260.29mm
Width:7001800mm( body stock),601800mm(end&tab stock)
Paint:ppg or valspar
Lubrication：henkel
Special specification is available on customer's requirement
Aluminu Can are dived to the can body and pull top, it can pack automatically with can filler to pack carbonated beverage like coloca, persi and hot filling juice for long shelf life. Now we have sizes as below:
Slick 330ml/310ml
270ml/200ml
Slim 250ml
Stubby 250ml
Size  62.5*37mm 
Diameter  62.5 mm 
Material  aluminum 
Thickness  0.25 mm 
Coating  Epoxyphenolic lacquer 
Usage  Used for packing dried food, farm production, such as, milk powder, coffee powder, tea, seasoning, seeds, air freshener, etc. 
Printing  as per your requirement 
Main Products  202#(d=52mm),206#(d=56mm),209#(d=62.5mm),211#(d=65.4mm),300#(d=72.9mm),307#(d=83.3mm),309#(d=90mm),401#(d= 99mm),502#(d=126.4mm),603#(d=153mm),701#(d= 177mm). 
Remarks  We can make new mould as per your requirement. 
 Q:why is aluminium used instead of copper for high voltage electrical cables?
 All of the earlier answers are correct and the first one is very good but nobody mentioned the fact that aluminium does not have much tensile strength, and that as a result of this, some high voltage aluminium cables have a steel cable in the centre to give them the required tensile strength. Aluminium busbars are very popular in low voltage work because they do not need to be flexible or possess much tensile strength but yet provide the benefits of being light and cheap.
 Q:How would you write the formula for aluminum sulfate and calcium chloride?
 AlSO4 CaCl
 Q:What kind of pattern aluminium plate has?
 Aluminum alloy decorative plate: aluminum alloy perforated sound absorption boardThe aluminum alloy perforated plate is made of various aluminum alloy plates and punched by machine. Pass a round hole, square hole, round hole, square hole, triangle hole according to the needs of small hole combination. This is a new product that reduces noise and has an ornamental effect. Aluminum Alloy perforated plate material is light, high temperature resistance, corrosion resistance, fire resistance, moistureproof, shock proof, good chemical stability, beautiful appearance, elegant color, threedimensional sense of strong, good decorative effect, and is convenient to assemble and can be used in hotels, restaurants, cinemas, studios and other public buildings.
 Q:Could you tell me the difference between the 6063 aluminum plate and the 5052 aluminum plate?
 6063 and 5052 alike: aluminum alloys with Mg Mg as the main alloying elementThe difference is that, in addition to the magnesium Mg element, there is also a silicon Si element in the 6063 aluminum plate. The strength is slightly better than that of 5052
 Q:What glue does stainless steel board and aluminium board glue well?
 Stainless steel plate and aluminum plate usually use acrylic structure glue.Acrylic structural adhesive twocomponent acrylate modified by synthesis and curing agent structural adhesive composition, A, B two agents are 100% solid components; fast curing, high strength, about 5 minutes to 30 minutes after the initial fixation strength up to 80%, 24 hours after the maximum strength; excellent resistance to acid and alkali, water, oil medium, high temperature resistance, better aging properties of metal items between strong adhesion, resistance strength is better than that of epoxy resin, excellent resistance to thermal shock, working at 30~150 DEG C; compliance with international environmental standards, has passed the European ROSH standard.
 Q:Analysis of a MagnesiumAluminum 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[(0.250a)/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:What oil does the aluminum plate stretch with?
 There are professional tensile oil, you can ask to pull the rice cooker liner manufacturers know, if not a lot of production, you can use peanut oil first try, the effect barely enough!
 Q:12ft aluminum boat weight limit?
 there okorder.com/ Weight Capacity 335 lb / 151kg Max. Person Capacity 2 Max. HP Capacity 3 hp / 2kw Max. Person Weight Capacity 295 lb / 133kg Hull Gauge .050 / 1.27mm Package Length 18' / 5.49m Seats 3 the 6.5hp ob will push the capacity weight of the boat with out a problem, but may not be able to bring the boat onto a plane. hope this helps
 Q:What is aluminum foil made of?
 Thermite or meth?
 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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