Non-Stick Aluminum Circles Disc Disk For Deep Draw Press Cookware Product

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Qingdao
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TT OR LC
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2 m.t.
Supply Capability:
5000 m.t./month

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Item specifice:

Grade: 1000 Series Surface Treatment: Mill Finish Shape: Round,Flat
Temper: O-H112,Soft Application: Kitchen Use Technique: DC/CC
Thickness: 0.05-4.0mm Width: 200mm---980mm Packaging: Wooden pallets
Application: Cookware Alloy: 1060/1050 Temper: O-H112
Surface Treatment: Mill finished Alloy Or Not: Is Alloy Diameter: 200-800mm

Product Description:

Aluminum Circles For Deep Draw Press Cookware Product 


Packaging & Delivery

Packaging Details:25kg/carton-boxes(35*32*19CM) &Fumigation wooden-pallets (110*110*15cm)
20GP can loading 22 Ton,total 21 pallets,20 CBM.
Delivery Detail:within 25 days

Specifications

1).Excellent Surface Quality for lighting reflectors.
2).Excellent deep drawing and spinning quality.
3).Anodizing Quality

 Aluminium circle/Aluminium discs/disks for cooking utensils Cookware

  • Aluminum disc Surface Finish: Bright & smooth surface, without flow lines, slightly oiled to avoid White rusting.Smooth,Brightly ,No Zone of Fracture ,No Scratched ,No oil Sludge

  • We enjoy a good reputation among our customers for our aluminum circle with high quality and competitive price, win high appreciation from our clients with the excellent workman-ship and professional ability. The quality matches with ISO 9001 quality management system.

  • It is equipped with 5 hot tandem rolling line, 4 cold mill production lines, 4 annealing surface machines and a complete finishing equipment.



  • Specification:


  • Aluminum disc chemical Properties(WT.%)


  • Alloy

    Si

    Fe

    Cu

    Mn

    Mg

    Cr

    Ni

    Zn

    Ca

    V

    Ti

    Other

    Min.A1

    1050

    0.25

    0.4

    0.05

    0.05

    0.05

    -

    -

    0.05

    -

    0.05

    0.03

    0.03

    99.5

    1070

    0.25

    0.25

    0.04

    0.03

    0.03

    -

    -

    0.04

    -

    0.05

    0.03

    0.03

    99.7

    3003

    0.6

    0.7

    0.05-0.2  

    1.0-1.5 

    -

    -

    -

    0.1

    -

    -

    -

    0.15

    96.95-96.75 



Non-Stick Aluminum Circles Disc Disk For Deep Draw Press Cookware Product


Non-Stick Aluminum Circles Disc Disk For Deep Draw Press Cookware Product

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Any element that allows the transfer of electrical flow to move across it usually has the characteristic of having free movement electrons in the outer shell. This is why metals are particularly good at electrical conduction. The best metal for transferring electrical charge is silver followed by copper and then aluminum. Much electrical wire is made from pure aluminum because it is cheaper than copper or silver and the resistance to flow is only slightly lower than copper. The electrons in the outer shells are still bound to the aluminum nucleus but they are free to accept a charge from neighboring aluminum atoms and transfer that charge to the next aluminum in the electrical flow. This is basically the definition of electrical resistance. Copper performs this transfer more efficiently than aluminum so it's electrical resistivity is less than aluminums. But in both base metals, the outer electrons never leave the atom, they simply transfer their charge to the next atom in line
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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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Keyboard would be more useful and practical. Game would be more fun.

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