Aluminium Foil Container for Food Packaging

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Loading Port:
Shanghai
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TT OR LC
Min Order Qty:
5 m.t
Supply Capability:
5000 m.t/month

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Product Description:

Aluminum foil, an eco-friendly & non-toxic material. The most effective use of the aluminium foil is as a packing material. Mainly used for household purpose, such as storing, packaging, roasting, freezing, heating in food factory, hotel, restaurant, airline etc.

  

Specifications:

Top Out:160x110mm

In Out:145x95mm

Bottom:125x75mm

Height:35mm

 

Applications:

Banquet,Cuisine,Roasting,Freezing,Baking,Storage,BBQ.

 

Advantages:

Environmental protection,recyclable,convenient,safe,healthy,no peculiar smell

Aluminium Foil Container for Food Packaging

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Q:Into which compound does aluminium nitrate decompose?
l would say you have mostly aluminum oxide as a product which probably contains some Nitrogen dioxide gas, a second product of the decomposition of aluminum nitrate. If I were you I would not heat nitrate salts as they can explode upon heating. they also need to be kept away from any kind of reducing agent such as powdered metals, or any fuels, reactions can be violent.
Q:An aluminum clock pendulum?
A for a swinging pendulum period T =2pi*√(L/g), where L is length of the pendulum, g=9.8m/s^2; metals shrink when cooled, so period will decrease, number of ticks per hour will increase, the clock will hurry up; B♣ period at t1°=20C° is T1=2pi*√(L1/g); ♣ period at t2°=-5C° is T2=2pi*√(L2/g); ♣ the law of linear extension says L2/L1= 1+s*Δt°, where s=23.1·10-6 (1/К°) is specific linear extension of aluminum, Δt°=t2°-t1°; ♦ thus T2/T1 =√(L2/L1) = √(1+s*Δt°) =f1/f2, where frequency f1= 1 Hz at 20C° or 3600 ticks per hour, hence frequency at -5C° is f2= f1/√(1+s*Δt°); therefore instead of 3600 ticks per hour the clock will do 3600/√(1+s*Δt°) ticks per hour; thus it will gain 3600*(1/√(1+s*Δt°) –1) = = 3600*(1/√(1 -23.1·10-6 *25°) –1) = 1.04 s/hour;
Q:what's the temperature tolerance of aluminum sheets?
aluminum is a kind of silver white metal,melting point:660.4℃, boiling point: 2467℃,density: 2.70 g/cm³,very light, about 1/4 of iron's density.it has low hardness and good ductility, which makes it suitable for being pulled into threadlets or pressed into aluminum foils, and the later is usually used for packaging candy and cigarette. it has good electrical and temperature conductivity. it's used for the manufacture of electric wire and cable in electric power industry, and for the manufacture of cooker in daily life.combined with the magnesium , copper,zinc,tin,manganese,chromium,zirconium,silicon and other elments,it can formulate various alloys that is broadly used for the manufacture of airplane,car,ship, materials of daily living equipment and the doors and windows of construction industry.aluminum is one of the best reflectors of heat and light, so it is used as thermal insulation material and the manufacture of reflecting mirror in the reflecting telescope.
Q:High quality Aluminum or Mediocre Carbon frames?
Personally I would go with the high quality Aluminum because it would be stiffer, but not nesaccarily lighter. A mediocre carbon frame are known to not take stress well and can break more easily than a high quality Aluminum. I would only go with Carbon if it was high quality.
Q:3mm heavy aluminum plate, multiple cubic meters per cubic meter?
Long by wide by high density is multiplied by the weight. The density of aluminum is different according to the different metal content, one to eight series aluminum density is between 2.71-2.75.
Q:Can aluminum plate be welded?
Q:Which statement describes a chemical property of aluminum?
The answer is choice 2. Why? It's the only one that describes a chemical interaction. The other three describe physical properties of aluminum, that is properties that do not involve a change in the chemical identity of aluminum metal. In choices 1, 3, and 4, the aluminum metal remains as aluminum metal. However, in choice 2, the metal reacts and becomes aluminum ion.
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
Q:How many atoms in cube of aluminum?
2.8cm * 2.8cm * 2.8cm= 21.95 cm^3 is volume of Aluminum present Density of Al = 2.7g/cm^3 Convert volume of aluminum into grams of Aluminum, make sure your units cancel 21.95 cm^3 * (2.7g/1 cm^3) = 59.26 g Al Next, find how many moles of Al you have. MW Al = 27g/mol 59.26/27 = 2.19 moles Al Use Avogadro's number to find number of atoms of Al - 2.19 moles Al * (6.022x10^23 atoms/1 mol Al) = 1.32x10^24 atoms Al
Q:Aluminium oxide drill bits?
It acts a lubricant.

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