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2 years ago

What technique is used to separate the Aluminum soda cans?

Chemistry

1 answer:

Diano4ka-milaya [45]2 years ago

8 0

Answer:

Commonly, Overband Magnets are used to separate steel cans and Eddy Current Separators are used to expel and separate aluminium cans; The Aluminium Cans are discarded into general waste, which is then processed.

Explanation:

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1. Some athletes have as little as 3.0% body fat. If such a person has a body mass of 65 kg, how many pounds of body fat does th

77julia77 [94]

1. 3.0% ----> 3.0 kg fat= 100 kg body weigh
also remember that 1 kg= 2.20 lbs

$65 kg \frac{3.0 kg}{100 kg} x \frac{2.20 lb}{1 kg} = 4.29 lbs$

2. 0.94 g/mL----> 0.94 grams= 1 mL
1 Liters= 1000 mL
1kg= 1000 grams

$3 Liters \frac{1000 mL}{1 L} x \frac{0.94 grams}{1 mL} x \frac{1 kg}{1000 g} x \frac{2.20 lbs}{1 kg} = 6.20 lbs$

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2 years ago

Read 2 more answers

The periodic table is useful because it is organized to show many patterns and trends. What property shows a decrease moving fro

Alex73 [517]

Atomic radius

Explanation:

A decrease in moving from the bottom of Group 3 to the top of the group shows an increase in trend from top to bottom.

The atomic radius of elements increases from top to bottom and decreases from the bottom to the top.

The elements in group 3 are B, Al , Ga, In and Tl

Atomic radius is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state of metals.
Downs a group from top to bottom, atomic radii increases progressively due to successive shells being added.
This compensates for the size reducing effect of the increased nuclear charge.

learn more:

Atomic radii brainly.com/question/5048216

#learnwithBrainly

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3 years ago

PbSO4 has a Ksp = 1.3 * 10-8 (mol/L)2.

Oduvanchick [21]

i. The dissolution of PbSO₄ in water entails its ionizing into its constituent ions:

$\mathrm{PbSO_{4}}(aq) \rightleftharpoons \mathrm{Pb^{2+}}(aq)+\mathrm{SO_4^{2-}}(aq).$

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ii. Given the dissolution of some substance

$xA{(s)} \rightleftharpoons yB{(aq)} + zC{(aq)}$ ,

the Ksp, or the solubility product constant, of the preceding equation takes the general form

$K_{sp} = [B]^y [C]^z$ .

The concentrations of pure solids (like substance A) and liquids are excluded from the equilibrium expression.

So, given our dissociation equation in question i., our Ksp expression would be written as:

$K_{sp} = \mathrm{[Pb^{2+}] [SO_4^{2-}]}$ .

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iii. Presumably, what we're being asked for here is the <em>molar </em>solubility of PbSO4 (at the standard 25 °C, as Ksp is temperature dependent). We have all the information needed to calculate the molar solubility. Since the Ksp tells us the ratio of equilibrium concentrations of PbSO4 in solution, we can consider either [Pb2+] or [SO4^2-] as equivalent to our molar solubility (since the concentration of either ion is the extent to which solid PbSO4 will dissociate or dissolve in water).

We know that Ksp = [Pb2+][SO4^2-], and we are given the value of the Ksp of for PbSO4 as 1.3 × 10⁻⁸. Since the molar ratio between the two ions are the same, we can use an equivalent variable to represent both:

$1.3 \times 10^{-8} = s \times s = s^2 \\s = \sqrt{1.3 \times 10^{-8}} = 1.14 \times 10^{-4} \text{ mol/L}.$

So, the molar solubility of PbSO4 is 1.1 × 10⁻⁴ mol/L. The answer is given to two significant figures since the Ksp is given to two significant figures.

8 0

2 years ago

Rank these species by their ability to act as an oxidizing agent. cu+ mg2+ fe2+

natita [175]

Explanation:

The species or elements which gain electrons and reduces itself are known as oxidizing agent or oxidant.

Ability of an element to act as an oxidizing agent depends on its electrode potential.

The electrode potential of $Cu^{+}$ is 0.52 V.

The electrode potential of $Fe^2+$ is -0.41 V.

The electrode potential of $Mg^{2+}$ is -2.38 V.

Greater is the value of electrode potential, stronger will be the oxidizing agent.

Therefore, rank of these species by their ability to act as an oxidizing agent are as follows.

$Cu^{+}$ > $Fe^2+$ > $Mg^{2+}$

7 0

2 years ago

Look at the equation for a displacement reaction below. What lost electrons in the reaction? Give the full name, not the symbols

garri49 [273]

Answer:

Cupric ions

Explanation:

In the single displacement reaction shown, the cupric ions lost two electrons.

Cu²⁺ + 2e⁻ → Cu

The replacement of a metallic ion in solution by a metal atom higher in the activity series than than the metal in solution falls into this category of reactions.

Since Zn higher in the series, it displacements the cupric ions.

5 0

2 years ago