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Vladimir79 [104]
3 years ago
12

Which properties make a metal a good material to use for electrical wires? malleability and reactivity conductivity and ductilit

y ductility and malleability reactivity and conductivity
Chemistry
2 answers:
Makovka662 [10]3 years ago
6 0

Answer:

conductivity and ductility

Explanation:

Metals are connected by crystalline lattices, each atom being surrounded by 8 or 12 other atoms of the same metal element, thus having equal attractions in all directions. In addition, since metal atoms have only 1, 2, or 3 electrons in the last electron layer (and this layer is usually well away from the nucleus, and therefore attracts little electrons); The result is that the electrons escape easily and move freely through the lattice.

This lattice structure and chemical bonding result in a number of properties that are characteristic of metal atoms. Two strong characteristics of metals are the conductivity and ductility that make metals great materials for the production of electrical wires.

  • Conductivity: Metals are great conductors of electricity and, because of this property, are widely used in electrical wires. This property is explained by the fact that since metals have a "sea" of free or delocalized electrons, these electrons allow the rapid transition of electricity through the metal. When subjected to an external voltage, these free electrons go to the positive pole of the external source. This movement of electrons is what we call electric current.
  • Ductility and Meability: Malleability is the ability to shape metals into thin blades by hammering the heated metal or passing it through rolling cylinders; and ductility is the transformation of wires by passing the metal through holes under heating. These two properties result from the fact that metal atoms can "slip" over each other.
Schach [20]3 years ago
3 0
The metal properties that make up electrical wires are conductivity and ductility. Conductivity is important for free electrons to flow, hence electricity. Ductility is as well integral for strips to be formed, but with ample strength that does not break the material.
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Why do we see different phases of the moon.
nadya68 [22]

Answer:

The Moon itself does not generate light; it is lit up by the Sun. As the Moon orbits the Earth, the portion of illuminated Moon that we see changes – giving rise to the phases of the Moon. ... Sometimes the Earth, Moon and Sun are aligned such that the Earth is directly between the Sun and the Moon.

Explanation:

5 0
3 years ago
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How many kilowatt-hours of electricity are used to produce 4.50 kg of magnesium in the electrolysis of molten mgcl2 with an appl
ratelena [41]
First, we need to determine the half reaction of magnesium. It would be expressed as:

Mg2+ + 2e- = Mg

Given the mass of magnesium metal that is produced, we calculate the total charge of the electrolysis by the relations we can get from the half reaction. We do as follows:

4.50 kg Mg ( 1000 g / 1 kg ) ( 1 mol / 24.305 g ) ( 2 mol e- / 1 mol Mg ) ( 96500 C / 1 mol e- ) = 35733388.2 C

We are given the applied EMF in units of V. This value is equal to J/C. So, 5 V is equal to 5 J/C.

35733388.2 C (5 J/C) = 178666941 J
178666941 J ( 1 kW-h / 3.6x10^6 J ) = 49.63 kW-h
3 0
3 years ago
Sulfur reacts with oxygen and creates two compounds. Compound A contains 1.34 g of sulfur for every 0.86 g of oxygen. Compound B
spayn [35]

Answer:

The mass ratio of oxygen rounded to the nearest whole no. 3 : 2

Explanation:

According to Law of Multiple proportion when two elements combine to make two or more different compounds, the mass ratio of the two element in the first compound, when divided by the mass ratio of the second compound , form a simple whole number ratio.

Compound A contains 1.34 g of sulfur for every 0.86 g of oxygen

        \frac{1.34}{0.86}= 1.5

Compound B contains 11.63 g of sulfur for every 10.49 g of oxygen

      \frac{11.643}{10.49}= 1.0

Ratio of oxygen in each compound

   always put the larger number over the smaller number.

\frac{CompoundA}{Compound B}=\frac{1.5}{1.0}=\frac{3}{2}

3 0
3 years ago
If the atomic weight of nitrogen is 14.01, what is the mass of the nitrogen atoms in one mole of cadmium nitrate, Cd(NOÀ)
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The answer is 28.02.

The chemical formula for cadmium nitrate is Cd(NO₃)₂.

There are in total 2 nitrogen atoms in Cd(NO₃)₂. If <span>the atomic weight of nitrogen is 14.01</span>, the mass <span>of two nitrogen atoms in one mole of cadmium nitrate is 28.02:
2 </span>· 14.01 = 28.02
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3 years ago
Conclusion: Summarize the results of this experiment in one or two sentences. What
GaryK [48]

Answer:

The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic color the salt turns the flame of a Bunsen burner.

4 0
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