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pochemuha
3 years ago
10

Match each section of the atomic structure

Chemistry
2 answers:
siniylev [52]3 years ago
6 0

Answer:

1e 2a 3d 4c 5d

Explanation:

1) Electrons

These particles spin around the nucleus of the atom, they are negatively charged. They have their orbits.

2) Protons

This another kind of subatomic particle, is located on the nucleus of the atom is positively  charged they constitute the atomic nuclei with the other main particle

3) Neutrons

They do not possess electric charge, all atoms but the Hydrogen have neutrons within. Neutrons and Protons form the Atomic Nuclei.

4) Electron Cloud

Even though, this illustration is transparent maybe a translucid one should be more appropriate. An Electron Cloud depicts an area and the electron is not depicted as a dot. So, again this is not the best picture of an Electron Cloud, the point is since we can't precisely say where the electron passes let's draw areas, instead of lines.

5) Nucleus

The mass of particles at the center whose constituent parts are mainly the protons and neutrons and other subatomic particles, not mentioned here as quarks, bosoms, etc.

Lesechka [4]3 years ago
4 0

<>"Atomic particles. Protons and neutrons are heavier than electrons and reside in the nucleus at the center of the atom. Electrons are extremely lightweight and exist in a cloud orbiting the nucleus. The electron cloud has a radius 10,000 times greater than the nucleus."<>

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2. The elements in this set are the group II alkaline earth metals, and they follow the same trend as the alkali metals. Of the elements here, beryllium (Be) would have the highest effective nuclear charge, and so it would be the least likely to lose its valence electrons. In fact, beryllium has a tendency not to lose (or gain) electrons, i.e., ionize, at all; it is unique among its congeners in that it tends to form covalent bonds.

3. While the alkali and alkaline earth metals would lose electrons to attain a noble gas configuration, the group VIIA halogens, as we have here, would need to gain a valence electron for an full octet. The trends in the group I and II elements are turned on their head for the halogens: The smaller the atomic number, the less shielding, and so the greater the pull by the nucleus to gain a valence electron. And as the atomic number increases (such as when you go down the group), the more shielding there is, the weaker the effective nuclear charge, and the lesser the tendency to gain a valence electron. Bromine (Br) has the largest atomic number among the halogens in this set, so an electron would feel the smallest pull from a bromine atom; bromine would thus be the least likely here to gain a valence electron.

4. The pattern for the elements in this set (the group VI chalcogens) generally follows that of the halogens. The greater the atomic number, the weaker the pull of the nucleus, and so the lesser the tendency to gain electrons. Tellurium (Te) has the highest atomic number among the elements in the set, and so it would be the least likely to gain electrons.
7 0
3 years ago
A slurry of flakes soybeans weighing a total of 100 kg contains 75 kg of inert solids and 25 kg of solution with 10 wt% oil and
lubasha [3.4K]

Answer:

the amounts and compositions of the overflow V1 and underflow L1 leaving the stage are 75kg and 125kg respectively.

Explanation:

Let state the given parameters;

Let A= solvent (hexane)

B= solid(inert soiid)

C= solvent(oil)

F_{solution} = mass of solvent + mass of oil (i.e A+C)

<u>Feed Phase:</u>

Total feed (i.e slurry of flakes soybeans)= 100kg

B= mass of solid =75 kg

F= mass of solvent + mass of oil (i.e A+C)

 = 25kg

Mass ratio of oil to solution Y_{F} =\frac{Mass C}{Mass (A+C)}

mass of oil (C) =25 × 0.1 wt = 2.5kg

mass of hexane  in feed = 25 ×  0.9 =22.5kg + 2.5 =25kg

therefore  Y_{F} = \frac{2.5}{25}

= 0.1

mass ratio of solid to solution Y_{A}  =  \frac{Mass A}{Mass (A+C)}=[tex]\frac{75}{25}

=3

<u>Solvent Phase:</u>

C= Mass of oil= 0(kg)

A= Mass of hexane = 100kg

mass of solutions = A+C = 0+100kg

solvent= 100kg

<u>Underflow:</u>

underflow = L₁ = (unknown) ???

L₁ = E₁ + B

the value of N for the outlet and underflow is 1.5 kg

i.e N₁ = \frac{mass B}{mass(A+C)}

solution in underflow E₁ = Mass (A+C)

<u>Overflow:</u>

Overflow = V₁ = (unknown) ???

solution in overflow V₁ = Mass (A+C)

This is because, B = 0 in overflow

Solid Balance: (since the solid is inert, then is said to be same in feed & underflow).

solid in feed = solid in underflow = 75

75=  E₁ × N₁

75 =  E₁ × 1.5

E₁ = 50kg

Underflow L₁ = E₁ × B

= 50 + 75

=125kg

The Overall Balance: Feed + Solvent = underflow + overflow

100 + 100 = 125 + V₁

V₁ = 75kg

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