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gayaneshka [121]
3 years ago
14

Any advice for a GCSE chemistry student sitting first chemistry paper?

Chemistry
1 answer:
julia-pushkina [17]3 years ago
8 0

Answer:

when you go to gtake exam.be quite and don't make noise..

Explanation:

don't be scared

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what is the pOH of a with a pH of 10 14

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study the diagram and answer the following questions. a. Name A and B. b. Which element is it? Write its name and B symbol. p*=
Oksana_A [137]

Ai. A is called nucleus

Aii. B is called electron shell

B. The name of the element is Neon and the symbol of the element is Ne.

Ci. The atomic mass of the element is 20.

Cii. The atomic number of the element is 10

Di. The element belongs to group 18.

Dii. The element belongs to period 2.

E. The element has a valency of 8

A. Determination of the name of A and B

In the diagram given above, A is called the nucleus because it contains the neutrons and the protons of the atom.

B is called the electron shell because it contains the electrons of the atoms

B. Determination of the name and symbol of the element.

From the diagram given above,

The element has 10 protons. This simply means that the atomic number of the element is 10.

Comparing the atomic number of the element with those in the periodic table, the element is Neon with a symbol of Ne.

C. Determination of the atomic mass and atomic number.

Ci. Determination of the mass number

Proton = 10

Neutron = 10

<h3>Mass number =?</h3>

Mass number = Proton + Neutron

Mass number = 10 + 10

<h3>Mass number = 20</h3>

Thus, the atomic mass of the element is 20.

Cii. Determination of the atomic number.

The atomic number of an element is simply the number of protons in the atom of the element.

Proton = 10

Therefore,

Atomic number = proton = 10

Thus, the atomic number of the element is 10

D. Determination of the group and period.

Di. The group to which an element belongs to can be obtained by simply calculating the number of electrons in the outermost shell of the atom.

From the diagram given above, the outermost shell has 8 electrons. This suggest that the element has completely filled outermost shell.

Therefore, the element belongs to group 18 (i.e the noble gas)

Dii. The number of electron shell talks about the period to which and element belongs to.

From the diagram given above, the element has 2 electron shells.

Therefore, the element belongs to period 2

E. Determination of the valence electron(s)

The valence electron(s) is the number of electrons in the outermost shell of the atom.

From the diagram given above,

The outermost shell has 8 electrons.

Therefore, the element has 8 valence electrons

Learn more: brainly.com/question/5352570

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2 years ago
Exactly 2.00 g of an ester A containing only C, H, and O was saponified with 15.00 mL of a 1.00 M NaOH solution. Following the s
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The sum of IE₁ through IE₄ for Group 4A(14) elements shows a decrease from C to Si, a slight increase from Si to Ge, a decrease
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The correct answer is IE decreases down the group.

On moving down the group, the size of the element increases due to the increase in the number of shells. The element with smallest size has less the attraction of the nucleus on the valance electron. It needs more energy to remove an electron from its valance shell. Hence, the IE decreases down the group.

Why does ionization energy increase down the group but decreases going across a period?

Because there are more protons with time, the ionization energy rises. As a result, there will be more attraction because the nuclear charge has increased.

Even if there is stronger attraction, one should be aware that the shielding effect and distance from the nucleus remain largely constant. The same primary quantum shell contains all of the valence electrons, which explains this.

Therefore, while distance from the nucleus and the shielding effect stay fairly constant, an increase in nuclear charge causes an increase in attraction and increases the energy required to remove an electron.

The ionization energy drops with each group. This is because the outside electrons acting as a shield or screen for the nucleus make the attraction between them weaker and make it easier for them to be withdrawn.

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