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DIA [1.3K]
3 years ago
10

4. Write the electronic configuration of first 20 elements in the periodic table.

Chemistry
1 answer:
DIA [1.3K]3 years ago
7 0

Answer:

Name Atomic Number Electron Configuration Period 1 Hydrogen 1 1s1 Helium 2 1s2 Period 2 Lithium 3 1s2 2s1 Beryllium 4 1s2 2s2 Boron 5 1s2 2s22p1 Carbon 6 1s2 2s22p2 Nitrogen 7 1s2 2s22p3 Oxygen 8 1s2 2s22p4 Fluorine 9 1s2 2s22p5 Neon 10 1s2 2s22p6 Period 3 Sodium 11 1s2 2s22p63s1 Magnesium 12 1s2 2s22p63s2 Aluminum 13 

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D. all of these can produce nuclear energy
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3 years ago
A solution of sodium hydroxide was titrated against a solution of sulfuric acid. How many moles of sodium hydroxide would react
jeka57 [31]

Answer:

2 mole of Sodium hydroxide reacts with 1 mole of Sulfuric acid

Explanation:

Write down the equation in the beginning with reactants and products:

NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Now try to balance it. Try with Na first:

2NaOH + H₂SO₄ → Na₂SO₄ + H₂0

Na atoms are balanced. There are 6 Oxygen atoms on the right and 5 on the left. Balance by increasing the H₂O moles:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

Check if H atoms are also balanced. They are. That means our final reaction is:

2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂0

2 Moles of NaOH reacts with 1 mole of H₂SO₄

5 0
3 years ago
The dark areas on the world map show places with climates that are alike. The table shows information for those places. A
vlabodo [156]

Answer:

Choice B

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6 0
2 years ago
Why do some elements form double and triple bonds during bonding
jekas [21]

Explanation:

Elements need a total of eight electrons to gain stability and look like a noble gas. So, they sometimes need sharing of two, four or even six electrons to complete their octate. So, they form double and triple covalent bonds. One more  the reason is the  interaction between the p orbitals of the combining atoms. for example  A double bond, as in ethene H2C=CH2, arises from one combination of the s orbitals and one combination of the p_y orbitals.

4 0
3 years ago
What is the minimum cation-to-anion radius ratio (rC/rA) for an octahedral interstitial lattice site?
mash [69]

Answer:

0. 414

Explanation:

Octahedral interstitial lattice sites.

Octahedral interstitial lattice sites are in a plane parallel to the base plane between two compact planes and project to the center of an elementary triangle of the base plane.

The octahedral sites are located halfway between the two planes. They are vertical to the locations of the spheres of a possible plane. There are, therefore, as many octahedral sites as there are atoms in a compact network.

The Octahedral interstitial void ratio range is 0.414 to 0.732. Thus, the minimum cation-to-anion radius ratio for an octahedral interstitial lattice site is 0. 414.

3 0
3 years ago
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