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Travka [436]
3 years ago
10

Match the element with its number of valence electrons.

Chemistry
2 answers:
cricket20 [7]3 years ago
8 0
<h2>Answer:</h2>

Valence electrons are that electrons that are present in the atom, outside the nucleus in orbits. The list of atoms with correct number of valance electrons is as follows;

A. Nitrogen: 5

B. Silicon: 4

C. Oxygen: 6

D. Magnesium: 2

marissa [1.9K]3 years ago
4 0

Answer :

Element            Valence electrons

Nitrogen                         5

Silicon                            4

Oxygen                          6

Magnesium                    2

Explanation :

Valence electrons : Valence electrons are the electrons which is present on the outermost shell of the electron.

The given element is nitrogen that belongs to group 15 and the atomic number 7.

The number of electrons present on nitrogen element are 7.

The electronic configuration of nitrogen element is,

1s^22s^22p^3

The number of valence electrons present on nitrogen element are, 5.

The given element is silicon that belongs to group 14 and the atomic number 14.

The number of electrons present on silicon element are 14.

The electronic configuration of silicon element is,

1s^22s^22p^63s^23p^2

The number of valence electrons present on silicon element are, 4.

The given element is oxygen that belongs to group 16 and the atomic number 8.

The number of electrons present on oxygen element are 8.

The electronic configuration of oxygen element is,

1s^22s^22p^4

The number of valence electrons present on oxygen element are, 6.

The given element is magnesium that belongs to group 2 and the atomic number 12.

The number of electrons present on magnesium element are 12.

The electronic configuration of magnesium element is,

1s^22s^22p^63s^2

The number of valence electrons present on magnesium element are, 2.

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sun

Explanation:

source means where it comes from and the source of energy comes from the sun

3 0
3 years ago
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if 1.0 L of a 2.0 M solution contains 1.204 * 10^24 molecules of a solute, how many molecules of solute will 1.0 L of a 0.5 M so
Rama09 [41]

The formula for molariy is nomber of moles/volume

0.5=n/1=>n=0.5 moles

one mole contains 6.023*10^23 molecules

so 0.5 contain 3.0115*10^23 molecules

4 0
3 years ago
The oxidation of sulfur dioxide by oxygen to sulfur trioxide has been implicated as an important step in the formation of acid r
Igoryamba

Answer:

The value of the equilibrium constant K_p at this temperature is 3.42.

Explanation:

Partial pressure of the sulfur dioxide =p_1=0.564 atm

Partial pressure of the oxygen gas =p_2=0.102 atm

Partial pressure of the sulfur trioxide =p_3=0.333

2 SO_2(g) + O_2(g)\rightleftharpoons 2 SO_3(g)

The expression of an equilibrium constant is given by :

K_p=\frac{(p_3)^2}{(p_1)^2\times p_2}

K_p=\frac{(0.333 atm)^2}{(0.564 atm)\times (0.102 atm)}=3.42

The value of the equilibrium constant K_p at this temperature is 3.42.

7 0
4 years ago
What do you predict will happen to the volume inside a container when pressure &amp; temperature remains constant but the number
Pani-rosa [81]

Answer:

<u>increase</u>

Explanation:

<u>The Ideal Gas Equation</u>

  • PV = nRT
  • V = nRT/P
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The volume of a gas is directly proportional to the number of moles it contains. Hence, if the number of particles is increased, then the volume inside the container will <u>increase</u>

7 0
2 years ago
If 152 grams of ethane (c2h6) are reacted with 231 grams of oxygen gas, what is the mass of the excess reactant leftover after t
Naya [18.7K]
Answer is: <span>the mass of the excess reactant (ethane) leftover is 90.135 grams.
</span>Chemical reaction: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O<span>(g).
m(</span>C₂H₆) = 152 g.
n(C₂H₆) = m(C₂H₆) ÷ M(C₂H₆).
n(C₂H₆) = 152 g ÷ 30 g/mol.
n(C₂H₆) = 5.067 mol.
m(O₂) = 231 g.
n(O₂) = 231 g ÷ 32 g/mol.
n(O₂) = 7.218 mol; limiting reactant.
From chemical reaction: n(O₂) : n(C₂H₆) = 7 : 2.
n(C₂H₆) = 2 · 7.218 mol ÷ 7.
n(C₂H₆) = 2.0625mol.
Δn(C₂H₆) = 5.067 mol - 2.0625 mol.
Δn(C₂H₆) = 3.0045 mol.
Δm(C₂H₆) = 3.0045 mol · 30 g/mol = 90.135 g.
6 0
4 years ago
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