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Serhud [2]
3 years ago
7

Which Of These Elements Has Four Valence Electrons? Hafnium Radon Silicon Sulfur

Chemistry
2 answers:
Free_Kalibri [48]3 years ago
8 0
Silicon has 4 valence Electron may be
zvonat [6]3 years ago
4 0

Answer : The elements that has four valence electrons are hafnium (Hf) and silicon (Si).

Explanation :

Electronic configuration : It is defined as the representation of electrons around the nucleus of an atom.  

Number of electrons in an atom are determined by the electronic configuration.

Atomic number : It is defined as the number of electrons or number of protons present in a neutral atom.

Valence electrons or outermost electrons : They are electrons in an atom that participate in bonding.

(a) Hafnium is a metal that belongs to group 4 and period 6. The atomic number of Hafnium is, 72.

The electronic configuration of hafnium will be,

[Xe]4f^{14}5d^26s^2

The total number of valance electrons = [2 + 2] = 4

(b) Radon is a noble gas that belongs to group 18 and period 6. The atomic number of radon is, 86.

The electronic configuration of radon will be,

[Xe]4f^{14}5d^{10}6s^26p^6

The total number of valance electrons = [2 + 6] = 8

(c) Silicon is a non-metal that belongs to group 14 and period 3. The atomic number of silicon is, 14.

The electronic configuration of silicon will be,

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

The total number of valance electrons = [2 + 2] = 4

(d) Sulfur is a non-metal that belongs to group 16 and period 3. The atomic number of sulfur is, 16.

The electronic configuration of sulfur will be,

1s^22s^22p^63s^23p^4

The total number of valance electrons = [2 + 4] = 6

Hence, from this we conclude that, the elements that has four valence electrons are hafnium (Hf) and silicon (Si).

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Answer:

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Explanation:

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The picture above represents which of the following?
PSYCHO15rus [73]
This is a plant cell because there are chloroplasts in this diagram which are present in plant cells
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My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h
Fed [463]

Answer:

P=atm

b=\frac{L}{mol}

Explanation:

The problem give you the Van Der Waals equation:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

First we are going to solve for P:

(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}

P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}

Then you should know all the units of each term of the equation, that is:

P=atm

n=mol

R=\frac{L.atm}{mol.K}

a=atm\frac{L^{2}}{mol^{2}}

b=\frac{L}{mol}

T=K

V=L

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

P=\frac{L.atm}{L-L}-atm

Then operate the fraction subtraction:

P=P=\frac{L.atm-L.atm}{L}

P=\frac{L.atm}{L}

And finally you can find the answer:

P=atm

Now solving for b:

(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT

(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}

b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}

Replacing units:

b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}

Multiplying and dividing units,(please see the second photo below), we have:

b=\frac{L-\frac{L.atm}{atm}}{mol}

b=\frac{L-L}{mol}

b=\frac{L}{mol}

7 0
3 years ago
Make a prediction about the relationship between electrons and molecular shapes
7nadin3 [17]
The electrons and the nuclei will settle into positions that minimize repulsion and maximize attraction.
4 0
3 years ago
How many moles are represented by 3.01 x10^24 oxygen atoms?
asambeis [7]
<h3>Answer:</h3>

5.00 mol O₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.<u> </u>

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.01 × 10²⁴ atoms O₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.01 \cdot 10^{24} \ atoms \ O_2(\frac{1 \ mol \ O_2}{6.022 \cdot 10^{23} \ atoms \ O_2})
  2. Multiply/Divide:                \displaystyle 4.99834 \ mol \ O_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

4.99834 mol O₂ ≈ 5.00 mol O₂

4 0
3 years ago
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