Which is likely to have the strongest metallic bonds

Chemistry

2 answers:

marshall27 [118]4 years ago

7 0

Answer: Option (B) is the correct answer.

Explanation:

Metallic bond is defined as the bonds formed by positively charged atoms due to sharing of delocalised electrons among the lattice of cations.

Formation of metallic bonds depend on two factors which are number of valence electrons released in metallic bonding and the number of protons present within the nucleus of an atom.

So, more is the number of valence electrons more they will be able to pull the protons of the cation as opposite charges attract each other. Hence, stronger will be the metallic bond.

Magnesium has 2 valence electrons and acquires a +2 charge only. And, iron can have either +2 or +3 charge. Whereas magnesium have just 12 protons and iron have 26 protons.

Therefore, iron will be able to form strongest metallic bond due to more valence electrons and more number of protons.

Lapatulllka [165]4 years ago

5 0

Iron would be the answer

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H2SO4 + 2NaNO2 → 2HNO2 + Na2SO4

dem82 [27]

<h3>Answer:</h3>

23.459 g NaNO₂

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

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Stoichiometry

Reading a Periodic Table
Using Dimensional Analysis

<h3>Explanation:</h3>

Step 1: Define

[RxN] H₂SO₄ + 2NaNO₂ → 2HNO₂ + Na₂SO₄

[Given] 24.14714 g Na₂SO₄

Step 2: Identify Conversions

[RxN] 1 mol Na₂SO₄ = 2 mol NaNO₂

Molar Mass of Na - 22.99 g/mol

Molar Mass of N - 14.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of Na₂SO₄ - 2(22.99) + 32.07 + 4(16.00) = 142.05 g/mol

Molar Mass of NaNO₂ - 22.99 + 14.01 + 2(16.00) = 69.00 g/mol

Step 3: Stoichiometry

Set up: $\displaystyle 24.14714 \ g \ Na_2SO_4(\frac{1 \ mol \ Na_2SO_4}{142.05 \ g \ Na_2SO_4})(\frac{2 \ mol \ NaNO_2}{1 \ mol \ Na_2SO_4})(\frac{69.00 \ g \ NaNO_2}{1 \ mol \ NaNO_2})$
Multiply/Divide: $\displaystyle 23.4587 \ g \ NaNO_2$