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Artist 52 [7]
4 years ago
14

Which element can bond with titanium (Ti) to form a polar covalent bond?

Chemistry
2 answers:
snow_lady [41]4 years ago
6 0

Answer: Your answer will D

Explanation: <u><em>The element which can bond with titanium to form a polar covalent bond is Nitrogen. Polar covalent bond is formed between two elements whose electronegativity difference is in the range of 0.4-1.7.</em></u>

<u><em /></u>

pogonyaev4 years ago
5 0

Answer: Option (A) is the correct answer.

Explanation:

A polar covalent bond is formed due to unequal sharing of electrons between the combining atoms.

Hence, when two metals combine together then there could occur sharing of electrons. But when a metal combines with a non-metal then there will occur transfer of electrons due to which there will be formation of an ionic compound.

So, more is the difference in electronegativity of combining metal atoms more will be unequal sharing of electrons.

Electronegativity of Ti is 1.54, Be is 1.57 and electronegativity of Mn is 1.55. As there is more difference between electronegativity of Ti and Be. So, they will form a polar covalent bond.

Thus, we can conclude that Be is the element which bond with titanium (Ti) to form a polar covalent bond.

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Why it is impossible for an isolated atom to exist in the hybridized state?
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Hybridization refers to the mixing of atomic orbitals in an atom. The number of hybrid orbitals needs to be equal to the number of orbitals that have involved in prior to mixing.  

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7 0
3 years ago
1. A student needs to make 80mL of a 4.5 x 10^-3 M solution of H_3PO_4 (mm = 98) from solid. Describe how the student would do t
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Answer:

1. 35 mg of H₃PO₄

2. 27 mol AlF₃; 82 mol F⁻

3. 300 mL of stock solution.

Explanation:

1. Preparing a solution of known molar concentration

Data:

V = 80 mL

c = 4.5 × 10⁻³ mol·L⁻¹

Calculations:

(a) Moles of H₃PO₄

Molar concentration = moles of solute/litres of solution

c = n/V

n = Vc = 0.080L × (4.5 × 10⁻³ mol/1 L) = 3.60 × 10⁻⁴ mol

(b) Mass of H₃PO₄  

moles = mass/molar mass

n = m/MM

m = n × MM = 3.60 × 10⁻⁴ mol × (98 g/1 mol) = 0.035 g = 35 mg

(c) Procedure

Dissolve 35 mg of solid H₃PO₄  in enough water to make 80 mL of solution,

2. Moles of solute.

Data:

V = 4900 mL

c = 5.6 mol·L⁻¹

Calculations:

Moles of AlF₃ = cV = 4.9 L AlF₃ × (5.6 mol AlF₃/1L AlF₃) = 27 mol AlF₃

Moles of F⁻ = 27 mol AlF₃ × (3 mol F⁻/1 mol AlF₃) = 82 mol F⁻.

3. Dilution calculation

Data:

V₁= 750 mL; c₁ = 0.80 mol·L⁻¹

V₂ = ?            ; c₂ = 2.0   mol·L⁻¹

Calculation:

V₁c₁ = V₂c₂

V₂ = V₁ × c₁/c₂ = 750 mL × (0.80/2.0) = 300 mL

Procedure:

Measure out 300 mL of stock solution. Then add 500  mL of water.

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