I think it would be minimize so u can have more friction
Answer:
The correct option is;
4 percent ionic, 96 percent covalent, 222 pm
Explanation:
The parameters given are;
Phosphorus:
Atomic radius = 109 pm
Covalent radius = 106 pm
Ionic radius = 212 pm
Electronegativity of phosphorus = 2.19
Selenium:
Atomic radius = 122 pm
Covalent radius = 116 pm
Ionic radius = 198 pm
Electronegativity of selenium= 2.55
The percentage ionic character of the chemical bond between phosphorus and selenium is given by the relation;
Using Pauling's alternative electronegativity difference method, we have;
![\% \, Ionic \ Character = \left [18\times (\bigtriangleup E.N.)^{1.4} \right ] \%](https://tex.z-dn.net/?f=%5C%25%20%5C%2C%20Ionic%20%5C%20Character%20%3D%20%5Cleft%20%5B18%5Ctimes%20%28%5Cbigtriangleup%20E.N.%29%5E%7B1.4%7D%20%20%5Cright%20%5D%20%5C%25)
Where:
Δ E.N. = Change in electronegativity = 2.55 - 2.19 = 0.36
Therefore;
![\% \, Ionic \ Character = \left [18\times (0.36)^{1.4} \right ] \% = 4.3 \%](https://tex.z-dn.net/?f=%5C%25%20%5C%2C%20Ionic%20%5C%20Character%20%3D%20%5Cleft%20%5B18%5Ctimes%20%280.36%29%5E%7B1.4%7D%20%20%5Cright%20%5D%20%5C%25%20%3D%204.3%20%5C%25)
Hence the percentage ionic character = 4.3% ≈ 4%
the percentage covalent character = (100 - 4.3)% = 95.7% ≈ 96%
The bond length for the covalent bond is found adding the covalent radii of both atoms as follows;
The bond length for the covalent bond = 106 pm + 116 pm = 222 pm.
The correct option is therefore, 4 percent ionic, 96 percent covalent, 222 pm.
Answer:
E° = -0.133 V
Explanation:
In the reaction:
X(s) + Y⁺(aq) → X⁺(aq) + Y(s)
<em>1 electron is transferred from X to Y</em>
Now, using Nernst equation:
E° = RT / nF ln K
<em>Where R is gas constant (8.314 J/molK), T is absolute temperature (Usually 298.15K), n are transferred electrons (1, for the reaction), F is faraday constant (96485C/mol) and K is equilibrium constant (5.59x10⁻³)</em>
Replacing:
E° = 8.314 J/molK*298.15K / 96485C/mol*1 ln 5.59x10⁻³
<em>E° = -0.133 V</em>
The volume (in liters) of CO₂ that can be consumed at STP by 435 g Na₂O₂ is 125 L of Co₂
<u><em>calculation</em></u>
2Na₂O₂(s) +2 CO₂ (g)→ 2 Na₂CO₃(s) + O₂(g)
Step 1 : find the moles of Na₂O₂
moles = mass÷ molar mass
from periodic table the molar mass of Na₂O₂ = (23 x2) +( 16 x2) = 78 g/mol
moles= 435 g÷ 78 g/mol = 5.58 moles
Step 2: use the mole ratio to determine the moles of CO₂
from given equation Na₂O₂ : CO₂ =2 :2 =1:1
Therefore the moles of CO₂ is also = 5.58 moles
Step 3: find the volume of CO₂ at STP
that is at STP 1 mole of a gas = 22.4 L
5.58 moles = ? l
<em>by cross multiplication</em>
= (5.58 moles x 22.4 L) / 1 mole = 125 L