Answer:
The answer is 0.023 moles of phosphorus
Explanation:
The 15-15-15 fertilizer is a fertilizer of great versatility, made with nitrogen, phosphorus and potassium, which makes it one of the fertilizers most used for fertilizer in the sowing plant, thus covering the crop requirements from planting. .
This fertilizer consists of 14.25% phosphorus pentoxide (P2O5). Therefore, we have to remove 14.25% at 10 grams of 15-15-15 fertilizer to calculate the moles of phosphorus. As follows:
Grams of P2O5 = 10 g x 0.1425 = 1.425 g
We calculate the molecular weight of phosphorus. We use the periodic table:
Phosphorus molecular weight = 2 x 30.97 = 61.94 g/mol
Now we calculate the moles of phosphorus in the fertilizer:
Phosphorus moles = 1,425 g/61.94 g/mol = 0.023 moles
A Campfire is an example, the solid wood becomes ash.
Answer:
B.
Explanation:
The up and down movement of gases and liquids caused by heat transfer is convection.
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.