Answer is: C) the fact that the number of lone pairs of electrons on the central atom is greater in the case of water.
Carbon(IV) oxide is nonpolar because CO₂ is linear molecule and the oxygen atoms are symmetrical (bond angles 180°).
Water is polar because of the bent shape of the molecule.
Oxygen atom in water molecule has sp3 hybridization. The bond angle between the two hydrogen atoms is approximately 104.45°.
Oxygen atom has atomic number 8, it means it has eight protons and eight electrons, so atom has neutral charge. Oxygen is a nonmetal.
Electron configuration of oxygen atom: ₈O 1s² 2s² 2p⁴.
Oxygen atom has six valence electrons
, two lone pairs and two electrons that form two sigma bonds with hydrogen atoms.
Carbon is a chemical element with symbol C and atomic number 6, which means it has 6 protons and six electrons. Four valence electrons are in 2s and 2p orbitals.
Electron configuration of carbon atom: ₆C 1s² 2s² 2p².
In carbon dioxide, carban has sp hybridization with no lone pairs.
Explanation:
At STP ,2.24 L contain 0.1 mole of N²
<h2>So,No. of molecules of N2 = 6.022*10²²</h2>
I believe you would just put a 2 in front of NH3 and keep the other ones as 1
The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
To learn more about enthalpy here
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Answer:
The law that suggests that at a constant pressure and the volume of gas directly proportional to its temperature is the Boyle's law.
Explanation:
