I believe your answer Is C. An ammonia molecule has a trigonometrical pyramidal shape. Figure C has a <span>has a trigonometrical pyramidal shape.</span>
I hope I help
Remember that any intersection of lines is a C, and that the number of hydrogens attached are the necessary to complet the 4 bonds.
1) CH3 - CH (OH) - CH (CH3) -CH3
2) CH3 - O - CH(CH3)-CH2 - CH3
I have used the parenthesis to indicate that the radical inside is in other branch, bonded by a single line -
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The ml is also called as the magnetic quantum number. The value
of ml can range from –l to +l including zero. Hence all of the possible values for ml given
that l = 2 are:
<span>-2, -1, 0, + 1, + 2</span>
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.
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