Start with Unbalanced Equation and balance it, so...
C7H16+O2--->CO2+H2O
There are 7 C atoms on the left-hand side, so we need 7 C atoms on the right-hand side. Add a 7 in front of the CO2...7CO2+H2O on right side now.
We have fixed 16 H atoms on the left-hand side, so we need 16 H atoms on the right-hand side. Add an 8 in front of H2O to make 16 (8x2)...7CO2+8H2O on right side now.
There are 22 O atoms on the right-hand side: 14 from the CO2 and 8 from the H2O. Add an 11 in front of the O2 on the left side to make 22 (11x2).
Every formula now has a fixed coefficient. You should have a balanced equation of...
C7H16+11O2--->7CO2+8H2O
Oxidation of D -Ribose in presence of hypobromous acid gives D-Ribonic acid
Answer:
In He2 molecule,
Atomic orbitals available for making Molecular Orbitals are 1s from each Helium. And total number of electrons available are 4.
Molecular Orbitals thus formed are:€1s2€*1s2
It means 2 electrons are in bonding molecular orbitals and 2 are in antibonding molecular orbitals .
Bond Order =Electrons in bonding molecular orbitals - electrons in antibonding molecular orbitals /2
Bond Order =Nb-Na/2
Bond Order =2-2/2=0
Since the bond order is zero so that He2 molecule does not exist.
Explanation:
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