Answer:
Options B and C
Explanation:
Let's take a look at the options and get our answer by way of elimination. The basic definition of a neutral solution is given as;
A neutral solution is a substance which is neither acid nor basic . it has a PH of 7. it will have equal amount of H+ AND OH- ions in it.
a) a neutral solution does not contain any H3O+ or OH- This is wrong because take water as an example, it is neutral but contains both ions.
b) a neutral solution contains [H2O] = [H3O+]. This option is correct cause it is in line with the definition above.
c) an acidic solution has [H3O⁺] > [OH⁻]. Acidic solutions are any solution that has a higher concentration of hydrogen ions than water. This option is correct.
d) a basic solution does not contain any H3O⁺. This option is wrong. Basic solutions are any solution that has a higher concentration of hydroxide ions than water. This means they contain H3O⁺ but [OH⁻] is greater.
Gallum: Z = 31
electron configuration: [Ar] 4s^2 3d10 4s2 4p1
Highest energy electron: 4p1
Quantum numbers:
n = 4, because it is the shell number
l = 1, it corresponds to type p orbital
ml = may be -1, or 0, or +1, depending on space orientation, they correspond to px, py, pz
ms = may be -1/2 or +1/2, this is the spin number.
The answer would be D. The only minerals harder than corundum is diamond.
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:
The equation is:
3 O₂ + 4 Co → 2 Co₂O₃
Oxidation half reaction:
Co → Co³⁺ + 3 e
Reduction half reaction:
O₂ + 4 e → 2 O²⁻
To balance the equation number of electrons lost must be equal to number or electrons gained so we must multiply oxidation half time 4 and reduction half times 3
