The reaction is as follow,
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + Al(OH)₃
Ca are balance on both sides,
There are 2 Al at left side and one at right so, multiply Al(OH)₃ by 2 to balance,
So,
Ca(OH)₂ + Al₂(SO₄)₃ → CaSO₄ + 2 Al(OH)₃
Now, Ca and Al are balanced, now balance SO₄, which is 3 at left hand side and one at right hand side, so multiply CaSO₄ on right side by 3, so,
Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
Again Ca got imbalance, so multiply Ca(OH)₂ by 3 to balance Ca, So,
3 Ca(OH)₂ + Al₂(SO₄)₃ → 3 CaSO₄ + 2 Al(OH)₃
The Equation is balance now with respect to every element.
Result:
The Ratio is 3 : 1 : 3 : 2, so, Option-G is correct.
Explanation:
It is given that total energy is -443 kJ/mol and formula to calculate the lattice energy is as follows.
Total energy = heat of sublimation + bond dissociation energies + ionization energy for Cs + EA of
+ lattice energy
-443 kJ/mol = 76 + 121 + 376 - 349 + Lattice energy
Lattice energy = (-443 - 76 -121 - 376 + 349) kJ
Lattice energy = -667 kJ
Therefore, we can conclude that -667 kJ is the magnitude of the lattice energy for CsCl.
Answer:
Zn2+ is colourless
Explanation:
We know that transition metal salts are usually coloured due to the possibility of d-d transition.
This d-d transition can only occur when there are vacant d-orbitals. The electronic configuration [Ar] 4s23d8 suggests the presence of vacant d-orbitals and the possibility of the compounds of Zn2+ being coloured.
However, the absence of colours in Zn2+ compounds shows that there is no d-d transition(electronic) spectra observed for Zn2+ because the d orbitals are completely filled. This means that the correct electronic configuration of the ion is [Ar] 3d10.
North magnets attract the opposite side in this case south would attract to north and vise versa if it was negative it would attract to positive and vise versa
Answer:
D: Chlorine( I may be wrong my chemistry year was a nightmare)