The valence electron configuration for antimony (Sb) is:
Sb = 5s²5p³5d⁰
In SbCl₅²⁻, antimony has a -2 charge i.e. it has 2 additional electrons
Sb²⁻ = 5s²5p⁵5d⁰
Following a two electron transition from p→d orbital we have:
Sb²⁻ = 5s²5p³5d²
There is a total of 5 unpaired electrons (3 in the p and 2 in the d) which can form five bonds with the 5 Cl atoms.
Thus the hybridisation of Sb in SbCl₅²⁻ is sp³d²
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there's no picture here but I guess the answer would be:
considering the constant temperature, if you double the volume, the pressure would be halved.
like: volume is 2, pressure is 4
if 2×2, then:4÷2
Answer:
A decrease in the total volume of the reaction vessel (T constant)
Explanation:
- Le Châtelier's principle predicts that the moles of H2 in the reaction container will increase with a decrease in the total volume of the reaction vessel.
- <em><u>According to the Le Chatelier's principle, when a chnage is a applied to a system at equilibrium, then the equilibrium will shift in a way that counteracts the effect causing it.</u></em>
- In this case, a decrease in volume means there is an increase in pressure, therefore the equilibrium will shift towards the side with the fewer number of moles of gas.
Answer:
2 moles
Explanation:
Let us first start by calculating the molecular mass of Al₂O₃.
The mass of a mole of any compound is called it's molar mass. 1 molar mass 6.02 X 10²³, or Avogadro's number, of compound entities.
Say, 1 mole of Al₂O₃ has 6.02 X 10²³ of Al₂O₃ molecules/atoms. It also has 2*6.02 X 10²³ number of Al atoms and 3*6.02 X 10²³ number of O atoms.
Molecular mass of Al : 26.981539 u
Molecular mass of O: 15.999 u
Therefore, molecular mass of Al₂O₃ is:
= 
u
= 101.960078 u
This can be approximated to 102 u.
1mole weighs 102 u
So, 2moles will weigh 2*102 = 204 u
