Hey there!
No of hybrid orbitals , H = ( V +S - C + A ) / 2
Where H = no . of hybrid orbitals
V = Valence of the central atom = 5
S = No . of single valency atoms = 4
C = No . of cations = 1
A = No . of anions = 0
For PCl4 +
Plug the values we get H = ( 5+4-1+0) / 2
H = 4 ---> sp3 hybridization
sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations
Answer C
Hope that helps!
The Plum Pudding model of the atom was like a modernchocolate chip cookie it had subatomic particles embedded inside it in no particular arrangement
the gold foil experiment proved that there was a lot of empty space in an atom Because of the launched Particles.mostly deflected but some went through the foil
The correct answer is 3) 2CO2(g) ⇄ 2CO(g) + O2(g)
this is the correct one because it is a decomposition reaction and all the number of atoms is equal on both sides.
there are 2 C atoms on both sides.
and 4 O atoms on both sides.
and 1) the atoms numbers are equal on both sides but not correct as it not a
correct number as it has 1/2 O2.
and 2) CO2(g) ⇆ CO(g) + O2
the number of O atoms is not equal on both sides of the equation.
we have 2 O atoms on the left side and 3 O atoms on the right side.
so, this not a balanced equation.
4) also not correct 2CO(g) + O2 ⇆ 2CO2
as it is not a decomposition reaction and the 2CO & O2 are as reactants not products.
so the correct answer is 3) 2CO2(g) ⇆ 2CO(g) + O2(g)
Gasoline is predominantly octane, C8H18. Something like soap would be a great homogenizer. Soap is composed of a long hydrocarbon chain with a tiny, highly polar tip on one end. Usually, the soap is the anion of a salt, NaX. This allows the polar end of the soap to stick to water, while the nonpolar end sticks to the oil.
Answer:
A). The complementary shapes of an enzyme and a substrate.
Explanation:
The Lock-and-key mechanism was proposed by Emil Fischer for the first time and characterized as the metaphor which helps in elucidating the specificity of the enzymatic reactions. In this metaphor, the lock is described as the enzyme while 'key' is characterized as the substrate which the enzyme acts upon. If the key is not appropriately sized, it will not fit into the active site i.e. the keyhole of the lock or enzyme and reaction will not take place. Thus, <u>option A</u> is the correct answer.
