Recall that your hypothesis is that these values are the fraction of atoms that are still radioactive after n half-life cycles.
Record in the appropriate blanks.
2 answers:
Answer : A= 0.5, B = 0.25 , C = 0.125, D = 0.015625 and E = 0.00390625
Explanation :
Half life of a substance is defined as the amount of time taken by the substance to reduce to half of its original amount.
Here n represents the number of half lives.
The amount of substance that remains after n half lives can be calculated using the given formula,
So when we have n =1,
Fraction of substance that remains = 0.5¹ = 0.5.
That means after first half life over, the amount of substance that remains is 0.5 times that of original.
Therefore we have A = 0.5
When n = 2, we have 0.5² = 0.25
So when 2 half lives are over, the amount of substance that remains is 0.25 times that of original
Therefore B = 0.25
When n = 3, we have 0.5³ = 0.125
So when 3 half lives are over, the amount of substance that remains is 0.125 times that of original.
Therefore we have C = 0.125
When n = 6 , we have 0.5⁶ = 0.015625
So D = 0.015625
When n = 8, we have 0.5⁸ = 0.00390625
Therefore E = 0.00390625
The values for A, B, C, D and E are 0.5, 0.25, 0.125, 0.015625 and 0.00390625 respectively.
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Answer :
(1) The hybridization of central atom beryllium in is,
(2) The hybridization of central atom nitrogen in is,
(3) The hybridization of central atom carbon in is,
(4) The hybridization of central atom xenon in is,
Explanation :
Formula used :
where,
V = number of valence electrons present in central atom
N = number of monovalent atoms bonded to central atom
C = charge of cation
A = charge of anion
Now we have to determine the hybridization of the following molecules.
(1) The given molecule is,
The number of electron pair are 2 that means the hybridization will be and the electronic geometry of the molecule will be linear.
(2) The given molecule is,
If the sum of the number of sigma bonds, lone pair of electrons and odd electrons present is equal to three then the hybridization will be, .
In nitrogen dioxide, there are two sigma bonds and one lone electron pair. So, the hybridization will be, .
(3) The given molecule is,
The number of electron pair are 4 that means the hybridization will be and the electronic geometry of the molecule will be tetrahedral.
(4) The given molecule is,
Bond pair electrons = 4
Lone pair electrons = 6 - 4 = 2
The number of electrons are 6 that means the hybridization will be and the electronic geometry of the molecule will be octahedral.
But as there are four atoms around the central xenon atom, the fifth and sixth position will be occupied by lone pair of electrons. The repulsion between lone and bond pair of electrons is more and hence the molecular geometry will be square planar.