A weak bond called a ______ bond results from an attraction between a slightly positive region in a molecule and a slightly nega
1 answer:
<u>Answer:</u> The weak bond having slightly positive region and a slightly negative region is polar covalent bond.
<u>Explanation:</u>
Covalent bond is defined as the bond which is formed by the sharing of electrons between the atoms. <u>For Example:</u> HCl, etc..
They are of two types:
- <u>Polar covalent bond:</u> This bond is formed when difference in electronegativity between the atoms is present. When atoms of different elements combine, it results in the formation of polar covalent bond.
In this bond, a dipole is created. More electropnegative atom will have a slight negative charge and less electronegative atom will have a slight positive charge. For Example: etc..
- <u>Non-polar covalent bond:</u> This bond is formed when there is no difference in electronegativity between the atoms. When atoms of the same element combine, it results in the formation of non-polar covalent bond. For Example:
etc..
Hence, the weak bond having slightly positive region and a slightly negative region is polar covalent bond.
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a) To find the mass after t years:
we will use this formula:
A = Ao / 2^n
when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)
b) Mass after 90 y :
by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mg
C) Time for 1 mg remaining:
when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
we will use this formula:
A = Ao / 2^n
when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)
b) Mass after 90 y :
by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mg
C) Time for 1 mg remaining:
when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
Answer:
1.33 Å
Explanation:
Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å
Also,
For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.
Thus,
.................1
Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.
Thus,
...................2
Given that:
To find,
Using 1 and 2 , we get:
<u>Size of the potassium ion = 1.33 Å</u>