Answer:
6
Explanation:
FCC is face centered cubic lattice. In FCC structure, there are eight atoms at the eight corner of the cubic unit cell and one atom centered in each of the faces. FCC unit cells consist of four atoms, (8/8) at the corners and (6/2) in the faces.
Given that, Cu has FCC structure and it contains a vacancy at origin (0, 0, 0). And there is no other vacancy directly adjacent to the vacancy at the origin. So, all the adjacent positions contain Cu atoms. Hence, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy.
the above FCC unit cell clearly indicates that there are six adjacent atoms adjacent to the vacancy at origin
So, the total number of adjacent atoms of the vacancy at origin can jump into this vacancy is 6.
If Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
<h3>
What is base dissociation constant?
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The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.
The dissociation reaction of hydrogen cyanide can be given as
HCN --- (H+) + (CN-)
Given,
The value of Ka for HCN is 2.8× 10^(-9)
The correlation between base dissociation constant and acid dissociation constant is
Kw = Ka × Kb
Kw = 10^(-14)
Substituting values of Ka and Kw,
Kb = 10^(-14) /{2.8×10^(-9) }
= 3.5× 10^(-6)
Thus, we find that if Ka for HBrO is 2. 8×10^−9 at 25°C, then the value of Kb for BrO− at 25°C is 3.5× 10^(-6).
DISCLAIMER: The above question have mistake. The correct question is given as
Question:
Given that Ka for HBrO is 2. 8×10^−9 at 25°C. What is the value of Kb for BrO− at 25°C?
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Answer:
a b
Explanation:
I think so take it with a grain of salt
Because in difrent materials atoms are more compact or less compact.if they are less compact then it will be easear for them to move
