Answer : The value of
for
is
.
Solution : Given,
Solubility of
in water = ![4.4\times 10^{-5}mole/L](https://tex.z-dn.net/?f=4.4%5Ctimes%2010%5E%7B-5%7Dmole%2FL)
The barium carbonate is insoluble in water, that means when we are adding water then the result is the formation of an equilibrium reaction between the dissolved ions and undissolved solid.
The equilibrium equation is,
![BaCO_3\rightleftharpoons Ba^{2+}+CO^{2-}_3](https://tex.z-dn.net/?f=BaCO_3%5Crightleftharpoons%20Ba%5E%7B2%2B%7D%2BCO%5E%7B2-%7D_3)
Initially - 0 0
At equilibrium - s s
The Solubility product will be equal to,
![K_{sp}=[Ba^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BBa%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)
![K_{sp}=s\times s=s^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3Ds%5Ctimes%20s%3Ds%5E2)
![[Ba^{2+}]=[CO^{2-}_3]=s=4.4\times 10^{-5}mole/L](https://tex.z-dn.net/?f=%5BBa%5E%7B2%2B%7D%5D%3D%5BCO%5E%7B2-%7D_3%5D%3Ds%3D4.4%5Ctimes%2010%5E%7B-5%7Dmole%2FL)
Now put all the given values in this expression, we get the value of solubility constant.
![K_{sp}=(4.4\times 10^{-5}mole/L)^2=19.36\times 10^{-10}mole^2/L^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%284.4%5Ctimes%2010%5E%7B-5%7Dmole%2FL%29%5E2%3D19.36%5Ctimes%2010%5E%7B-10%7Dmole%5E2%2FL%5E2)
Therefore, the value of
for
is
.
Answer:
The correct answer is AMP+H2O→ Adenosine + pi
Explanation:
The above reaction is least energetic because there is no phosphoanhydride bond present with adenosine mono phosphate.Phospho anhydride bond is an energy rich bond.
As a result hydrolysis of AMP generates very little amount of energy in comparison to the hydrolysis of ATP and ADP.
<u>Answer:</u> The edge length of the unit cell is 0.461 nm
<u>Explanation:</u>
We are given:
Atomic radius of iridium = 0.163 nm
To calculate the edge length, we use the relation between the radius and edge length for FCC lattice:
![a=2\sqrt{2}R](https://tex.z-dn.net/?f=a%3D2%5Csqrt%7B2%7DR)
Putting values in above equation, we get:
![a=2\sqrt{2}\times 0.163=0.461nm](https://tex.z-dn.net/?f=a%3D2%5Csqrt%7B2%7D%5Ctimes%200.163%3D0.461nm)
Hence, the edge length of the unit cell is 0.461 nm