4
Explanation:
I am not positive but it makes more sense
Answer : The solubility of this compound in g/L is
.
Solution : Given,
![K_{sp}=2.42\times 10^{-11}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D2.42%5Ctimes%2010%5E%7B-11%7D)
Molar mass of
= 114.945g/mole
The balanced equilibrium reaction is,
![MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3](https://tex.z-dn.net/?f=MnCO_3%5Crightleftharpoons%20Mn%5E%7B2%2B%7D%2BCO%5E%7B2-%7D_3)
At equilibrium s s
The expression for solubility constant is,
![K_{sp}=[Mn^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BMn%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)
Now put the given values in this expression, we get
![2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L](https://tex.z-dn.net/?f=2.42%5Ctimes%2010%5E%7B-11%7D%3D%28s%29%28s%29%5C%5C2.42%5Ctimes%2010%5E%7B-11%7D%3Ds%5E2%5C%5Cs%3D0.4919%5Ctimes%2010%5E%7B-5%7D%3D4.919%5Ctimes%2010%5E%7B-6%7Dmoles%2FL)
The value of 's' is the molar concentration of manganese ion and carbonate ion.
Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.
![s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L](https://tex.z-dn.net/?f=s%3D4.919%5Ctimes%2010%5E%7B-6%7Dmoles%2FL%5Ctimes%20114.945g%2Fmole%3D565.414%5Ctimes%2010%5E%7B-6%7Dg%2FL)
Therefore, the solubility of this compound in g/L is
.
Answer:
D)Gases usually weigh less than solids.
Answer:
Explanation:pounds or kilograms
Answer:
ΔG = -1366KJ/mol
Explanation:
The detailed step by step calculation is as shown in the attachment.
The relationship between ΔG, Temperature, gas constant and the reaction quotient was applied.