Answer: Equilibrium concentration of
at
is 4.538 M
Explanation:
Initial concentration of
= 0.056 M
Initial concentration of
= 4.60 M
The given balanced equilibrium reaction is,
![COCl_2+2Cl^-\rightleftharpoons [CoCl_4]^{2-}+6H_2O](https://tex.z-dn.net/?f=COCl_2%2B2Cl%5E-%5Crightleftharpoons%20%5BCoCl_4%5D%5E%7B2-%7D%2B6H_2O)
Initial conc. 0.056 M 4.60 M 0 M 0 M
At eqm. conc. (0.056-x) M (4.60-2x) M (x) M (6x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CoCl_4]^{2-}\times [H_2O]^6}{[CoCl_2]^2\times [Cl^-]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCoCl_4%5D%5E%7B2-%7D%5Ctimes%20%5BH_2O%5D%5E6%7D%7B%5BCoCl_2%5D%5E2%5Ctimes%20%5BCl%5E-%5D%5E2%7D)
Given : equilibrium concentration of
=x = 0.031 M
Concentration of
= (4.60-2x) M =
=4.538 M
Thus equilibrium concentration of
at
is 4.538 M
Cr2(SO4)3(aq) + 3(NH4)2CO3(aq) → 3(NH4)2SO4(aq) + Cr2(CO3)3(s)
<span>Ionic: 2Cr+3 + 3SO4^-2 + 6NH4+ + 3CO3^-2 ----> 6NH4+ + 3SO4^-2 + Cr2(CO3)3 (spectator ions are NH4+, SO4^-2) </span>
<span>Net Ionic: 2Cr^+3(aq) + 3CO3^-2(aq) -------> Cr2(CO3)3(s) </span>
You have to use everything that is given since you have to know which is the limiting reactant. We find the limiting reactant by calculating the number of moles of each reactant and compare the number of moles. The limiting reactant would be the one that is consumed fully by the reaction.
Answer:
Carbohydrates,Monosaccharides,and disaccharides
Explanation: I'm good at science