Answer: Equilibrium concentration of ![[Cl^-]](https://tex.z-dn.net/?f=%5BCl%5E-%5D)
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 ![[CoCl_4]^{2-}](https://tex.z-dn.net/?f=%5BCoCl_4%5D%5E%7B2-%7D)
=x = 0.031 M
Concentration of = (4.60-2x) M = 
=4.538 M
Thus equilibrium concentration of at is 4.538 M
Answer: Volume – How much space an object or substance takes up. Mass – Measurement of the amount of matter in an object or substance.
Explanation:
Answer:
a) AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Explanation:
a) AgNO3 + KI → Ag+ + NO3- + K+ + I-
Ag+ + NO3- + K+ + I- → AgI + KNO3
AgNO3 + KI → AgI + KNO3
b) Ba(OH)2 + 2HNO3 → Ba^2+ + 2OH- + 2H+ + 2NO3-
Ba^2+ + 2OH- + 2H+ + 2NO3- → Ba(NO3)2 + 2H2O
Ba(OH)2 + 2HNO3 → Ba(NO3)2 + 2H2O
c) 2Na3PO4 + 3Ni(NO3)2 → 6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3-
6Na+ + 2PO4^3- + 3Ni^2+ + 6NO3- → Ni3(PO4)2 + 6NaNO3
2Na3PO4 + 3Ni(NO3)2 → Ni3(PO4)2 + 6NaNO3
d) 2Al(OH)3 + 3H2SO4 → 2Al^3+ + 6OH- + 6H+ + 3SO4^2-
2Al^3+ + 3OH- + 3H+ + 3SO4^2- → Al2(SO4)3 + 6H2O
2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O
Answer:
0.071L
Explanation:
From the question given, we obtained the following data:
Molarity of HCl = 2.25 M
Mass of HCl = 5.80g
Molar Mass of HCl = 36.45g/mol
Number of mole of HCl =?
Number of mole = Mass /Molar Mass
Number of mole of HCl = 5.8/36.45 = 0.159mole
Now, we can obtain the volume required as follows:
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.159mole/ 2.25
Volume = 0.071L
