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
It's adenosine triphosphate !
it has Penrose sugar and phosphate as backbone !
and nitrogenous base ... adenine.... in the middle !
We can calculate years by using the half-life equation. It is expressed as:
A = Ao e^-kt
<span>where A is the amount left at t years, Ao is the initial concentration, and k is a constant.
</span>From the half-life data, we can calculate for k.
1/2(Ao) = Ao e^-k(1620)
<span>k = 4.28 x 10^-4
</span>
0.125 = 1 e^-<span>4.28 x 10^-4 (</span>t)
t = 4259 years
Answer: 35.4 grams
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

where,
Molality = 2.65
n= moles of solute =?
= volume of solution in ml = 445 ml
Putting in the values we get:


Mass of solute in g=
Thus 35.4 grams of
is needed to prepare 445 ml of a 2.65 m solution of
.
The replacing of sodium hydroxide with potassium hydroxide
(KOH) to the reaction will least affect the organic product that forms.
Potassium hydroxide is an
inorganic compound with the formula KOH, and is commonly called caustic potash. Along with sodium hydroxide, this colorless
solid is a prototypical strong base.