<span>H2C2O4(aq) + 2OH- --> C2O4^2- + 2H2O(l)</span>
Answer:
1.7 ppm
Explanation:
Original amount N' = 2.6 ppm
time to testing t = 24 hr
final amount N = 2.1 ppm
Using exponential inhibited decay, we have
N = N'e^(-kt)
Where
N is the new reading
N' is the original reading
t is the decay time
k is the decay constant
Substituting, we have
2.1 = 2.6 x e^(-k x 24)
2.1 = 2.6 x e^(-24k)
0.808 = e^(-24k)
We take the natural log of both sides of the equation
Ln 0.808 = Ln (e^(-24k))
-0.213 = - 24k
K = 0.213/24 = 0.00886
After 48 hrs, the reading of free chlorine will be
N = 2.6 x e^(-0.00886 x 48)
N = 2.6 x e^(-0.425)
N = 2.6 x 0.654
N = 1.7 ppm
Answer:
Water splitting is the chemical reaction in which water is broken down into oxygen and hydrogen: 2 H2O → 2 H2 + O. Efficient and economical photochemical water splitting would be a technological breakthrough that could underpin a hydrogen economy.
Explanation: 2 H2O → 2 H2 + O.
Answer:
At equilibrium, the concentration of 
is going to be 0.30M
Explanation:
We first need the reaction.
With the information given we can assume that is:
+ 
⇄ 2
If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no nor present. Immediately, and are going to be produced until equilibrium is reached.
By the ICE (initial, change, equilibrium) analysis:
I: []=0 ; [ ]= 0 ; []=0.60M
C: []=+x ; [ ]= +x ; []=-2x
E: []=0+x ; [ ]= 0+x ; []=0.60-2x
Now we can use the constant information:
![K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }](https://tex.z-dn.net/?f=K_%7Bc%7D%3D%5Cfrac%7B%5Bproducts%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D%7B%5Breactants%5D%5E%7Bstoichiometric%20coefficient%7D%20%7D)
= 
= 
= 
At equilibrium, the concentration of is going to be 0.30M
