Answer:
The new pH of the solution 
Explanation:
Given
pH of bottle of Soda is 
As we know that
pH 
log [H+]
We will derive the hydrogen ion concentration when pH is
log [H+]
Taking anti-log on both the sides, we get -
H+ ion is equal to 
When the hydronuim ion concentration reduces by 
of the original concentration, the new hydrogen ion concentration becomes
The new pH is
![= - log [ 10^{-6}]\\= 6](https://tex.z-dn.net/?f=%3D%20-%20log%20%5B%2010%5E%7B-6%7D%5D%5C%5C%3D%206)
Answer:
rate = k[A][B] where k = k₂K
Explanation:
Your mechanism is a slow step with a prior equilibrium:
![\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrcl%7D%5Ctext%7BStep%201%7D%3A%26%20%5Ctext%7BA%20%2B%20B%7D%20%26%20%5Cxrightarrow%20%5Bk_%7B-1%7D%5D%7Bk_%7B1%7D%7D%20%26%20%5Ctext%7BC%7D%5C%5C%5Ctext%7BStep%202%7D%3A%20%26%20%5Ctext%7BC%20%2B%20A%7D%20%26%20%5Cxrightarrow%20%5B%20%5D%7Bk_%7B2%7D%7D%20%26%20%5Ctext%7BD%7D%5C%5C%5Ctext%7BOverall%7D%3A%20%26%20%5Ctext%7B2A%20%2B%20B%7D%20%26%20%5Clongrightarrow%20%5C%2C%20%26%20%5Ctext%7BD%7D%5C%5C%5Cend%7Barray%7D)
(The arrow in Step 1 should be equilibrium arrows).
1. Write the rate equations:
![-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]](https://tex.z-dn.net/?f=-%5Cdfrac%7B%5Ctext%7Bd%5BA%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-%5Cdfrac%7B%5Ctext%7Bd%5BB%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20%2B%20k_%7B1%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BC%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20-%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BD%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D)
2. Derive the rate law
Assume k₋₁ ≫ k₂.
Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.
In an equilibrium, the forward and reverse rates are equal:
k₁[A][B] = k₋₁[C]
[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)
rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]
The rate law is
rate = k[A][B] where k = k₂K
Answer:
Diphosphorus pentoxide
Carbon dichloride
BCl3
N2H4
Explanation:
These are all covalent compounds. To name covalent compounds, you add prefixes to the beginning of their names depending on what the subscript is of each element. The prefixes are:
1: Mono
2: Di
3: Tri
4: Tetra
5: Penta
6: Hexa
7: Hepta
8: Octa
9: Nona
10: Deca
For example, since the first one is Phopsphorus with a 2 next to it, you add the prefix Di to it.
If the first element in the compound only has one, meaning no number next to it, you do not say mono. This is why we just say "Carbon" for the second one instead of "Monocarbon."
Finally, you always have to end the second element in the compound with "ide." So, "chlorine" becomes "chloride," "oxygen" becomes "oxide," and so on.
During the transamination process, the enzyme transaminases use Pyridoxal pyrophosphate as a cofactor.
All transamination reactions, as well as several amino acid oxylation and deamination processes, involve the coenzyme pyridoxal phosphate. The aminotransferase enzyme's epsilon-amino group of a particular lysine group forms a Schiff-base bond with the aldehyde group of pyridoxal phosphate.
The epsilon-amino group of the lysine residue in the active site is replaced by the alpha-amino group of the amino acid substrate. The ensuing intermediate, a quinoid, undergoes deprotonation to become an aldimine, which is then protonated to become a ketimine by accepting a proton in a different position. Ketamine undergoes hydrolysis, leaving the amino group on the protein complex intact.
Know more about Pyridoxal pyrophosphate at:
brainly.com/question/14117818
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The known constant of water is that it freezes at 0 degrees Celsius and boils at 100 degrees Celsius. it can be determined that because of the solute added the water becomes a compound meaning it no longer has the constants of pure water hence the difference in freezing point
