Answer:
The answer to your question is None of your answers is correct, maybe the data are wrong.
Explanation:
Data
Concentration 1 = C1 = 1 M
Volume 2 = 5 ml
Concentration 2 = 0.05 M
Volume 1 = x
To solve this problem use the dilution formula
Concentration 1 x Volume 1 = Concentration 2 x Volume 2
Solve for Volume 1
Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1
Substitution
Volume 1 = (0.05 x 5) / 1
Simplification
Volume 1 = 0.25/1
Result
Volume 1 = 0.25 ml
Answer:
H₂SO₄ (aq) + H₂O (l) → HSO₄⁻ (aq) + H₃O⁺ (aq)
HSO₄⁻ (aq) + H₂O (l) ⇄ SO₄⁻²(aq) + H₃O⁺ (aq) Ka
Explanation:
The sulfuric acid is a dyprotic acid.
It is a considered a strong acid but only the first deprotonation is strong.
Second deprotonation is totally weak
That's why we have equilibrium when it release the second proton.
Since a deprotonation is treated for a weak acid, the acid is considered not to completely dissociate, that's why the equilibrium
The equilibrium constant K and the forward rate constant
k1 and backward rate constant k2 has the following relation:
K = k1 / k2
So from the equation, we can say that yes it is possible
to have large K even if k1 is small given that k2 is very small compared to k1:
(k2 very less than 1)
<span>k2 << k1</span>
Since the temperature is constant, therefore, this problem can be solved based on Boyle's law.
Boyle's law states that: " At constant temperature, the pressure of a certain mass of gas is inversely proportional to its pressure".
This can be written as:
P1V1 = P2V2
where:
P1 is the initial pressure = 1 atm
V1 is the initial volume = 3.6 liters
P2 is the final pressure = 2.5 atm
V2 is the final volume that we need to calculate
Substitute with the givens in the above mentioned equation to get the final volume as follows:
P2V1 = P2V2
1(3.6) = 2.5V2
3.6 = 2.5V2
V2 = 3.6 / 2.5 = 1.44 liters
Answer:
Non competitive inhibition
Explanation:
Hello,
During enzymatic catalysis, the active sites could be occupied by the very same products' molecules turning out into an inhibition (the reaction starts to slow down since to active places are available for the reagents to react). Nonetheless this inhibition is not competitive as long as the product does not react due to the active sites it is occupying.
Best regards.
