This problem could be solved easily using the Henderson-Hasselbach equation used for preparing buffer solutions. The equation is written below:
pH = pKa + log[(salt/acid]
Where salt represents the molarity of salt (sodium lactate), while acid is the molarity of acid (lactic acid).
Moles of salt = 1 mol/L * 25 mL * 1 L/1000 mL = 0.025 moles salt
Moles of acid = 1 mol/L* 60 mL * 1 L/1000 mL = 0.06 moles acid
Total Volume = (25 mL + 60 mL)*(1 L/1000 mL) = 0.085 L
Molarity of salt = 0.025 mol/0.085 L = 0.29412 M
Molarity of acid = 0.06 mol/0.085 L = 0.70588 M
Thus,
pH = 3.86 + log(0.29412/0.70588)
pH = 3.48
Answer:
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Explanation:
The HI donates a proton to the water, converting it to a hydronium ion
HI(aq) + H₂O(ℓ) ⟶ H₃O⁺(aq) + I⁻(aq)
Thus, the HI is behaving like a Brønsted acid.
The circulatory system is the system that transports blood through out the system❤️
So,
Formate has a resonating double bond.
In molecular orbital theory, the resonating electrons are actually delocalized and are shared between the two oxygens. So the carbon-oxygen bonds can be described as 1.5-bonds (option B). I'm not sure if option C is correct, however, because the likelihood of both delocalized electrons being in the area of one oxygen atom is less than 50%.<span />
Answer:
The answers to the question are
1. 2nd and above order order
2. 2nd order
3. 1/2 order
4. 1st order
5. 0 order
Explanation:
We have 
1. For nth order reaction half life 
∝ ![\frac{1}{[A_{0} ]^{n-1} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_%7B0%7D%20%5D%5E%7Bn-1%7D%20%7D)
Therefore for a 0 order reaction increasing concentration of the reactant there will increase
First order reaction is independent [A₀].
Second order reaction [A₀] decrease, increase.
Similarly for a third order reaction
1. 2nd order
2. 2nd order reaction
3. Order of reaction is 1/2.
4. 1st order reaction.
5. Zero order reaction.
