Answer:
Use a ratio of 0.44 mol lactate to 1 mol of lactic acid
Explanation:
John could prepare a lactate buffer.
He can use the Henderson-Hasselbalch equation to find the acid/base ratio for the buffer.
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\3.5 = 3.86 + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}\\\\\log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 3.5 - 3.86 = -0.36\\\\\dfrac{\text{[A$^{-}$]}}{\text{[HA]}} = 10^{-0.36} = \mathbf{0.44}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C3.5%20%3D%203.86%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%5C%5C%5C%5C%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%203.5%20-%203.86%20%3D%20-0.36%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%3D%2010%5E%7B-0.36%7D%20%3D%20%5Cmathbf%7B0.44%7D)
He should use a ratio of 0.44 mol lactate to 1 mol of lactic acid.
For example, he could mix equal volumes of 0.044 mol·L⁻¹ lactate and 0.1 mol·L⁻¹ lactic acid.
True energy passes from animal to animal or plant to plant
Answer:
Molecular formula = C₁₂H₁₂O₄
Empirical formula is C₃H₃O.
Explanation:
Given data:
Mass of C = 91.63 g
Mass of H = 7.69 g
Mass pf O = 40.81 g
Molar mass of compound = 220 g/mol
Empirical formula = ?
Molecular formula = ?
Solution:
Number of gram atoms of H = 7.69 / 1.01 = 7.61
Number of gram atoms of O = 40.81 / 16 = 2.55
Number of gram atoms of C = 91.63 / 12 = 7.64
Atomic ratio:
C : H : O
7.64/2.55 : 7.61 /2.55 : 2.55/2.55
3 : 3 : 1
C : H : O = 3 : 3 : 1
Empirical formula is C₃H₃O.
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
Empirical formula mass = 3×12+ 3×1.01 +16 = 55.03
n = 220 / 55.03
n = 4
Molecular formula = 4 (empirical formula)
Molecular formula = 4 (C₃H₃O)
Molecular formula = C₁₂H₁₂O₄
Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
We know that,
The relation between the 
for an ideal gas are :
As we are given :
Now we have to calculate the entropy change of the gas.
(b) As we know that, the work done for isochoric (constant volume) is equal to zero. 
(C) Heat during the process will be,
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Answer:
C₃H₆O₂
Explanation:
Propionic acid is a colorless liquid with a sharp rancid odor. Produces irritating vapor. (USCG, 1999)
