(a) The heat generated in the process is 28 kJ.
(b) The work done in the process is determined as -28 kJ.
(c) The change in the internal energy is 0.
<h3>
Heat of the isothermal compression </h3>
The heat generated in the process is negative done in the process.
W = -PΔV
W = -P(V₂ - V₁)
<h3>From A to B</h3>
W = -P(VB - VA)
W = -11(7 - 12.5)
W = 60.5 L.atm = 60.5 x 101.325 J/L.atm = 6,130.16 J
<h3>From C to D</h3>
W = -25(20.5 - 7)
W = -337.5 L.atm = -34,197.18 J
Total work , w = -34,197.18 J + 6,130.16 J = -28 kJ
q = - w
q = 28 kJ
<h3>Change in internal energy</h3>
ΔE = q + w
ΔE = 28 kJ - 28 kJ = 0
Learn more about change in internal energy here: brainly.com/question/17136958
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Liquid will no pour out. At the bottom the separating funnel consists of a valve it allow the liquid to flow out.The lid is on the top to pour in liquid from the top,it is air tight. Therefore as long as the lid is put, the liquid will not flow out at the bottom. The reason is that ,the incoming air has no way to get into the flask.
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.