(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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The littoral zone of a lake is the area closest to the shore. It has very little biological activity but includes a lot of oxygen. The water in the lake's littoral zone is freshwater, free of living organisms such as plants and fish.
For the answer to the question above, <span>Hydrophobic regions and hydrophilic regions in the molecules of the b-globin. The replacement causes these hemoglobin molecules to be stickies which gives the cell its sickle shape.
I hope this helps. Have a nice day!</span>
Answer:
You can
Explanation:
The concentration doesn't change when its equals out.
Answer:
a) After the balloon inflated after 440 uL of dropwise due to the reaction of 1-Decene and the solution in the conical vial. b) ⇒ 16 c) No was not the limiting reactant.
Explanation:
Generally, hydrogenation is the chemical reaction between a compound or element and molecular hydrogen in the presence of catalysts such as platinum.
a) After the balloon inflated after 440 uL of dropwise 1-Decene solution was added due to the reaction between 1-Decene and the solution in the conical vial.
b) ⇒ 16
c) was not the limiting reactant based on the mol to mol ratio of and decane which is 1:1. Therefore, if 0.8 mol of decane was produced then 0.8 mol of would also be produced.