Electrostatics is <span> a branch of physics that deals with the phenomena and properties of stationary or slow-moving electric charges. :) Hope this helps </span>
(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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An Arrhenius acid by definition dissociates in water to form H3O+ (or H+) ions while an arrhenius base dissociates in water to form OH- ions.
NH4+(aq) can be categorised as an arrhenius acid since it releases H3O+ ions in aqueous media
NH4+(aq) + H2O (aq) ↔ NH3 (aq) + H3O+(aq)
When it comes into contact with It will turn purple
Answer : The pH of the solution is, 2.67
Explanation :
The equilibrium chemical reaction is:
![Al^{3+}+H_2O\rightarrow Al(OH)^{2+}+H^+](https://tex.z-dn.net/?f=Al%5E%7B3%2B%7D%2BH_2O%5Crightarrow%20Al%28OH%29%5E%7B2%2B%7D%2BH%5E%2B)
Initial conc. 0.450 0 0
At eqm. (0.450-x) x x
As we are given:
![K_a=1.00\times 10^{-5}](https://tex.z-dn.net/?f=K_a%3D1.00%5Ctimes%2010%5E%7B-5%7D)
The expression for equilibrium constant is:
![K_a=\frac{(x)\times (x)}{(0.450-x)}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%280.450-x%29%7D)
Now put all the given values in this expression, we get:
![1.00\times 10^{-5}=\frac{(x)\times (x)}{(0.450-x)}](https://tex.z-dn.net/?f=1.00%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%280.450-x%29%7D)
![x=0.00212M](https://tex.z-dn.net/?f=x%3D0.00212M)
The concentration of
= x = 0.00212 M
Now we have to calculate the pH of solution.
![pH=-\log [H^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH%5E%2B%5D)
![pH=-\log (0.00212)](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%280.00212%29)
![pH=2.67](https://tex.z-dn.net/?f=pH%3D2.67)
Therefore, the pH of the solution is, 2.67