Answer:
1.65 L
Explanation:
The equation for the reaction is given as:
A + B ⇄ C
where;
numbers of moles = 0.386 mol C (g)
Volume = 7.29 L
Molar concentration of C = 
= 0.053 M
A + B ⇄ C
Initial 0 0 0.530
Change +x +x - x
Equilibrium x x (0.0530 - x)
![K = \frac{[C]}{[A][B]}](https://tex.z-dn.net/?f=K%20%3D%20%5Cfrac%7B%5BC%5D%7D%7B%5BA%5D%5BB%5D%7D)
where
K is given as ; 78.2 atm-1.
So, we have:
![78.2=\frac{[0.0530-x]}{[x][x]}](https://tex.z-dn.net/?f=78.2%3D%5Cfrac%7B%5B0.0530-x%5D%7D%7B%5Bx%5D%5Bx%5D%7D)
Using quadratic formula;
where; a = 78.2 ; b = 1 ; c= - 0.0530
= 
or 
= 
or 
= 0.0204 or -0.0332
Going by the positive value; we have:
x = 0.0204
[A] = 0.0204
[B] = 0.0204
[C] = 0.0530 - x
= 0.0530 - 0.0204
= 0.0326
Total number of moles at equilibrium = 0.0204 + 0.0204 + 0.0326
= 0.0734
Finally, we can calculate the volume of the cylinder at equilibrium using the ideal gas; PV =nRT
if we make V the subject of the formula; we have:
where;
P (pressure) = 1 atm
n (number of moles) = 0.0734 mole
R (rate constant) = 0.0821 L-atm/mol-K
T = 273.15 K (fixed constant temperature )
V (volume) = ???
V = 1.64604
V ≅ 1.65 L
Answer: Bacterial species where observed Typical number on cell Distribution on cell surface
Escherichia coli (common pili or Type 1 fimbriae) 100-200 uniform
Neisseria gonorrhoeae 100-200 uniform
Streptococcus pyogenes (fimbriae plus the M-protein) ? uniform
Pseudomonas aeruginosa 10-20 polar
Explanation:
Pili are structures that extend from the surface of some bacterial cells.
These are hollow, non-helical, filamentous appendages.
Hope it helps you
B. White Dwarf.
<h3>Explanation</h3>
The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.
As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:
Chandrasekhar Limit: 
.
A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 
, which is much smaller than the Chandrasekhar Limit.
As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.
