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Softa [21]
3 years ago
7

Consider the diagram of the basic structure of a bacterium. A diagram of a bacterium is labeled. Part A is the cell membrane, pa

rt B is the flagellum, part C is the nucleoid, and part D is the pilus. Which of the labeled structures in the image allows bacteria to exchange genetic material and thus evolve to be better pathogens? A B C D HURRY!!! 100 POINTS!!!!
Chemistry
2 answers:
Vesnalui [34]3 years ago
7 0

Answer:

It is part c

Explanation:

butalik [34]3 years ago
7 0

Answer:

the answer is D

Explanation:

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I think this is Chemistry but it might be physics
arlik [135]

Answer:

No it is not a physics because in this there is about classification of element so it is in chemistry that's why it is not physics.

8 0
2 years ago
Find the pH of the equivalence point(s) and the volume (mL) of 0.0372 M NaOH needed to reach the point(s) in titrations of(a) 42
Neko [114]

The pH of the equivalence points is 8.54

Concept of pH

A solution's acidity or alkalinity can be determined based on the concentration of hydrogen ions in the solution, or pH. Acidic aqueous solutions at 25 °C have a pH under 7, while basic or alkaline aqueous solutions have a pH above 7. Since the concentration of H3O+ is equal to the concentration of OH in pure water, a pH level of 7.0 at 25°C is referred to as "neutral". Strong bases may have a pH above 14, while very strong acids may have a pH below 14.

0.0520 M CH3COOH in 42.2 mL of moles is as follows:

2.194x103 mol CH3COOH = 0.0422L (0.0520mol / L)

that react with NaOH, resulting in:

NaOH + CH3COOH = CH3COO + Na+ + H2O

Thus, 1 mole of acetic acid and 1 mole of NaOH react.

As a result, 2.194x103 mol of NaOH are required to reach the equivalence point in volume:

To attain the equivalency point, 2.194x103 mol (1L / 0.0372mol) = 0.05899L 58.99mL of 0.0372 M NaOH

You will only have CH3COO at the equivalency point because it is in equilibrium with water, so:

H2O(l) + CH3COO(aq) CH3COOH(aq) + OH (aq)

A definition of equilibrium is:

Kb = 5.6x1010 = [OH] / [CH3COO] / [CH3COOH]

2.194x103mol of CH3COO has a molarity of (0.05899L + 0.0422L) = 0.02168M.

Therefore, equilibrium concentrations are:

[CH3COO]=0.02168M-X [CH3COOH]=X [OH]=X

5.6x1010 = [X] [X] / [0.02168M - X] converts to Kb.

1.214x1011 - 5.6x1010X = X2 X2 + 5.6x1010X - 1.214x1011 = 0 Finding the value of X:

False response; there are no negative concentrations. X: -3.48x106

As [OH] = X, [OH] = 3.484x106M, X is 3.484x106.

As 14 = pOH + pH pH = 8.54, pOH = -log [OH], or 5.46.

To know more about pH visit :

brainly.com/question/12546875

#SP4

3 0
1 year ago
Calculate the cell potential, E, for the following reactions at 26.29 °C using the ion concentrations provided. Then, determine
yanalaym [24]

Answer:

I will work only one of the listed equations ... you follow the given example for the remaining reactions. Thank you :-)

Rxn 1: Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s)

a) E(Pt⁺²/Fe°) = - 1.668v

b) Process is Non-spontaneous if E(cell) < 0

Explanation:

Pt°(s) + Fe⁺²(aq) ⇄ Pt⁺²(aq) + Fe°(s) ⇔

Pt°(s)|Pt⁺²[0.057M]║Fe⁺²[0.006M]|Fe°(s)

As written, Pt° is shown undergoing oxidation with Fe⁺² undergoing reduction. Applying the reduction potentials to the analytical equations for E(cell) and ΔG(cell) gives E(Pt/Fe⁺²) < 0 and ΔG(Pt/Fe⁺²) > 0 which indicate a non-spontaneous process. The following supports this conclusion.

E°(Fe⁺²) = -0.44v

E°(Pt⁺²) = +1.20v

E°(Pt/Fe⁺²) =E°(Redn) - E°(Oxidn) =E°(Fe⁺²) - E°(Pt⁺²)

= -0.44v - (+1.20v) = - 1.64v

[Fe⁺²] = 0.0066M

[Pt⁺²] = 0.057M

n = electrons transferred = 2

E(nonstd) = E°(std) - (0.0592/n)logQ);

Q = [Pt⁺²]/[Fe⁺²]

= -1.64v - (0.0592/2)log[0.057M]/[0.006M]v = -1.668v

Also, if ΔG(cell) > 0 => indicates non-spontaneous process

ΔG(Pt/Fe⁺²) = - nFE = -(2)(96,500Coulombs)((-1.664v) > 0 Kj => nonspontaneous rxn. (1 Coulomb-volt = 1 Kilojoule)

7 0
3 years ago
Urea makes up a significant amount of urine, and is the chemical that the body uses to rid cells of unwanted carbon dioxide and
Ira Lisetskai [31]
Urea is highly soluble in water. When it is allowed to dissolve in water in the presence of heat, it will yield ammonia and carbon dioxide. The reaction is shown below:

<span>NH2-CO-NH2 + H2O </span>⇒ 2 NH3 + CO2

As you can observe in the stoichiometric equations, 1 molecule of water can dissolve with 1 mole of urea.
3 0
3 years ago
If you have a pitcher of lemonade and the recipe calls for 2 lemons and you decided to use 4, would this be considered diluted o
pantera1 [17]

Answer: This would be considered concentrated because if you're upping the recipe on your own accord, it would be way more sour, causing the lemonade to be more concentrated. It would be diluted if you added less than 2 lemons.

7 0
3 years ago
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