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Alexeev081 [22]
3 years ago
11

Anyone can help me please

Chemistry
1 answer:
Vlad1618 [11]3 years ago
5 0

x=3

Use the pythagorean theorem:

a^2 + b^2 = c^2

4^2 + x^2 = 5^2

16 + x^2 = 25

x^2 = 25-16

x^2 = 9

√x^2 = √9

x = 3

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Since he was able to drag the keys along the table by moving the magnet, it shows that magnetic forces acts a distance.

Magnetism is a property exhibited by a wide range of materials.

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Lipid A, present as part of a Lipopolysaccharide complex in the outer membrane of gram negative bacteria cell walls, should be the answer, but it does not appear in the options included.  As option C states “Amino terminal triglyceride”, and triglycerides are lipids, then we could explore this option. However, nothing is said about lipoproteins linked to Lipid A (that acts as an endotoxin) to mention carboxyl- or amino- terminals (options A and C), so I would consider the core oligosaccharide option as more probable (see below*).

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Murein and peptidoglycan are names used to refer bacterial cell walls main component.  This complex is mainly composed by disaccharide units composed of alterning N-acetyl-muramic acid (that contains the same structure of N-acetylglucosamine, plus a tetrapeptide) and N-acetyl-glucosamine, which  form the backbone of the wall. These are responsible for the strength and shape of the cell.  In Gram negative bacteria, an outer layer called outer membrane, as it contains an important amount of lipids, linked to other molecules, is also present.  There, Lipid A is associated to a core oligosaccharide, and subsequently to the Antigen O (polysaccharide) forming Liposaccharides, wich stabilize and give strength to gram negative cell walls.  

*Lipid A is the main responsible molecule for toxicity in these cell walls. As in question answer options, is included in B option “Oligosaccharide core”, which is closely linked to Lipid A, it could be the option to choose. Moreover, oligo saccharides are involved in toxicity responses in several microorganisms .

6 0
3 years ago
Read 2 more answers
How to solve for K when given your anode and cathode equations and voltage
Aleks04 [339]

Answer:

See Explanation

Explanation:

In thermodynamics theory the Free Energy (ΔG) of a chemical system is described by the expression ΔG = ΔG° + RTlnQ. When chemical system is at equilibrium ΔG = 0. Substituting into the system expression gives ...

0 = ΔG° + RTlnKc, which rearranges to ΔG° = - RTlnKc.  ΔG° in electrochemical terms gives ΔG° = - nFE°, where n = charge transfer, F = Faraday Constant = 96,500 amp·sec and E° = Standard Reduction Potential of the electrochemical system of interest.

Substituting into the ΔG° expression above gives

-nFE°(cell) = -RTlnKc => E°(cell) = (-RT/-nF)lnKc = (2.303·R·T/n·F)logKc

=> E°(cell) = (0.0592/n)logKc = E°(Reduction) - E°(Oxidation)

Application example:

Calculate the Kc value for a Zinc/Copper electrochemical cell.

Zn° => Zn⁺² + 2e⁻  ;    E°(Zn) = -0.76 volt  

Cu° => Cu⁺² + 2e⁻ ;    E°(Cu) =  0.34 volt

By natural process, charge transfer occurs from the more negative reduction potential to the more positive reduction potential.

That is,

           Zn° => Zn⁺² + 2e⁻ (Oxidation Rxn)

Cu⁺² + 2e⁻ => Cu°             (Reduction Rxn)

E°(Zn/Cu) = (0.0592/n)logKc

= (0.0592/2)logKc = E°(Cu) - E°(Zn) = 0.34v - (-0.76v) = 1.10v

=> logKc = 2(1.10)/0.0592 = 37.2

=> Kc = 10³⁷°² = 1.45 x 10³⁷

3 0
3 years ago
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