<em>Suppose a mixture contains two isotopes of fictitious element "A" shown below. 45A and 46A. In the mixture 33% are 45A and 67% are 46A. What is the relative atomic mass for element "A"? Show your work below.</em>
The element A, with 2 isotopes, 45A (33%) and 46A (67%), has a relative atomic mass of 45.67 u.
The fictitious element A has 2 isotopes. 45A with an atomic mass of 45 u and an abundance of 33%, and 46 A with an atomic mass of 46 u and an abundance of 67%. The relative atomic mass (am) for element A is a weighted average that takes into account the mass of each isotope (
) and its abundance (
).
The element A, with 2 isotopes, 45A (33%) and 46A (67%), has a relative atomic mass of 45.67 u.
You can learn more about relative atomic mass here: brainly.com/question/21690748
Answer:
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*).
Explanation:
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
.
Ans: Final volume = 25.0 ml
<u>Given:</u>
Initial volume V1 = 50.0 ml
Initial pressure P1 = 20.0 atm
Final pressure P2 = 40.0 atm
<u>To determine:</u>
The final volume V2
<u>Explanation:</u>
Ideal gas equation: PV = nRT
under constant temperature, T and number of moles n we have:
PV = constant
or, P1V1 = P2V2
V2 = P1V1/P2 = 20*50/40 = 25 ml.
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