Answer:
Ca
2+
<K + <Ar<Cl − <S 2−
Explanation:
Ar,K +
,Cl −
,S 2−
,Ca 2+
have the same number of electrons. Their radii would be different because of their different nuclear charges. The cation with the greater positive charge will have a smaller radius because of the greater attraction of the electrons to the nucleus. Anion with the greater negative charge will have the larger radius. In this case, the net repulsion of the electrons will outweigh the nuclear charge and the ion will expand in size. Hence the correct order will be Ca
2+ <K + <Ar<Cl − <S 2−
Answer:
Multiply the number of moles in the product by the molecular weight of the product to determine the theoretical yield.
Explanation:
For example:
If you created 0.5 moles of Aluminium Oxide the molecular weight of Aluminium Oxide is 101.96g/mole, so you would get 50.98g as the theoretical yield.
So multiply,..
101.96x0.5= 50.98
This is the correct way to calculate the theoretical yield
......
Hello! The answer is D
A good note for these is when there are three elements, one being a singular element and another a compound and there’s a single switch, this could show a single-displacement
Have a good day gamer.
Answer:
Four possible isomers (1–4) for the natural product essramycin. The structure of compound 1 was attributed to essramycin by 1H NMR, 13C NMR, HMBC, HRMS, and IR experiments.
Explanation:
Three synthetic routes were used to prepare all four compounds (Figure 2A). All three reactions utilize 2-(5-amino-4H-1,2,4-triazol-3-yl)-1-phenylethanone (5) as the precursor, whereas each uses different esters (6–8) to construct the pyrimidinone ring. Isomer 1 was prepared by reaction A, which used triazole 5 and ethyl acetoacetate (6) in acetic acid. This was the reaction used in syntheses of essramycin by the Cooper and Moody laboratories.3,4 Reaction B produced compound 2 (minor product) and compound 3 (major product), which were separated chromatographically. This reaction allowed reagent 5 to react with ethyl 3-ethoxy-2-butenoate (7) in the presence of sodium in methanol, under reflux for 24 h. Compound 4 was prepared by reaction C, which was obtained by reflux of 5 and methyl 2-butynoate (8) in n-butanol.
2.49 x 10^46 is the answer