What are periods and groups on the periodic table?
2 answers:
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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.
If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.
<h3>What is Stoichiometry ?</h3>Stoichiometry helps us use the balanced chemical equation to measure quantitative relationships and it is to calculate the amounts of products and reactants that are given in a reaction.
<h3>What is Balanced Chemical Equation ?</h3>The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now we have to write the balanced equation
Mg + 2HNO₃ → Mg(NO₃)₂ + H₂
According to Stoichiometry
= 3.3 g H₂
Thus from the above conclusion we can say that If 40.0 grams of magnesium is reacted with an excess of nitric acid. 3.3 g of hydrogen gas will be produced.
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Answer: The volume for 0.850 mol of from a
solution is 1700 mL.
The volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.
Explanation:
Molarity is the number of moles of solute present in a liter of solution.
- As given moles of
are 0.850 mol and molarity of
Therefore, the volume for 0.850 mol of from a
solution is 1700 mL.
- As given mass of LiOH are 30.0 g from a 2.70 M LiOH (molar mass = 23.95 g/mol) solution. Hence, its number of moles are calculated as follows.
So, volume for LiOH solution is calculated as follows.
Therefore, volume of 30.0 g of LiOH from a 2.70 M LiOH solution is 464 mL.