Answer:
3.15 × 10⁻⁶ mol H₂/L.s
1.05 × 10⁻⁶ mol N₂/L.s
Explanation:
Step 1: Write the balanced equation
2 NH₃ ⇒ 3 H₂ + N₂
Step 2: Calculate the rate of production of H₂
The molar ratio of NH₃ to H₂ is 2:3. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of H₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 3 mol H₂/2 mol NH₃ = 3.15 × 10⁻⁶ mol H₂/L.s
Step 3: Calculate the rate of production of N₂
The molar ratio of NH₃ to N₂ is 2:1. Given the rate of decomposition of NH₃ is 2.10 × 10⁻⁶ mol/L.s, the rate of production of N₂ is:
2.10 × 10⁻⁶ mol NH₃/L.s × 1 mol N₂/2 mol NH₃ = 1.05 × 10⁻⁶ mol N₂/L.s
Answer:
Arsenic.
Explanation:
Hello there!
In this case, since insecticides are substances that act as poisons to get rid of insects in order to prevent their presence and/or reproduction in houses, companies, crops and others, a substance that has been widely used is the metalloid arsenic due to its direct affection of the insect's body (movement, performance, cellular functions).
In addition, high levels of arsenic in food could cause arsenic poisoning in humans as well, that is why such practice must be properly performed and by using the correct security protocol.
Best regards!
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.
Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
The hydronium ion concentration at pH = 5.
![5=-\log [H_3O^+]](https://tex.z-dn.net/?f=5%3D-%5Clog%20%5BH_3O%5E%2B%5D)
..............(1)
The hydronium ion concentration at pH = 3.
![3=-\log [H_3O^+]](https://tex.z-dn.net/?f=3%3D-%5Clog%20%5BH_3O%5E%2B%5D)
................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
![\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_3O%5E%2B%5D_%7Boriginal%7D%7D%7B%5BH_3O%5E%2B%5D_%7Bfinal%7D%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-5%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D%3D%5Cfrac%7B1%7D%7B100%7D)
![100\times [H_3O^+]_{original}=[H_3O^+]_{final}](https://tex.z-dn.net/?f=100%5Ctimes%20%5BH_3O%5E%2B%5D_%7Boriginal%7D%3D%5BH_3O%5E%2B%5D_%7Bfinal%7D)
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.