Answer:The crystal structures of five 6-mercaptopurine derivatives, viz. 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(3-methoxyphenyl)ethan-1-one (1), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-methoxyphenyl)ethan-1-one (2), C16H14N4O3S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-chlorophenyl)ethan-1-one (3), C15H11ClN4O2S, 2-[(9-acetyl-9H-purin-6-yl)sulfanyl]-1-(4-bromophenyl)ethan-1-one (4), C15H11BrN4O2S, and 1-(3-methoxyphenyl)-2-[(9H-purin-6-yl)sulfanyl]ethan-1-one (5), C14H12N4O2S. Compounds (2), (3) and (4) are isomorphous and accordingly their molecular and supramolecular structures are similar. An analysis of the dihedral angles between the purine and exocyclic phenyl rings show that the molecules of (1) and (5) are essentially planar but that in the case of the three isomorphous compounds (2), (3) and (4), these rings are twisted by a dihedral angle of approximately 38°. With the exception of (1) all molecules are linked by weak C—H⋯O hydrogen bonds in their crystals. There is π–π stacking in all compounds. A Cambridge Structural Database search revealed the existence of 11 deposited compounds containing the 1-phenyl-2-sulfanylethanone scaffold; of these, only eight have a cyclic ring as substituent, the majority of these being heterocycles.
Keywords: crystal structure, mercaptopurines, supramolecular structure
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Chemical context
Purines, purine nucleosides and their analogs, are nitrogen-containing heterocycles ubiquitous in nature and present in biological systems like man, plants and marine organisms (Legraverend, 2008 ▸). These types of heterocycles take part of the core structure of guanine and adenine in nucleic acids (DNA and RNA) being involved in diverse in vivo catabolic and anabolic metabolic pathways.
6-Mercaptopurine is a water insoluble purine analogue, which attracted attention due to its antitumor and immunosuppressive properties. The drug is used, among others, in the treatment of rheumathologic disorders, cancer and prevent
Step-by-step explanation:
If you cannot read it, let me know.
Given compound inequality 89 – 9.8t < 10.6 or 89 – 9.8t > 20.4.
Let us solve them one by one.
89 – 9.8t < 10.6
Subtracting 89 from both sides, we get
89-89 – 9.8t < 10.6 - 89
-9.8t < -78.4
Dividing both sides by -9.8, we get
t > 8.
<em>Note: Inequality sign get flip on dividing both side by a negative number.</em>
89 – 9.8t > 20.4
Subtracting 89 from both sides, we get
89-89 – 9.8t < 20.4 - 89
-9.8 t < - 68.6.
Dividing both sides by -9.8, we get
t < 7
<em>Note: Inequality sign get flip on dividing both side by a negative number.</em>
<h3><em> The solution to the inequality 89 – 9.8t < 10.6 is </em>t > 8. </h3><h3> The solution to the inequality 89 – 9.8t > 20.4 is t < 7.</h3>
When the projectile hits the ground:
89 – 9.8t = 0
Subtracting 89 from both sides, we get
89 - 89 – 9.8t = 0 -89.
-9.8t = -89.
Dividing both sides by 9.8, we get
<h3>to the nearest whole second t is about 9 seconds.</h3><h3>Therefore, variable solution set is {t < 7, t > 8, t = 9}.</h3><h3 />
Answer:
about 48 adults , because you have to divide the 477 by 10
Step-by-step explanation:
I got this question on my test and i chose this and i got it right.
I hope this helps!! :)
Answer:
Step-by-step explanation:
Sample standard deviation is found by steps below.
<u>Find the mean:</u>
- μ = (9 + 15 + 13 + 9 + 15)/5 = 12.2
<u>Find deviations from the mean and their squares:</u>
- 9 - 12.2 = - 3.2 ⇒ (-3.2)² = 10.24
- 15 - 12.2 = 2.8 ⇒ 2.8² = 7.84
- 13 - 12.2 = 0.8 ⇒ 0.8² = 0.64
- 9 - 12.2 = - 3.2 ⇒ (-3.2)² = 10.24
- 15 - 12.2 = 2.8 ⇒ 2.8² = 7.84
<u>Sum the squares:</u>
- 10.24*2 + 7.84*2 + 0.64 = 36.8
<u>Divide by N - 1 (N - number of samples):</u>
<u>Square root is the sample standard deviation:</u>
