I have attached the picture ot the H<span>aworth structure of glucose.
You can count in total 6 carbon atoms but only 5 are in the ring portion of the structure.
Therefore, the answer is 5.
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Answer:
Here's what I get
Explanation:
CH₃CH₂CH₂CH₂CH₂CH₃ — hexane
CH₂=CHCH₂CH₂CH₂CH₃ — hex-1-ene is the preferred IUPAC name (PIN). 1-Hexene is accepted
CH₃C≡CCH₃ — but-2-yne (PIN); 2-butyne is accepted
CH₃CH(CH₃)CH₂CH₂CH₃ — 2-methylpentane
CH₃CH₂CHCICH₂CH₃ — 3-chloropentane
Answer:
V= 12mL
Explanation:
you had the right idea with your Significant figures however, when we divide we see that it requires 2 significant figures as our least amount. this is because when looking at our division, 62 has 2 sig. fig. while 5.35 has a total 3. when looking at your answer we see that you had a total of 3 sig. figures. so in actuakity you had to round up to 12 and not to the tenths because the decimal makes .6 count as your third sig fig.
