Given 7.20 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100
1 answer:
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Answer : The final temperature of the metal block is,
Explanation :
As we know that,
.................(1)
where,
q = heat absorbed or released
= mass of aluminum = 55 g
= mass of water = 0.48 g
= final temperature = ?
= temperature of aluminum =
= temperature of water =
= specific heat of aluminum =
= specific heat of water=
Now put all the given values in equation (1), we get
Thus, the final temperature of the metal block is,
Answer: 4-allylanisole
Explanation: The doublets behind the 7 ppm belongs to the
para-substituted benzene ring. The three single-proton multi-plets around 5−6 ppm predicts that there has to be a single subsituted alkene group
A single plus a doublet around 3-4 ppm belongs to CH3 and CH2 Groups as they could be attached to the subsituted alkene group.
Moreover the interpretation of the NMR that there is no peak with a higher intensity for >180 ppm represents an absence of Carbonyl group.
The Predicted Number is attached from a chemical database along with their peaks information
