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AlexFokin [52]
3 years ago
6

Draw the structure of the compound identified by the following simulated 1H and 13C NMR spectra. The molecular formula of the co

mpound is C10H12O. (Blue numbers next to the lines in the 1H NMR spectra indicate the integration values.)

Chemistry
1 answer:
Luba_88 [7]3 years ago
6 0

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

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A solution of sodium hydroxide (NaOH) was standardized against potassium hydrogen phthalate (KHP). A known mass of KHP was titra
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Answer:

See explanation below

Explanation:

In order to calculate this, we need to use the following expression to get the concentration of the base:

MaVa = MbVb (1)

We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:

Atomic weights of the elements to be used:

K = 39.0983 g/mol;  H = 1.0078 g/mol;  C = 12.0107 g/mol;  O = 15.999 g/mol

MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol

Now, let's calculate the mole of KHP:

moles = 0.5053 / 204.2189 = 0.00247 moles

With the moles, we also know that:

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Replacing in (1):

n = MbVb

Now, solving for Mb:

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