The nahco3 is the limiting reactant and the hcl is the excess reactant in this experiment. determine the theoretical yield of th
1 answer:
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Complete Question
The complete question is shown on the first uploaded image
Answer:
a) Splitting pattern for Ha= 2+1 , Triplet
For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3
Splitting pattern for Hb and Hc= 3+1 , Quartet
For Proton Hd, number of non equivalent protons n= 0
Splitting pattern for Hd= 0+1 , Singlet
b) Splitting pattern for Ha= 2+1 , Triplet
For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3
Splitting pattern for Hb and Hc= 3+1 , Quartet
For Proton Hd, number of non equivalent protons n= 0
Splitting pattern for Hd= 0+1 , Singlet
c) The IUPAC name is Butan-2-ol
Explanation:
Considering the first question the rule used for prediction of splitting pattern is n+1 (Pascal's Triangle rule), where n is number of H atom on the adjacent carbon which are non equivalent.
According to that for molecule 1 as shown on the second uploaded image
For Proton Ha, number of non equivalent protons n= 2
Splitting pattern for Ha= 2+1 , Triplet
For Proton Hb and Hc both are equivalent to each other. non equivalent protons n= 3
Splitting pattern for Hb and Hc= 3+1 , Quartet
For Proton Hd, number of non equivalent protons n= 0
Splitting pattern for Hd= 0+1 , Singlet
Considering the second question for Molecule 2 as shown on the third uploaded image
For Proton Ha, number of non equivalent protons n= 1
Splitting pattern for Ha= 1+1=2 , Doublet
For Proton Hb, number of non equivalent protons n= 3
Splitting pattern for Hb= 3+1=4 , Quartet
For Proton Hc, number of non equivalent protons n= 3
Splitting pattern for Hc= 3+1=4 , Quartet
For Proton Hd, number of non equivalent protons n= 1
Splitting pattern for Ha= 1+1=2 , Doublet
Considering the third question
The name of the given molecule is gotten according to longest carbon chain = 4 (Prefix 'Butan')
Functional group = -OH (Suffix 'ol') at C-2
The IUPAC name is Butan-2-ol
Answer:
1.13 × 10⁶ g
Explanation:
Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.
Al³⁺ + 3 e⁻ → Al
We can establish the following relations:
- 1 Ampere = 1 Coulomb / second
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons circulate
- The molar mass of Al is 26.98 g/mol.
The mass of aluminum produced under these conditions is: