Which of these solids are most likely amorphous solids? Select all that apply. rubber sugar plastic glue stick candle wax bronze
2 answers:
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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 : The energy required to melt 58.3 g of solid n-butane is, 4.66 kJ
Explanation :
First we have to calculate the moles of n-butane.
Given:
Molar mass of n-butane = 58.12 g/mole
Mass of n-butane = 58.3 g
Now put all the given values in the above expression, we get:
Now we have to calculate the energy required.
where,
Q = energy required
= enthalpy of fusion of solid n-butane = 4.66 kJ/mol
n = moles = 1.00 mol
Now put all the given values in the above expression, we get:
Thus, the energy required to melt 58.3 g of solid n-butane is, 4.66 kJ