At what temperature (°C) will a 10.00 g sample of neon gas exert a pressure of 96.7 kPa in a
1 answer:
Answer : The temperature of neon gas will be, 221.0 K
Explanation :
To calculate the temperature of neon gas we are using ideal gas equation.
where,
P = pressure of neon gas = 96.7 kPa = 0.955 atm
Conversion used : (1 atm = 101.3 kPa)
V = volume of neon gas = 9.50 L
T = temperature of neon gas = ?
R = gas constant = 0.0821 L.atm/mole.K
w = mass of neon gas = 10.00 g
M = molar mass of neon gas = 20 g/mole
Now put all the given values in the ideal gas equation, we get:
Therefore, the temperature of neon gas will be, 221.0 K
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Answer:
As we are not provided with ¹H-NMR spectrum therefore we can not conclude a single result for the minor chlorination of propane with Molecular formula C₃H₆Cl₂.
Hence, we will draw all the possible isomers with molecular formula C₃H₆Cl₂ and predict the ¹H-NMR spectrum for each of them separately.
Below is the structures of possible minor products and the NMR spectrum respectively.
Possible Product A:
If the minor product formed is 2,2-Dichloropropane then it will give only one singlet peak in NMR spectrum.
Possible Product B:
If the minor product is 1,1-Dichloropropane then it will exhibit two triplets and one multiplet as shown in figure.
Possible Product C:
If the minor product is 1,2-Dichloropropane then it will give four peaks in NMR. The extra peak is formed due to two enantiomers about the chiral carbon.
Possible Product D:
If the minor product is 1,3-Dichloropropane then it will exhibit one triplet and one quintet signal as shown in figure.
