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lilavasa [31]
2 years ago
11

1. why is HCl not a good choice as the acid to catalyze a dehydration reaction?

Chemistry
1 answer:
Len [333]2 years ago
7 0

Answer:

1) HCl contains the Cl^- which is a good nucleophile

2) 2-methyl-2- heptanol > 2-heptanol > 1-heptanol

3) see image attached

Explanation:

If the dehydration of alcohols is carried out using HCl, the chloride ion which is a good nucleophile will attack the substrate to yield an undesirable product.

The dehydration of alcohols is an E1 reaction. Recall that the ease of E1 reaction increases in the order 3°> 2°> 1°. Hence, 2-methyl-2- heptanol forms a tertiary carbocation intermediate during dehydration and has the greatest ease of dehydration.

The three products formed during the dehydration of 3,3-dimethyl-2-butanol are shown in the image attached. Two out of the three are formed by rearrangement reactions.

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Which of the following statements is/are correct?
7nadin3 [17]

Answer:

All are correct

Explanation:

1) The angular momentum quantum number, l, are the subshells within a shell (principle quantum number) it talks about the "form" of an orbital, the number itself tells you about the number of angular nodes (a plane without electronic density). It starts at l=0 where you don't see any nodes and it takes the form of an sphere, and we knowing it bu another name an s-orbital. It takes values up to n-1.

l=0 (sphere - s-orbital)

l=1 (p-orbital)

l=2 (d-orbital)

2) The magnetic quatum number, ml relates to the number of orbitals within a subshell then it is related with l, taking values form -l to l incluing 0.

For l=0 (s-orbital) ml=0

For l=1 (p-orbital) ml=1,0,-1

For l=2 (d-orbital) ml=2,1,0,-1,-2

3) In every shell we are restricted by the total number of nodes of any orbital. Then if we want a d-orbital with l=3 we need at least 3 plane nodes only achievable with n=3 at least.

8 0
3 years ago
For water at 30C and 1 atm: a= 3.04x10^-4 k^-1 , k =4.52x10^-5 atm ^-1 = 4.46 x10^-10 m^2/N, cpm= 75.3j/9molk), Vm =18.1cm^3/mol
Lapatulllka [165]

Answer:

C_{vm} of water at 30C and 1 atm is 256.834 J/mol·K.

Explanation:

To solve the question, we note the Maxwell relation such as

C_{pm}-C_{vm}=\frac{9T\alpha ^2 V }{K}

Where:

C_{pm} = Specific heat of gas at constant pressure = 75.3 J/mol·K

C_{vm} = Specific heat of gas at constant volume = Required

T = Temperature = 30 °C = 303.15 K

α = Linear expansion coefficient = 3.04 × 10⁻⁴ K⁻¹

K = Volume comprehensibility = 4.52 × 10⁻⁵ atm⁻¹

Therefore,

75.3 - C_v = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} }

C_{vm}  = \frac{9\times 303.15 \times (3.04 \times 10^{-1} 1.81  \times 10^{-5}  }{4.52 \times 10^{-5} } - 75.3 = 256.834 J/mol·K.

8 0
3 years ago
4K + O2 + 2K20<br> Which statement is true?
Jet001 [13]

Answer:

yes 4K + O2 ------>2K20 is true.

5 0
2 years ago
Is the chemical equation balanced?<br><br> Yes or no
Strike441 [17]
I think it is not so no
6 0
2 years ago
Read 2 more answers
What is the percentage of lithium in lithium carbonate (Li2CO3)?
ale4655 [162]
Molar mass Li2CO3 = 73.89 g/mol
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% LI = 13.88/73.89*100 = 18.78% perfectly correct.
6 0
3 years ago
Read 2 more answers
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