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Margaret [11]
3 years ago
13

The C2-C3 single bond in butane freely rotates at room temperature, see equation (1) below. In contrast, the C2-C3 single bond i

n the butenyl cation, equation (2), has a high energy barrier for rotation and does not rotate at room temperature. Use structures to explain why the C2-C3 rotational barrier in butenyl cation is so high.
Chemistry
1 answer:
Murrr4er [49]3 years ago
8 0

Answer: Rotation occurs at single bonds that are sigma bonds. Rotational barrier is the amount of  activation energy required to covert rotamer to another by rotation that occurs around the sigma bond(C-C single bond). Due to the presence of steric hindrance that is the nonbonding interaction effects the reactivity of ions and molecules, activation energy increases. So the rotational barrier in butenyl cation is high.

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Where would you find water on the ph scale?
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1 year ago
A solution of malonic acid, H2C3H2O4 , was standardized by titration with 0.1000 M NaOH solution. If 21.17 mL of the NaOH soluti
MrRa [10]

Answer: The molarity of the malonic acid solution is 0.08335 M

Explanation:

H_2C_3H_2O_4 +2NaOH\rightarrow Na_2C_3H_2O_4+2H_2O

To calculate the molarity of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2C_3H_2O_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=?\\V_1=12.70mL\\n_2=1\\M_2=0.1000M\\V_2=21.17mL

Putting values in above equation, we get:

2\times M_1\times 12.70=1\times 0.1000\times 21.17\\\\M_1=0.08335M

Thus the molarity of the malonic acid solution is 0.08335 M

5 0
3 years ago
Answer with explanation
creativ13 [48]
The answer should be:

KOH (aq) + HCl (aq) --> KCl (aq) + H20 (l).

KOH is a base, because OH can accept a H+.
HCl is an acid because it can donate a H+.

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8 0
2 years ago
It is found that a gas undergoes a first-order decomposition reaction. If the rate constant for this reaction is 8.1 x 10-2 /min
kap26 [50]

Answer:

28.43 min

Explanation:

Using integrated rate law for first order kinetics as:

[A_t]=[A_0]e^{-kt}

Where,  

[A_t] is the concentration at time t

[A_0] is the initial concentration

Given that:

The rate constant, k = 8.1\times 10^{-2} min⁻¹

Initial concentration [A_0] = 0.1 M

Final concentration [A_t] = 1.0\times 10^{-2} M

Time = ?

Applying in the above equation, we get that:-

1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}

0.1e^{-8.1\times \:10^{-2}t}=10^{-2}

e^{-8.1\times \:10^{-2}t}=\frac{1}{10}

\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)

t=28.43\ min

3 0
3 years ago
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