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3 years ago

Adding more than one equivalent of HCl to pent-1-yne will lead to which product:______.

Chemistry

1 answer:

Ket [755]3 years ago

8 0

Answer:

c. 2,2-dichloropentane.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to firstly draw the structure of the reactant, pent-1-yne:

$CH\equiv C-CH_2-CH_2-CH_2$

Now, we infer the halogen is added to the carbon atom with the most carbon atoms next to it, in this case, carbon #2, in order to write the following product:

$CH\equiv C-CH_2-CH_2-CH_2+2HCl\rightarrow CH_3- CCl_2-CH_2-CH_2-CH_2$

Whose name is 2,2-dichloropentane.

Regards!

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How many carbon atoms are there in a diamond (pure carbon) with a mass of 55 mg ?

kari74 [83]

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Given that a for HCN is 6. 2×10^−10 at 25 °C. What is the value of b for cn− at 25 °C?

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If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).

<h3>What is base dissociation constant? </h3><h3 />

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 6.2× 10^(-10)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{6.2×10^(-10) }

= 1.6× 10^(-5)

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learn more about base dissociation constant :

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xeze [42]

Answer:

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The reaction C 4 H 8 ( g ) ⟶ 2 C 2 H 4 ( g ) C4H8(g)⟶2C2H4(g) has an activation energy of 262 kJ / mol. 262 kJ/mol. At 600.0 K,

ludmilkaskok [199]

Answer: $4.3\times 10^{-13}s^{-1}$

Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $600.0K$ = $6.1\times 10^{-8}s^{-1}$

$K_2$ = rate constant at $775.0$ = $?$

$Ea$ = activation energy for the reaction = 262 kJ/mol = 262000J/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = $600.0K$

$T_2$ = final temperature = $775.0K$

Now put all the given values in this formula, we get

$\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}]$

$\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150$

$(\frac{6.1\times 10^{-8}}{K_2})=141253.8$

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