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3 years ago

What happens in this reaction? butan-1-amine + CH3I -->

Chemistry

1 answer:

Galina-37 [17]3 years ago

7 0

Here we have to get the product between the reaction of butane-1-amine with methyl iodide (CH₃I).

The reaction between 1 mole of butan-1-amine and 1 mole of methyl iodide produces Methyl-butamine which is a secondary amine.

However, In presence of 2 moles of methyl iodide the reaction proceed to N, N-di-methylbutamine. The reaction is shown in the figure.

This is one of the effective reaction method to generate secondary and tertiary amine from primary amine.

The primary amine reacts with alkyl iodide to form secondary to tertiary amine. The final product depends upon the quantity of the alkyl iodide present in the reaction.

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Why would you ask a question if you didnt have a question?

Just get someone to report it, and itll be deleted

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2 years ago

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The notation for the nuclide 13755Cs gives information about

k0ka [10]

Answer is (3) both mass number and atomic number.

The notation is ₅₅¹³⁷Cs. The Cs represents the chemical symbol of Caesium element. The subscript number at the left hand side of the symbol indicates the atomic number. Hence, atomic number of Cs is 55. The superscript number at the left hand side of the symbol shows the mass number. Hence, the mass number of the Cs is 137.

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3 years ago

The gas-phase reaction follows an elementary rate law and is to be carried out first in a PFR and then in a separate experiment

astraxan [27]

Answer:

The activation energy is $=8.1\,kcal\,mol^{-1}$

Explanation:

The gas phase reaction is as follows.

$A \rightarrow B+C$

The rate law of the reaction is as follows.

$-r_{A}=kC_{A}$

The reaction is carried out first in the plug flow reactor with feed as pure reactant.

From the given,

Volume "V" = $10dm^{3}$

Temperature "T" = 300 K

Volumetric flow rate of the reaction $v_{o}=5dm^{3}s$

Conversion of the reaction "X" = 0.8

The rate constant of the reaction can be calculate by the following formua.

$V= \frac{v_{0}}{k}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]$

Rearrange the formula is as follows.

$k= \frac{v_{0}}{V}[(1+\epsilon )ln(\frac{1}{1-X}-\epsilon X)]............(1)$

The feed has Pure A, mole fraction of A in feed $y_{A_{o}}$ is 1.

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacte.

$=1(2-1)=1$

Substitute the all given values in equation (1)

$k=\frac{5m^{3}/s}{10dm^{3}}[(1+1)ln \frac{1}{1-0.8}-1 \times 0.8] = 1.2s^{-1}$

Therefore, the rate constant in case of the plug flow reacor at 300K is $1.2s^{-1}$

The rate constant in case of the CSTR can be calculated by using the formula.

$\frac{V}{v_{0}}= \frac{X(1+\epsilon X)}{k(1-X)}.............(2)$

The feed has 50% A and 50% inerts.

Hence, the mole fraction of A in feed $y_{A_{o}}$ is 0.5

$\epsilon =y_{A_{o}}\delta$

$\delta$ = change in total number of moles per mole of A reacted.

$=0.5(2-1)=0.5$

Substitute the all values in formula (2)

$\frac{10dm^{3}}{5dm^{3}}=\frac{0.8(1+0.5(0.8))}{k(1-0.8)}=2.8s^{-1}$

Therefore, the rate constant in case of CSTR comes out to be $2.8s^{-1}$

The activation energy of the reaction can be calculated by using formula

$k(T_{2})=k(T_{1})exp[\frac{E}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})]$

In the above reaction rate constant at the two different temperatures.

Rearrange the above formula is as follows.

$E= R \times(\frac{T_{1}T_{2}}{T_{1}-T_{2}})ln\frac{k(T_{2})}{k(T_{1})}$

Substitute the all values.

$=1.987cal/molK(\frac{300K \times320K}{320K \times300K})ln \frac{2.8}{1.2}=8.081 \times10^{3}cal\,mol^{-1}$

$=8.1\,kcal\,mol^{-1}$

Therefore, the activation energy is $=8.1\,kcal\,mol^{-1}$

8 0

3 years ago

The equilibrium constant (Kp) for the decomposition of phosphorus pentachloride (PCl5) to phosphorus trichloride (PCl3) and mole

Oksanka [162]

Answer: The partial pressure of $Cl_2$ is 1.86 atm

Explanation:

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as $K_c$

The given balanced equilibrium reaction is,

$PCl_5(g)\rightleftharpoons PCl_3(g)+Cl_2(g)$

Pressure at eqm. 0.973 atm 0.548atm x atm

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{(p_{PCl_3}\times (p_{Cl_2})}{(p_{PCl_5})}$

Now put all the given values in this expression, we get :

$1.05=\frac{(0.548)\times (x)}{(0.973)}$

By solving the term 'x', we get :

x = 1.86 atm

Thus, the partial pressure of $Cl_2$ is 1.86 atm

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3 years ago

Most continuous spectra are examples of what we also call thermal radiation spectra. Why do we call them "thermal" spectra?

lisov135 [29]

Answer:

They are emitted from heated objects

Explanation:

When objects are heated, they emit light at all wavelengths thereby forming a continuous spectrum. Electromagnetic radiation of all wavelengths and colours are usually represented in such spectrum. A thermal spectrum is quite a simple spectrum since it depends on temperature.

7 0

2 years ago