Explain preparation of alkanes C--C from industrial source for industrial preparation of alkane

1 answer:

4 0

Alkanes can be prepared from carboxylic acid via the removal of carbon dioxide. This process is known as decarboxylation. It produces alkane with a carbon atom lesser than that present in the carboxylic acid.

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Which element has the same number of energy levels as krypton (Kr) and the same number of valence electrons as nitrogen (N)?

Alex

The correct answer is indeed Arsenic (As). This is the element that fits the description of being in period 4 and group 5, just like Krypton and Nitrogen.

5 0

4 years ago

Calculate the standard molar enthalpy of formation, in kJ/mol, of NO(g) from the following data:

Paraphin [41]

The above question is incomplete, here is the complete question:

Calculate the standard molar enthalpy of formation of NO(g) from the following data at 298 K:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o = 66.4 kJ$

$2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o= -114.1 kJ$

Answer:

The standard molar enthalpy of formation of NO is 90.25 kJ/mol.

Explanation:

$N_2(g) + 2O_2 \rightarrow 2NO_2(g), \Delta H^o_{1} = 66.4 kJ$

$2NO(g) + O_2\rightarrow 2NO_2(g),\Delta H^o_{2} = -114.1 kJ$

To calculate the standard molar enthalpy of formation

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$ ...[3]

Using Hess’s law of constant heat summation states that the amount of heat absorbed or evolved in a given chemical equation remains the same whether the process occurs in one step or several steps.

[1] - [2] = [3]

$N_2+O_2\rightarrow 2NO(g),\Delta H^o_{3} = ?$

$\Delta H^o_{3} =\Delta H^o_{1} - \Delta H^o_{2}$

$\Delta H^o_{3}=66.4 kJ - [ -114.1 kJ] = 180.5 kJ$

According to reaction [3], 1 mole of nitrogen gas and 1 mole of oxygen gas gives 2 mole of nitrogen monoxide, So, the standard molar enthalpy of formation of 1 mole of NO gas :

= $\frac{\Delta H^o_{3}}{2 mol}$

$=\frac{180.5 kJ}{2 mol}=90.25 kJ/mol$

3 0

3 years ago

For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The

Alik [6]

Answer : The rate for trial 5 will be $4.25\times 10^{-2}Ms^{-1}$

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

$A+B+C\rightarrow D+E$