Which of the options below show typical bonding patterns for a neutral nitrogen atom in a molecule or polyatomic ion?

Respuesta:

Un avión fabricado con aluminio puede transportar una mayor cantidad de pasajeros comparado con uno de acero.

Explicación:

La masa total que el avión es capaz de levantar es:

Para el aluminio:

y

donde:

• L es longitud
• D es diámetro
• e es grosor

Para el acero (mismo procedimiento):

Sabiendo que la masa total que el avión puede levantar es constante y que el aluminio tiene una densidad menor que la del acero, podemos afirmar que el avión de aluminio puede levantar un mayor número de pasajeros.

También es posible estimar un peso promedio de los pasajeros para calcular cuántos podría soportar.

Assuming we have a 100g sample, the mass of each element is as follows:
C: 74 g
H: 7.4 g
N: 8.6 g
O: 10 g
Next, we calculate the moles of each by dividing the mass of each element by its molar mass:
C: (74 / 12) = 6.17
H: (7.4 / 1) = 7.4
N: (8.6 / 14) = 0.61
O: (10 / 16) = 0.625
Now, we take the smallest value to determine the ratio:
C: 10
H: 12
N: 1
O: 1
Thus, the empirical formula can be expressed as
C10H12NO

Answer:

Explanation:

Diethyl malonate possesses greater acidity compared to monocarbonyl substances (pKa=13) because its alpha hydrogens are linked to two carbonyl groups. Consequently, the malonic ester can be readily changed into its enolate ion by reacting it with sodium ethoxide in ethanol. When the malonic ester undergoes alkylation, a hydrogen atom in the alpha position becomes acidic, permitting another round of alkylation to yield a dialkylated malonic ester.

In this scenario, when diethyl malonate interacts with urea in the presence of sodium ethoxide base, the second alkylation step occurs within the molecule, producing a cyclic compound known as barbituric acid.

Answer: The right choice is (c) application of both a mobile phase and a stationary phase.

Explanation:

Chromatography: This refers to a technique for separating a mixture where the mixture is distributed between two phases at varying rates, one being stationary and the other moving.

Mobile phase: The component in which the mixture is dissolved is referred to as the mobile phase.

Stationary phase: This is an adsorbent medium that remains in place while a liquid or gas passes over its surface, thus remaining stationary.

Consequently, a key characteristic of any chromatography technique involves utilizing both a mobile and a stationary phase.

Answer:

The mass of 22-Na included in the sample amounts to 0.0599 g

Explanation:

The total mass of the isotope mixture is 1.8385g.

It has an apparent mass of 22.9573 u.

For 23-Na, the relative atomic mass is 22.9898 u, while for 22-Na it is 21.9944 u.

Let the relative abundance of 23-Na be denoted as X.

This means that the relative abundance of 22-Na can be expressed as (1-X).

The equation formed is 21.9944 (1-X) + 22.9898 X = 22.9573.

Rearranging gives: 21.9944 - 21.9944X + 22.9898X = 22.9573.

Which simplifies to 22.9898X - 21.9944X = 22.9573 - 21.9944.

Hence, 0.9954X = 0.9639, leading to X = 0.9674.

The relative abundance of 23-Na is now identified as 0.9674.

Consequently, the relative abundance of 22-Na is 1 - 0.9674 = 0.0326.

Now, the mass of 22-Na contained within the 1.8385g sample is determined by

Relative abundance of 22-Na multiplied by the mass of the total sample = 0.0326 × 1.8385g = 0.0599 g.