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neonofarm [45]
4 years ago
12

Which one of the following best represents the predicted approximate chemical shift and coupling for the hydrogen(s) indicated w

ith the arrow?1.00 ppm, quartet 2.40 ppm, singlet 2.40 ppm, quartet 3.00 ppm, quartet 2.40 ppm, triplet

Chemistry
1 answer:
melomori [17]4 years ago
5 0

Answer:

The answer is quartet 2.40 ppm.

Note: Kindly find an attached image below for the part of the solution to this question

Sources: The image was researched from Course hero platform

Explanation:

Solution

Multiplicity or (n+1) rule:

It helps in determination of multiplicity of an individual proton or individual types of proton which are available in the molecule.

Multiplicity =(n+1)

Thus

The non equivalent protons which are attached from adjacent atom is denoted by n.

Now because there are three non-equivalent protons are present at adjacent carbon of methylene group, hence the multiplicity of methylene hydrogen is given as follows:

The multiplicity will be the same for the two hydrogen's. thus we compute multiplicity only for one  hydrogen atom stated below:

Non- equivalent = 3

Multiplicity = (3 +1)

= 4

= Quartet for 2H

A quartet for 2H indicates that the hydrogen atoms attached from the carbon, which is attached one side from a methyl group and the other side form an atom that have no any hydrogens.

Now due +I effect of carbonyl group, chemical shift value is high for these two hydrogens which is exactly at 2.40 ppm or 2.40 Quartet.

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Enter your answer in the provided box. Sodium stearate (C17H35COONa) is a major component of bar soap. The Ka of the stearic aci
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Answer:

11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.

Explanation:

Concentration of sodium stearate acid : c

Moles of sodium stearate = \frac{3.96 g}{306 g/mol}=0.01294 mol

Volume of the solution = 10.0 mL = 0.010 L

c=\frac{0.01294 mol}{0.010 L}=1.294 M

C_{17}H_{35}COONa\rightleftharpoons C_{17}H_{35}COO^-+Na^+

[C_{17}H_{35}COO^-]=c=1.294 M

C_{17}H_{35}COO^-+H_2O\rightleftharpoons C_{17}H_{35}COOH +OH^-

initially c

c           0    0

At equilibrium

(c-x)       x    x

Dissociation constant of an acid = K_a=1.3\times 10^{-5}

Expression of a dissociation constant of an acid is given by:  

K_a=\frac{[C_{17}H_{35}COOH][OH^-]}{[C_{17}H_{35}COO^-]}

K_a=\frac{(x)^2\times x}{(c -x)}

1.3\times 10^{-5}=\frac{x^2}{1.294-x}

Solving for x;

x = 0.0041 M

[OH^-]=0.0041 M

The pOH of the solution:

pOH=-\log[OH^-]=-\log[0.0041 M]=2.39

pH = 14 -pOH

pH = 14 - 2.39 = 11.61

11.61 is the pH of 10.0 mL of a solution containing 3.96 g of sodium stearate.

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A partir de 140 g de n2 y una cantidad de h2 se obtuvo 153G de NH³. Cual es el % de rendimiento de la reacción química
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Respuesta:

90.0 %

Explicación:

Paso 1: Escribir la ecuación química balanceada

N₂ + 3 H₂ ⇒ 2 NH₃

Paso 2: Calcular el rendimiento teórico de NH₃ a partir de 140 g de N₂

En la ecuación balanceada, participan de N₂: 1 mol × 28.01 g/mol = 28.01 g y de NH₃: 2 mol × 17.03 g/mol = 34.06 g.

140 g N₂ × 34.06 g NH₃ /28.01 g N₂ = 170 g NH₃

Paso 3: Calcular el rendimiento porcentual de NH₃

El rendimiento experimental de NH₃ es 153 g. Podemos calcular el rendimiento porcentual usando la siguiente fórmula.

R% = rendimiento experimental / rendimiento teórico × 100%

R% = 153 g / 170 g × 100% = 90.0 %

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