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Lerok [7]
3 years ago
11

Predict the aldol product when the following ketone undergoes self-condensation in the presence of NaOH. Do not show the dehydra

ted product. Be sure to include all lone pairs in your product.

Chemistry
1 answer:
Anit [1.1K]3 years ago
5 0

Answer:

β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal.

Explanation:

When acetaldehyde is treated with dil.NaOH it undergoes self condensation as it contains alpha-hydrogen atom in its compound forming β-hydroxyaldehyde (an aldol) namely 3-Hydroxy butanal. This compound upon further heating will eliminate a molecule of water forming aldol condensation product namely Crotonaldehyde Or But-2-en-al. see the diagram attached.

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a friend is trying to describe a material to you. which of the following is not a description of a physical property of the subs
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The answer for it is C. when you mix it with water it fizzes

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Cuántos mililitros de agua hay en 3 litros de una botella de naranjas y las solucion tiene una concentracion de 20% v/v
sertanlavr [38]

Respuesta:

2400 mL

Explicación:

Paso 1: Información dada

  • Volumen de solución: 3 L (3000 mL)
  • Concentración de naranja: 20 % v/v

Paso 2: Calcular el volumen de naranja

La concentración de naranja es de 20 % v/v, es decir, cada 100 mL de solución hay 20 mL de naranja.

3000 mL Sol × 20 mL Naranja/100 mL Solución = 600 mL Naranja

Paso 3: Calcular el volumn de agua

El volumen de soluciónes igual a la suma de los volúmenes de naranja y agua.

VSolución = VNaranja + VAgua

VAgua = VSolución - VNaranja

VAgua = 3000 mL - 600 mL = 2400 mL

8 0
3 years ago
Sediments are almost always deposited in flat layers. Does it appear that forces in the Earth affected the rock layers in this r
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I think so but what region do you live in
4 0
3 years ago
A 20.0g sample of metal with a specific heat of 5 J/(g°C) raised it's initial temperature to 40.0 when 500J heat was added. What
slavikrds [6]

The initial temperature of the metal = 35 °C

<h3>Further explanation</h3>

Heat can be formulated :

Q = m . c . ΔT

Q = heat, J

c = specific heat, J/g C

ΔT = temperature, °C

m = 20 g

c = 5 J/(g°C)

Q = 500 J

T₁ = 40 C

the initial temperature :

\tt \Delta t(T_2-T_1)=\dfrac{Q}{m.c}\\\\40-T_1=\dfrac{500}{20.5}\\\\40-T_1=5\\\\T_1=35^oC

3 0
2 years ago
Identify the missing coefficient in the following equation:<br><br>3<br><br>2<br><br>1<br><br>0
Pie

Answer:

2

Explanation:

In balancing nuclear reactions the mass number and atomic numbers are usually conserved. This implies that from the given equation, the sum of the number of the subscript on the right hand side must be equal to that on the left hand side. This also applies to the superscript:

   For the mass numbers(superscript):

   235 + 1 = 1 + 139 + 95

       236 = 235

This is not balanced

  For the atomic number:

  92 + 0 = 0 + 53 + 39

      92 = 92

This is balanced.

We simply inspect to see how to balance the mass number.

By putting a coefficient of 2 behind the neutron atom, the equation becomes balanced.

7 0
3 years ago
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