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Oliga [24]
3 years ago
8

Chapter 15 of your textbook discusses conjugation and various reactions of 1,3-dienes. In CHE 321, we discussed various ways to

make alkenes, but we did not discuss methodology for making dienes. Fortunately, we can apply some of our standard CHE 321 reactions to this important problem. Choose the major product (compound A) of the following reaction sequence.
Chemistry
1 answer:
xxMikexx [17]3 years ago
5 0

Answer:

Dienes are alkenes that contain two carbon-carbon double bonds, so they have the same properties as these hydrocarbons.

In the attached file are the two reactions of dienes production.

Explanation:

Two ways to obtain dienes are as follows:

-Reaction of oxidative dehydrogenation of an alkane, is an exothermic process and occurs at lower temperatures, diene and water are formed, generating greater conversion at lower temperature levels.

-Dehydration of primary alcohols. The treatment of alcohols with acid at elevated temperatures produces dienes due to water loss. For example, heating ethanol in the presence of sulfuric acid produces ethene by the loss of a water molecule.

Download pdf
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Write the net ionic equation:. . A solution of diamminesilver(I) chloride is treated with dilute nitric acid
zimovet [89]
<span> Ag(NH3)2Cl + 3HNO3 = AgNO3 +2NH4NO3 + HCl </span>
<span>or
 Ag(NH3)2Cl + HNO3 = Ag(NH3)2NO3 + HCl  this the complete balanced equation
now remove spectator ions to get net ionic equation
so 
</span>
<span> 2H+ + 2NO3- + [Ag(NH3)2]+ Cl- -> AgCl + 2NH4+ + 2NO3-  2NO3-  2H+  [Ag(NH3)2]+ + Cl- -> AgCl + 2NH4+
</span>hope it helps
5 0
3 years ago
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In multicellular organismscells that become specialized to perform a specific function are called?
Zolol [24]
The cell proliferates to produce many cells that result in multicellar organism.

answer
8 0
2 years ago
* The cost of table salt and table sugar is Rs 15 per kg.
Vilka [71]

Cost per mole

Table salt : Rs 0.878

Table sugar : Rs 23.63

<h3>Further explanation</h3>

Given

Cost table salt (NaCl) = 15/kg

Cost table sugar(sucrose-C12H22O11) = 69/kg

Required

cost per mole

Solution

mol of 1 kg Table salt(NaCl ,MW= 58.5 g/mol) :

\tt mol=\dfrac{1000~g}{58.5}=17.09~mol=Rs~15\rightarrow 1~mol=Rs~0.878

mol of 1 kg Table sugar(C12H22O11 ,MW= 342 g/mol) :

\tt mol=\dfrac{1000}{342}=2.92~mol=Rs~69\rightarrow 1~mol=Rs~23.63

6 0
2 years ago
According to the kinetic molecular theory, which statement describes the particles of an ideal gas? -1)The gas particles are arr
avanturin [10]

Answer: option # 4, the collisions of the gas particles may result in the transfer of energy.

Explanation:

The kinetic molecular theory (KMT) explains the behavior and properties of gases in terms of the energy, the size, and the motion of the gas particles.

In terms of size, according to the KMT the gases are formed by small particles separated from each other in a vacuum. The volume of the particles is negligible and it is considered that they do not occupy any space.

Since the particles are separated they do not exert either attraction or repulsion to each other.

Regarding the motion, the particles are in constant random motion. They move in straight lines until collide with other particles or with the walls of the veseel. The collisions are elastic (the total kinetic energy is conserved). The kinetic energy may be trasferred between the particles, but the total kinetic energy does not change.

The kinetic energy and the temperature are related: the temperature is a measure of the average kinetic energy of the particles of gas. At a given temperature all the gases have the same average kinetic energy.

Now, check every choice:

1)The gas particles are arranged in a regular pattern:

False. The particles occupy all the volumen and are in random motion.

2) The force of attraction between the gas particles are strong:

False. The particles are separated and they do not exert any force on each other.

3) The gas particles are hard spheres in continuous circular motion.

False. The particles travel in straight until they collide.

4) The collisions of the gas particles may result in the transfer of energy.

True. When particles collide they may transfer kinetic energy but the total kinetic energy is conserved.

4 0
3 years ago
Hcl and nh3 react to form a white solid, nh4cl. if cotton plugs saturated with aqueous solutions of each are placed at the ends
IgorLugansk [536]

24.4 cm.

<h3>Explanation</h3>

HCl and NH₃ reacts to form NH₄Cl immediately after coming into contact. Where NH₄Cl is found is the place the two gases ran into each other. To figure out where the two gases came into contact, you'll need to know how fast they move relative to each other.

The speed of a HCl or NH₃ molecule depends on its <em>kinetic energy</em>.

E_\text{k} = 1/2 \; m \cdot v^{2}

Where

  • E_\text{k} is the <em>kinetic energy</em> of the molecule,
  • m its mass, and
  • v^{2} the square of its speed.

Besides, the <em>kinetic theory</em> <em>of gases</em> suggests that for an ideal gas,

E_\text{k} \propto T

where \text{T} its temperature in degrees kelvins. The two quantities are directly proportional to each other. In other words, the <em>average kinetic energy</em> of molecules shall be the same for <em>any ideal gas </em>at the same<em> temperature</em>. So is the case for HCl and NH₃

E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3)

m(\text{HCl}) \cdot v^{2}(\text{HCl}) = E_\text{k} (\text{HCl}) = E_\text{k} (\text{NH}_3) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)

Where

  • m(\text{HCl}), v(\text{HCl}), and E_\text{k}(\text{NH_3}) the mass, speed, and kinetic energy of an HCl molecule;
  • m(\text{NH}_3), v(\text{NH}_3), and E_\text{k}(\text{NH}_3) the mass, speed, and kinetic energy of a NH₃ molecule.

The ratio between the mass of an HCl molecule and a NH₃ molecule equals to the ratio between their <em>molar mass</em>. HCl has a molar mass of 35.45; NH₃ has a molar mass of 17.03. As a result, m(\text{HCl}) = 36.45 / 17.03 \; m(\text{NH}_3). Therefore:

36.45 /17.03\; m(\text{NH}_3) \cdot v^{2}(\text{HCl}) = m(\text{HCl}) \cdot v^{2}(\text{HCl}) = m(\text{NH}_3) \cdot v^{2}(\text{NH}_3)

36.45 /17.03\; v^{2}(\text{HCl}) = v^{2}(\text{NH}_3)

\sqrt{36.45 /17.03}\; v(\text{HCl}) = v(\text{NH}_3)

The <em>average </em>speed NH₃ molecules would be  \sqrt{36.45/17.03} \approx 1.463 <em>if</em>  the <em>average </em>speed of HCl molecules v(\text{HCl}) is 1.

\text{Time before the two gases meet} = \frac{\text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}

\text{Distance from the HCl end} = v(\text{HCl}) \times \text{Time before the two gases meet}\\\phantom{\text{Distance from the HCl end}} = v(\text{HCl}) \times \frac{ \text{Length of the Tube}}{v(\text{HCl}) + v(\text{NH}_3)}\\\phantom{\text{Distance from the HCl end}} = \frac{v(\text{HCl})}{v(\text{HCl}) + v(\text{NH}_3)} \times \text{Length of the Tube}\\\phantom{\text{Distance from the HCl end}} = \frac{1}{1 + 1.463} \times 60.0\; \text{cm} \\\phantom{\text{Distance from the HCl end}} = 24.4 \; \text{cm}

8 0
3 years ago
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