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2 years ago

Two common methods to generate an aldehyde is by oxidation of an alcohol and through ozonolysis.

Chemistry

2 answers:

leonid [27]2 years ago

8 0

Answer:

A. True

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

kotykmax [81]2 years ago

5 0

Answer:

a. True.

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

 ${ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}$ 

 ${ \sf{CH _{3} CHO \: \: is \: ethanal}}$ 

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

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What does it mean for energy to be conserved

marysya [2.9K]

Answer: It means that the energy is being stored or preserved to be used at a later time.

Your question was a little vague, so if this isn't the answer you were looking for, just let me know and I'll fix it.

Hope this helps! Good luck! :D

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2 years ago

Calculate %ic of the interatomic bonds for the intermetallic compound al6mn. on the basis of this result, what type of interatom

pashok25 [27]

Answer : % ionic character is 0.20%

Bonding between the two metals will be purely metallic.

Explanation: For the calculation of % ionic character, we use the formula

$\% \text{ ionic character}= [1-e^{\frac{-(X_A-X_B)^2)}{4}}]\times(100\%)$

where $X_A$ & $X_B$ are the Pauling's electronegativities.

The table attached has the values of electronegativities, by taking the values of Al and Mn from there,

$X_{Al}=1.61$

$X_{Mn}=1.55$

Putting the values in the electronegativity formula, we get

$\% \text{ ionic character}= [1-e^{\frac{-(1.61-1.55)^2)}{4}}]\times(100\%)$

= 0.20%

Now, there are 3 types of inter atomic bonding

1) Ionic Bonding: It refers to the chemical bond in which there is complete transfer of valence electrons between atoms

2) Covalent Bonding: It refers to the chemical bond involving the sharing of electron pairs between 2 atoms.

3) Metallic Bonding: It refers to the chemical bond in which there is an electrostatic force between the positively charged metal ions and delocalised electrons.

In $Al_6Mn$ compound, the % ionic character is minimal that is $\sim$ 0.20% and there are two metal ions present, therefore this compound will have metallic bonding.

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3 years ago

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$

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