<span>The lattice structure in a metal is weaker than the lattice structure of an ionic compound. This is because the metals have free electrons which can freely move around while ionic compounds are strongly bonded. Hope this answers the question.</span>
Answer:
The rate of consumption of 
is 2.0 mol/L.s
Explanation:
Applying law of mass action to this reaction-
![-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D-%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7B%5CDelta%20%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BN_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
where ![-\frac{\Delta [NH_{3}]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D)
represents rate of consumption of , ![-\frac{\Delta [O_{2}]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D)
represents rate of consumption of 
, ![\frac{\Delta [N_{2}]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BN_%7B2%7D%5D%7D%7B%5CDelta%20t%7D)
represents rate of formation of 
and ![\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
represents rate of formation of 
.
Here rate of formation of is 3.0 mol/(L.s)
From the above equation we can write-
![-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
Here ![\frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D%3D3.0%20mol%2F%28L.s%29%29)
So, ![-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B4%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
Hence, ![-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B4%7D%7B6%7D%5Ctimes%203.0%20mol%2F%28L.s%29%3D2.0%20mol%2F%28L.s%29)
write an equation to represent the oxidation of an alcohol.
identify the reagents that may be used to oxidize a given alcohol.
identify the specific reagent that is used to oxidize primary alcohols to aldehydes rather than to carboxylic acids.
identify the product formed from the oxidation of a given alcohol with a specified oxidizing agent.
identify the alcohol needed to prepare a given aldehyde, ketone or carboxylic acid by simple oxidation.
write a mechanism for the oxidation of an alcohol using a chromium(VI) reagent.
The reading mentions that pyridinium chlorochromate (PCC) is a milder version of chromic acid that is suitable for converting a primary alcohol into an aldehyde without oxidizing it all the way to a carboxylic acid. This reagent is being replaced in laboratories by Dess‑Martin periodinane (DMP), which has several practical advantages over PCC, such as producing higher yields and requiring less rigorous reaction conditions. DMP is named after Daniel Dess and James Martin, who developed it in 1983.
This page looks at the oxidation of alcohols using acidified sodium or potassium dichromate(VI) solution. This reaction is used to make aldehydes, ketones and carboxylic acids, and as a way of distinguishing between primary, secondary and tertiary alcohols.
Oxidizing the different types of alcohols
The oxidizing agent used in these reactions is normally a solution of sodium or potassium dichromate(VI) acidified with dilute sulfuric acid. If oxidation occurs, the orange solution containing the dichromate(VI) ions is reduced to a green solution containing chromium(III) ions. The electron-half-equation for this reaction is
Cr2O2−7+14H++6e−→2Cr3++7H2O
Answer: 7.79 grams of ethanol were put into the beaker.
Explanation:
To calculate the mass of ethanol, we use the equation:
Density of ethanol = 0.779 g/mL
Volume of water = 10.00 mL
Putting values in above equation, we get:
Thus 7.79 grams of ethanol were put into the beaker.
