Answer:
The ion's sizes and their radius ratio.
Explanation:
Ionic compounds take the form of three dimentional arrays by alternating anions and cations in order to minimize the potential energy of the system by maximizing the attractive force between opposite charges.
<em>The resultant geometric structures are known as crystal lattices and their arrangements depend on the ion's sizes and, since they are bound though electrostatic attraction, their radius ratio.</em>
I hope you find this information useful and interesting! Good luck!
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:
12 more oxygen
Explanation:
An atom is the smallest particles that can take part in a chemical reaction
The given compounds are:
3Mg₃(PO₄)₂
Number of oxygen atoms = 3[2 x 4] = 24 oxygen atoms
For;
4Al₂O₃;
Number of oxygen atoms = 4 x 3 = 12 oxygen atoms
In 3Mg₃(PO₄)₂ , there are 24 - 12 = 12 more oxygen atoms than in 4Al₂O₃;
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Answer : The Lewis-dot structure of
is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, ![NH_3](https://tex.z-dn.net/?f=NH_3)
As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in
= 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :
![\text{Formal charge}=\text{Valence electrons}-\text{Non-bonding electrons}-\frac{\text{Bonding electrons}}{2}](https://tex.z-dn.net/?f=%5Ctext%7BFormal%20charge%7D%3D%5Ctext%7BValence%20electrons%7D-%5Ctext%7BNon-bonding%20electrons%7D-%5Cfrac%7B%5Ctext%7BBonding%20electrons%7D%7D%7B2%7D)
![\text{Formal charge on N}=5-2-\frac{6}{2}=0](https://tex.z-dn.net/?f=%5Ctext%7BFormal%20charge%20on%20N%7D%3D5-2-%5Cfrac%7B6%7D%7B2%7D%3D0)
![\text{Formal charge on }H_1=1-0-\frac{2}{2}=0](https://tex.z-dn.net/?f=%5Ctext%7BFormal%20charge%20on%20%7DH_1%3D1-0-%5Cfrac%7B2%7D%7B2%7D%3D0)
![\text{Formal charge on }H_2=1-0-\frac{2}{2}=0](https://tex.z-dn.net/?f=%5Ctext%7BFormal%20charge%20on%20%7DH_2%3D1-0-%5Cfrac%7B2%7D%7B2%7D%3D0)
![\text{Formal charge on }H_3=1-0-\frac{2}{2}=0](https://tex.z-dn.net/?f=%5Ctext%7BFormal%20charge%20on%20%7DH_3%3D1-0-%5Cfrac%7B2%7D%7B2%7D%3D0)
Hence, the Lewis-dot structure of
is shown below.