<u>Answer:</u> The mass of iron in the ore is 10.9 g
<u>Explanation:</u>
We are given:
Mass of iron (III) oxide = 15.6 g
We know that:
Molar mass of Iron (III) oxide = 159.69 g/mol
Molar mass of iron atom = 55.85 g/mol
As, all the iron in the ore is converted to iron (III) oxide. So, the mass of iron in iron (III) oxide will be equal to the mass of iron present in the ore.
To calculate the mass of iron in given mass of iron (III) oxide, we apply unitary method:
In 159.69 g of iron (III) oxide, mass of iron present is 
So, in 15.6 g of iron (III) oxide, mass of iron present will be = 
Hence, the mass of iron in the ore is 10.9 g
Answer:Iodine:
Electron configuration:
[Kr]4d10 5s2 5p5
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p5
Explanation:
Iodine is an element in group 17. It has only on umpired electron in the 5p orbital it combines with fluorine which also has one unpaired electron in its outermost orbital to form different complex interhalogen compounds by covalent bonding. These interhalogen compounds exhibit various shapes and properties.
Answer:
https://zo-om.us/j/3349111797?pwd=V3F5NCtEQThCTGtCQ3VENlBIemx3dz09
take the dash out
Explanation:
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<h2>
Answer</h2>
Bromination:
Any reaction or process in which bromine (and no other elements) are introduced into a molecule.
Bromonium Ion:
The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.(2)
Mechanism:
Step 1:
In the first step of the reaction, a bromine molecule approaches the electron-rich alkene carbon–carbon double bond. The bromine atom closer to the bond takes on a partial positive charge as its electrons are repelled by the electrons of the double bond. The atom is electrophilic at this time and is attacked by the pi electrons of the alkene [carbon–carbon double bond]. It forms for an instant a single sigma bond to both of the carbon atoms involved (2). The bonding of bromine is special in this intermediate, due to its relatively large size compared to carbon, the bromide ion is capable of interacting with both carbons which once shared the π-bond, making a three-membered ring. The bromide ion acquires a positive formal charge. At this moment the halogen ion is called a "bromonium ion".
Step 2:
When the first bromine atom attacks the carbon–carbon π-bond, it leaves behind one of its electrons with the other bromine that it was bonded to in Br2. That other atom is now a negative bromide anion and is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side. As it attacks and forms a bond with one of the carbons, the bond between the first bromine atom and the other carbon atoms breaks, leaving each carbon atom with a halogen substituent.
In this way the two halogens add in an anti addition fashion, and when the alkene is part of a cycle the dibromide adopts the trans configuration.
