Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is 
.
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ---- 
atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.
One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
Answer:
This is because, within a period or family of elements, all electrons are added to the same shell.
Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
<span>Answer:
K because it is metal and typically forms ionic bonds. Ar is also unlikely to form any bonds because it has a full outer shell of electrons, but it can form covalent bonds.</span>
