The answer above is correct:)
Answer:
2K +F₂→ 2KF
Explanation:
When we balance an equation, we are trying to ensure that the number of atoms of each element is the same on both sides of the arrow.
On the left side of the arrow, there is 1 K atom and 2 F atoms. On the right, there is 1 K and 1 F atom.
Since the number of K atoms is currently balanced, balance the number of F atoms.
K +F₂→ 2KF
Now, that the number of F atoms is balanced on both sides, check if the number of K atoms are balanced.
<u>Left</u>
K atoms: 1
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The number of K atoms is not balanced.
2K +F₂→ 2KF
<u>Left</u>
K atoms: 2
F atoms: 2
<u>Right</u>
K atoms: 2
F atoms: 2
The equation is now balanced.
Answer:
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g
Explanation:
The equilibrium constant will remain the same irrespective of the amount of reactant taken.
Let us calculate the equilibrium constant of the reaction.
Kc=![\frac{[PCl_{3}][Cl_{2}]}{[PCl_{5}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BPCl_%7B3%7D%5D%5BCl_%7B2%7D%5D%7D%7B%5BPCl_%7B5%7D%5D%7D)
Let us calculate the moles of each present at equilibrium

molar mass of PCl₅=208
molar mass of PCl₃=137
molar mass of Cl₂=71
moles of PCl₅ = 
moles of PCl₃= 
moles of Cl₂ = 
the volume is 5 L
So concentration will be moles per unit volume
Putting values
Kc = 
Now if the same moles are being transferred in another beaker of volume 2L then there will change in the concentration of each as follow

Initial 0.02 0.06 0.04
Change -x +x +x
Equilibrium 0.02-x 0.06+x 0.04+x
Conc. (0.02-x)/2 (0.06+x)/2 (0.04+x)/2
Putting values
0.024 = 
Solving



x = -0.01
so the new moles of
PCl₅ = 0.02 + 0.01 =0.03
PCl₃ = 0.06-0.01 = 0.05
Cl₂ = 0.04-0.01 = 0.03
mass of each will be:
mass= moles X molar mass
PCl₅ = 0.03 X 208 = 6.24g
PCl₃ = 0.05 X 137 =6.85 g
Cl₂ = 0.03X71 = 2.13 g
Answer:
Aluminum, because it is in group 13 and has three valence electrons.
I have provided two images to help with this question. The first image is the reaction that is taking place. The γ-pyran is treated with the hydride acceptor triphenylmethyl perchlorate. A hydride is a hydrogen atom containing a lone pair of electrons giving it a negative charge. The triphenylmethyl cation is a positively charged carbocation that greatly wants to accept an electron pair to stabilize its charge. Therefore, it abstracts a hydride from the γ-puran starting material. It grabs one of the hydrogen atoms that is drawn in the reaction scheme. This results in the formation of triphenylmethane and a pyrylium perchlorate salt with the formula C₅H₅ClO₅. The important aspect of the structure is shown in the attached images. The most stable resonance form of the pyrylium cation is shown with a positive charge on the oxygen.
The reason this pyrylium ion is the most stable resonance form is because the formation of the oxonium ion (positive charged oxygen with 3 bonds) leads to an aromatic compound. There are 6 pi electrons in conjugation in this ring similar to a benzene ring and this results in the most stable structure.