Answer:
K = 0.2
Explanation:
Based on the chemical dissociation of N₂O₄:
N₂O₄ ⇄ 2NO₂
The equilibrium constant, K, of the reaction is:
K = [NO₂]² / [N₂O₄]
Now, if 20% of N₂O₄ is dissociated, 80% remains as N₂O₄ = 0.8mol/L = 0.8M
as 20% is dissociated, 0.2moles of N₂O₄ were dissociated and 0.2*2 = 0.4mol/L of NO₂ are produced.
Replacing in K:
K = [0.4M]² / [0.8M]
<h3>K = 0.2</h3>
Answer:
The van't hoff factor of 0.500m K₂SO₄ will be highest.
Explanation:
Van't Hoff factor was introduced for better understanding of colligative property of a solution.
By definition it is the ratio of actual number of particles or ions or associated molecules formed when a solute is dissolved to the number of particles expected from the mass dissolved.
a) For NaCl the van't Hoff factor is 2
b) For K₂SO₄ the van't Hoff factor is 3 [it will dissociate to give three ions one sulfate ion and two potassium ions]
Out of 0.500m and 0.050m K₂SO₄, the van't hoff factor of 0.500m K₂SO₄ will be more.
c) The van't Hoff factor for glucose is one as it is a non electrolyte and will not dissociate.
Answer:
molarity of acid =0.0132 M
Explanation:
We are considering that the unknown acid is monoprotic. Let the acid is HA.
The reaction between NaOH and acid will be:

Thus one mole of acid will react with one mole of base.
The moles of base reacted = molarity of NaOH X volume of NaOH
The volume of NaOH used = Final burette reading - Initial reading
Volume of NaOH used = 22.50-0.55= 21.95 mL
Moles of NaOH = 0.1517X21.95=3.33 mmole
The moles of acid reacted = 3.33 mmole
The molarity of acid will be = 
Answer: 9.9 grams
Explanation:
To calculate the moles, we use the equation:

a) moles of 

b) moles of 


According to stoichiometry :
1 mole of
combine with 1 mole of
Thus 0.33 mole of
will combine with =
mole of
Thus
is the limiting reagent as it limits the formation of product.
As 1 mole of
give = 1 mole of 
Thus 0.33 moles of
give =
of 
Mass of 
Thus theoretical yield (g) of
produced by the reaction is 9.9 grams
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