Answer:
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Explanation:
The solubility constant give us the molar solubilty of ionic compounds. In general for a compound AB the ksp will be given by:
Ksp = (A) (B) where A and B are the molar solubilities = s² (for compounds with 1:1 ratio).
It follows then that the higher the value of Ksp the greater solubilty of the compound if we are comparing compounds with the same ionic ratios:
Comparing AgBr: Ksp = 5.4 x 10⁻¹³ with AgCl: Ksp = 1.8 x 10⁻¹⁰, AgCl will be more soluble.
Comparing Ag2CO3: Ksp = 8.0 x 10⁻¹² with AgCl Ksp = AgCl: Ksp = 1.8 x 10⁻¹⁰ we have the complication of the ratio of ions 2:1 in Ag2CO3, so the answer is not obvious. But since we know that
Ag2CO3 ⇄ 2 Ag⁺ + CO₃²₋
Ksp Ag2CO3 = 2s x s = 2 s² = 8.0 x 10-12
s = 4 x 10⁻12 ∴ s= 2 x 10⁻⁶
And for AgCl
AgCl ⇄ Ag⁺ + Cl⁻
Ksp = s² = 1.8 x 10⁻¹⁰ ∴ s = √ 1.8 x 10⁻¹⁰ = 1.3 x 10⁻⁵
Therefore, AgCl is more soluble than Ag₂CO₃
The order of solubility is AgBr < Ag₂CO₃ < AgCl
Answer:
116.3 grCO2
Explanation:
1st - we balance the equation so that it finds the same amount of elements of the product side and of the reagent side
C6H6 +15/2 O2⟶ 6CO2 +3 H2O
2nd - we calculate the limiting reagent
39.2gr C6H6*(240grO2/78grC6H6)=120 grO2
we don't have that amount of oxygen so this is the excess reagent and oxygen the limiting reagent
3rd - we use the limiting reagent to calculate the amount of CO2 in grams
105.7grO2*(264grCO2/240grO2)=116.3 grCO2
LBr is ionic compound because k for potassium is metal which means it’s on the left side of the periodic table and Br which is bromine is a non metal which means is on the right side of the periodic table.
In Conclusion when a non metal and metal come together the are called ionic compound
The chemical formula for table sugar is C12H22O11. From the formula, you can say that it is an organic molecule where in every molecule contains 12 atoms of carbon, 22 atoms of hydrogen and 11 atoms of oxygen. There are 45 atoms in total in one molecule of sugar.
Answer:
Explanation:
For transitions:
So, 
and 
(As the hydrogen has to ionize)
Thus,
