Answer:
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓
Ksp = [2s]² . [s] → 4s³
Explanation:
Ag₂CrO₄ → 2Ag⁺ + CrO₄⁻²
Chromate silver is a ionic salt that can be dissociated. When we have a mixture of both ions, we can produce the salt which is a precipitated.
2Ag⁺ (aq) + CrO₄⁻² (aq) ⇄ Ag₂CrO₄ (s) ↓ Ksp
That's the expression for the precipitation equilibrium.
To determine the solubility product expression, we work with the Ksp
Ag₂CrO₄ (s) ⇄ 2Ag⁺ (aq) + CrO₄⁻² (aq) Ksp
2 s s
Look the stoichiometry is 1:2, between the salt and the silver.
Ksp = [2s]² . [s] → 4s³
Answer:
6.02 x 10²³ atoms
Explanation:
Given parameters:
Number of moles CaSO₄ = 1 mole
Unknown:
Number of Ca atoms in the given compound = ?
Solution:
The given compound is:
CaSO₄
1 mole of CaSO₄ is made up of 1 mole of Ca atoms
Now;
1 mole of any substance contains 6.02 x 10²³ atoms
1 mole of Ca atoms will also contain 6.02 x 10²³ atoms
Answer: 1.46moles
Explanation:
Applying PV= nRT
P= 1atm, V= 32.6L, R= 0.082, T = 273K
Substitute and simplify
1×32.6/(0.082×273)=n
n= 1.46moles
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
They have the same number of protons
