A saturated solution is one in which no more solute is able to dissolve in a given solvent at a particular temperature. Some amount of the solute is left undissolved in the solution.
Unsaturated solution has solute in lower proportions than required to form a saturated solution.
Supersaturated solution has solute in amounts greater than a saturated solution.
We can take the help of solubility curve in order to find out the amount of a salt required to prepare a saturated solution of that salt at a particular temperature.
The solubility of KI at 10
is 136 g/ 100 mL water
The solubility of
at
is 21 g/100 mL water.
The solubility of
at
is 80 g/100 mL water.
The solubility of NaCl at
is 38 g/ 100 mL water.
So the correct answer will be KI, as it would need 136 g KI / 100 mL water to form a saturated solution at
.So, if we have 80g KI/ 100mL water it would be an unsaturated solution.
Answer:
100 C or 212 F
Explanation:
That is the boiling point for liquid so both gas and liquid are presented.
Calculate the mass defect and nuclear binding energy of an atom ... To convert to joules per nucleon, simply divide by the number of nucleons.
Answer:

Explanation:
We know we will need an equation with masses and molar masses, so let’s gather all the information in one place.
M_r: 58.12 44.01
2C₄H₁₀ + 13O₂ ⟶ 8CO₂ + 10H₂O
m/g: 9.511
1. Moles of C₄H₁₀

2. Moles of CO₂
The molar ratio is 8 mol CO₂:2 mol C₄H₁₀

3. Mass of CO₂
