A. Less than 1% of the alpha particles went un-deflected through the gold foil.
Answer:
A solution in which no more solute can be dissolved in is referred to as SATURATED. In such a solution, the concentration of solute is called SOLUBILITY . When that concentration is reported in moles per liter, it is more specifically called MOLAR SOLUBILITY. A special equilibrium constant called the SOLUBILITY PRODUCT constant is calculated from the molar concentrations of the aqueous components of the dissolution equation.
Explanation:
The solubility of a solute in a solvent is the maximum amount of solute in moles that will be dissolved in 1dm3 of the solvent at a specified temperature. Once the maximum number or concentration has been reached, the solvent can no longer take in solutes and this point in the reaction, the solution is said to be saturated. That is the composition of the saturated solution is not affected by the presence of excess solute. An unsaturated solution has a lower concentration of solute and can dissolve more solutes if added until it becomes saturated.
Solubility when reported in moles per liter is called molar solubility of the solution and it gives a more accurate measurement of yh solubility of a solution. The solubility product constant is calculated from the molar concentrations of the aqueous components of the dissolution equation. This solubility product constant explains the balance between dissolved ions from the salt and undissolved salt in a dissolution equation.
Conditions:
Low pressure and low temperature
Low pressure and high temperature
High pressure and low temperature
High pressure and high temperature
Answer:
Molar heat of solution of KBr is 20.0kJ/mol
Explanation:
Molar heat of solution is defined as the energy released (negative) or absorbed (Positive) per mole of solute being dissolved in solvent.
The dissolution of KBr is:
KBr → K⁺ + Br⁻
In the calorimeter, the temperature decreases 0.370K, that means the solution absorbes energy in this process. The energy is:
q = 1.36kJK⁻¹ × 0.370K
q = 0.5032kJ
Moles of KBr in 3.00g are:
3.00g × (1mol / 119g) = 0.0252moles
Thus, molar heat of solution of KBr is:
0.5032kJ / 0.0252moles = <em>20.0kJ/mol</em>
