Answer:
After the solution is heated, but before additional solute is added
Explanation:
An unsaturated solution is a solution that contains less solute than it can normally hold at a given temperature. Hence an unsaturated solution can still dissolve more solute.
When the solution is heated, the saturated cold solution becomes an unsaturated hot solution which is capable of dissolving more solute at this point.
Once more solute is dissolved, the solution becomes saturated again just before it begins to cool since no more solute dissolves in the solution at some point before cooling and addition of seed crystals.
Answer:
Two factors that might have a affect of which copper sulphate mineral will occur at a given location is:
A. Copper sulphate high solubility in water
B. Also it binds nicely with the sediments or the crystal.
Explanation:
As it is mentioned here that copper sulphate can be crystallized as an anhydrate which means that their is no waterin those crystals or can be as of those three different hydrates whose crystal structure varies with the amount of water present in it.
The four forms are also given of the copper sulphate are:
- Bonatite
- Boothite
- Chalcanthite
- Chalcocyanite
So, the two factors that might give an affect which type of copper sulphate mineral willoccur at a given location is:
A. The copper sulphate high solubility in water.
B. It binds extremely nicely with the sediments or say to the crystal. It is also regulated by plants.
Using the Henderson-Hasselbalch equation on the solution before HCl addition: pH = pKa + log([A-]/[HA]) 8.0 = 7.4 + log([A-]/[HA]); [A-]/[HA] = 4.0. (equation 1) Also, 0.1 L * 1.0 mol/L = 0.1 moles total of the compound. Therefore, [A-] + [HA] = 0.1 (equation 2) Solving the simultaneous equations 1 and 2 gives: A- = 0.08 moles AH = 0.02 moles Adding strong acid reduces A- and increases AH by the same amount. 0.03 L * 1 mol/L = 0.03 moles HCl will be added, soA- = 0.08 - 0.03 = 0.05 moles AH = 0.02 + 0.03 = 0.05 moles Therefore, after HCl addition, [A-]/[HA] = 0.05 / 0.05 = 1.0 Resubstituting into the Henderson-Hasselbalch equation: pH = 7.4 + log(1.0) = 7.4, the final pH.
Answer:An increase in temperature commonly will increase the rate of reaction. An growth in temperature will improve the common kinetic electricity of the reactant molecules.
Explanation:
