Answer:
25.6mL NaOH
Explanation:
We are given the Molarity of the solution () and the volume of the solution (.02L).
By multiplying the two together, we can find the moles of solution that are reacted with HCl.
This gives us .0082 moles of HCl.
We then find the moles of NaOH that are needed to react with the HCl using the equation.
As HCl and NaCl have a 1:1 ratio, we need .0082 mol of NaOH.
Dividing this value by the Molarity of the solution
Gives us the answer, in Liters (.0256), which we can then divide by 100 convert to mL.
Answer:
Meristematic Tissues. Tissues where cells are constantly dividing are called meristems or meristematic tissues. These regions produce new cells.
Explanation:
thank me later
The answer is D; Mercury-194
All of the others are not when I looked them up
Because the atoms are separating bc of the heat. A solid has really compact atoms so if it was melted its atoms would become looser and form a liquid
Answer:
The spilling of part of the salt by students while measuring the solubility of a salt will make the calculated solubility by the student to be higher than the exact value.
Explanation:
Solubility is the maximum amount in gram or moles of a particular salt that can completely dissolve in a given solvent at a particular temperature. Therefore, in calculation of the solubility is requires three major factors:
1. The amount of salt,
2. The temperature, and
3. The volume of the solvent used.
This experiment and the mistake done by this student only caused an alteration in the amount of salt used while the volume of solvent was held constant. The student would have therefore thought that she/he had added a particular amount of salt; let's say: x moles, which we should assume to be the maximum amount that can be dissolved by the given solvent at that temperature. The spilled salt will generally reduce the moles of the salt that is eventually added by let's say: y moles. Therefore, the actual amount of salt that would have been added will be x - y moles. The solvent will still have room to accommodate more salt since the solubility is constant at this unaltered temperature. The addition of more salt, let"s say: m moles to make up for the loss will make the student record x + m moles of the salt per unit volume of the solvent instead of x - y + m which was eventually added.
Therefore, since the assumed x + m moles will always be higher than the actual x + m - y moles of the salt, the recorded solubility by the student will always be higher than the actual value.