In titration, the moles of acid equal moles of base. You were given that 22.75ml of 0.215M NaOH is used, so calculate the number of moles of that base the experiment used in total. After that because you know mol base = mol acid, whatever amount of base you use must be the total amount of acid present in the solution. You were given the volume of the acid, and you have just found the total mols of acid. Using these two information, solve for the concentration. And one more thing, even though I'm pretty sure it won't affect your answer, you should always convert things to the proper units. Since the concentration we're talking about in this problem is molarity, which has the unit mol/L, you should always have all of your numbers in these units. It just make it simpler and will not confuse you
Answer: -
Solubility of a substance depend on the balance of intermolecular forces between the solvent and solute, and the entropy change that accompanies this process.
Temperature and pressure also plays a role in solubility.
A solution having Group 1 cations like lithium, sodium, potassium etc are always soluble.
A solution having NH₄⁺ is soluble.
All salts with anion as nitrates, acetates, chlorates, and perchlorates are soluble in water.
Answer:
i believe the answer is supersaturated
Explanation:
Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
- 15 molecules H₂ *
= 10 molecules NH₃
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.
