Answer: 10
Explanation:
The detailed solution is contained in the image attached. The molar mass of hydrated and anhydrous salts are obtained and the number of moles of hydrated and hydrated salts are equated. The masses of hydrated and anhydrous salts are gives in the question and are simply substituted accordingly. This can now be used to obtain the number of molecules of water of crystallization as required in the question.
Answer:
[H+] = 1.74 x 10⁻⁵
Explanation:
By definition pH = -log [H+]
Therefore, given the pH, all we have to do is solve algebraically for [H+] :
[H+] = antilog ( -pH ) = 10^-4.76 = 1.74 x 10⁻⁵
Answer:
What happens if the solvent-solute attraction is greater than the solute-solute attraction when two substances are mixed? The solute particles are pulled apart and dispersed throughout the particles of the solvent, which holds the particles of the solute in solution.
Explanation:
Answer:
7.04 g
Explanation:
Let's consider the reaction in the last step of the Ostwald process.
3 NO₂(g) + H₂O(l) → 2 HNO₃(aq) + NO(g)
The molar mass of HNO₃ is 63.01 g/mol. The moles corresponding to 6.40 g are:
6.40 g × (1 mol/63.01 g) = 0.102 mol
The molar ratio of NO₂ to HNO₃ is 3:2. The reacting moles of NO₂ are:
0.102 mol HNO₃ × (3 mol NO₂/2 mol HNO₃) = 0.153 mol NO₂
The molar mass of NO₂ is 46.01 g/mol. The mass corresponding to 0.153 moles is:
0.153 mol × (46.01 g/mol) = 7.04 g
.2135 mol of Ba(OH)2 are needed to neutralize .427 mol of 2HC2H3O2
