Answer:
12.8 g
Explanation:
Step 1: Write the balanced neutralization reaction
Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O
Step 2: Calculate the moles corresponding to 17.9 g of HCl
The molar mass of HCl is 36.46 g/mol.
17.9 g × 1 mol/36.46 g = 0.491 mol
Step 3: Calculate the moles of Al(OH)₃ that completely react with 0.491 moles of HCl
The molar ratio of Al(OH)₃ to HCl is 1:3. The reacting moles of Al(OH)₃ are 1/3 × 0.491 mol = 0.164 mol
Step 4: Calculate the mass corresponding to 0.164 moles of Al(OH)₃
The molar mass of Al(OH)₃ is 78.00 g/mol.
0.164 mol × 78.00 g/mol = 12.8 g
The reaction between KOH and HBr is as follows ;
KOH + HBr ---> H₂O + KBr
Stoichiometry of base to acid is 1:1 molar ratio
Both are strong acid and strong base therefore complete ionization takes place
The number of KOH moles added - 0.50 M / 1000 mL/L x 22 mL = 0.011 mol
the number of HBr moles - 0.25 M /1000 mL/L x 44 mL = 0.011 mol
the number of H⁺ ions and OH⁻ ions are equal therefore the whole amount of acid has been completely neutralised by base.
No remaining acid nor base, therefore solution is neutral.
pH = 7
thats the pH value for a neutral solution
Answer : The correct option is, (C) 1.1
Solution : Given,
Initial moles of 
= 1.0 mole
Initial volume of solution = 1.0 L
First we have to calculate the concentration .
The given equilibrium reaction is,
Initially c 0
At equilibrium 
The expression of 
will be,
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
where,
= degree of dissociation = 40 % = 0.4
Now put all the given values in the above expression, we get:
Therefore, the value of equilibrium constant for this reaction is, 1.1
