Answer:
The final solution is 1.5 times more acidic than the initial solution
Explanation:
Answer: i need that answer to
Explanation:
Answer: 4.55x10^-7
Explanation:
H2CO3 + H20 <==> H3O+ + HCO3-
Desigining an ICE table, we have:
Initial conc. of H2CO3 = 0.29 M
Initial conc. of H3O+ = 0
Initial conc. of HCO3- = 0
Change in conc. of H2CO3 = - x
Change in conc. of H3O+ = x
Change in conc. of HCO3- = x
Equilibrium conc. of H2CO3 = 0.29 - x
Equilibrium conc. of H3O+ = x
Equilibrium conc. of HCO3- = x
but pH = 3.44
pH = - log[H30+]
[H30+] = 10^-3.44 = 3.63x10^-4
[H30+] = 3.63x10^-4
Ka = [H3O+].[HCO3-] / [H2CO3]
[H30+] = x = 3.63x10^-4
[HCO3-] = 3.63x10^-4
[H2CO3] = 0.29 - x = 0.29 - 3.63x10^-4 = 0.289637
Ka = (3.63x10^-4)(3.63x10^-4)/0.289637 = 4.55x10^-7
Answer:
Option D is correct = 58 g
Explanation:
Data Given:
mass of LiOH = 120 g
Mass of Li3N= ?
Solution:
To solve this problem we have to look at the reaction
Reaction:
Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)
1 mol 3 mol
Convert moles to mass
Molar mass of LiOH = 24 g/mol
Molar mass of Li₃N = 35 g/mol
So,
Li₃N (s) + 3H₂0 (l) -----------► NH₃ (g) + 3LiOH (l)
1 mol (35 g/mol) 3 mol (24 g/mol)
35 g 72 g
So if we look at the reaction 35 g of Li₃N react with water and produces 72 g of LiOH , then how many g of Li₃N will be react to Produce by 120 g of LiOH
For this apply unity formula
35 g of Li₃N ≅ 72 g of LiOH
X of Li₃N ≅ 120 g of LiOH
By Doing cross multiplication
Mass of Li₃N = 35 g x 120 g / 72 g
mass of Li₃N = 58 g
120 g of LiOH will produce from 58 g of Li₃N
So,
Option D is correct = 58 g