Aluminium Hydroxide on decomposition produces Al₂O₃ and Water vapors.
<span> 2 Al(OH)</span>₃ → Al₂O₃ + 3 H₂O
According to equation at STP,
67.2 L (3 moles) of H₂O is produced by = 78 g of Al(OH)₃
So,
65.0 L of H₂O will be produced by = X g of Al(OH)₃
Solving for X,
X = (65.0 L × 78 g) ÷ 67.2 L
X =
75.44 g of Al(OH)₂Result: 75.44 g of Al(OH)₂ is needed to decompose in order to produce 65.0 L of water at STP in stoichiometry
It would be 23, s choice C.
Answer:
pH =6.70
Explanation:
When Kw = 1x10^-14
[H3O+] = [OH-] --> 1x10-7
Kw =3.98x^10-14
3.98x^10-14 = [H3O+] . [OH-]
√ (3.98x^10-14) = [H3O+] = [OH-]
[H3O+] = 1.99×10^-7
- log [H3O+] = pH ---> -log 1.99x10^-7 = 6.70
Answer:
4.99 × 10³ g/mol
Explanation:
Step 1: Given and required data
- Mass of the covalent compound (m): 62.4 g
- Volume of the solution (V): 1.000 L
- Osmotic pressure (π): 0.305 atm
- Temperature (T): 25°C = 298 K
Step 2: Calculate the molarity (M) of the solution
The osmotic pressure is a colligative pressure. For a covalent compound, it can be calculated using the following expression.
π = M × R × T
M = π / R × T
M = 0.305 atm / (0.0821 atm.L/mol.K) × 298 K
M = 0.0125 M
Step 3: Calculate the moles of solute (n)
We will use the definition of molarity.
M = n / V
n = M × V
n = 0.0125 mol/L × 1.000 L = 0.0125 mol
Step 4: Calculate the molar mass of the compound
0.0125 moles of the compound weigh 62.4 g. The molar mass is:
62.4 g/0.0125 mol = 4.99 × 10³ g/mol
Answer:
BaF2(s) ------> Ba2 (aq) + 2F- (aq)
Explanation:
Entropy refers to the degree of disorderliness in a system. Processes that lead to greater disorderliness in a system are said to increase the entropy of the system or lead to a positive value of ΔS.
If we consider the process, BaF2(s) ------> Ba2 (aq) + 2F- (aq), we will notice that ions were produced in solution thereby increasing the disorderliness of the system.
