Answer:
- (0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄)
Explanation:
A buffer solution is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid.
Also, a buffer solution is a solution which resists changes in pH when acid or alkali is added to it.
- (0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄) when dissolved in 1 L H₂O will produce a buffer because NaH₂PO₄ is considered as weak acid while Na₂HPO₄ is its conjugate base. 2.
- (0.1 mol H₃O⁺ + 0.1 mol Cl⁻) is not a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid.
- (0.1 mol HCl + 0.1 mol NaoH) HCL is a strong acid and NaOH is a strong base so it will not form a buffer when dissolved in water.
- (0.1 mol H₃O⁺ + 0.1 OH⁻) is not a mixture of a weak acid and its conjugate base, or weak base and its conjugate acid
- (0.1 mol NaCl+ 0.1 mol KCl) NaCL and KCL are salts so it will not form a buffer when dissolved in water.
So the right choice is
- (0.1 mol NaH₂PO₄ + 0.1 mol Na₂HPO₄)
Answer: -
The first compound is 3-Methylheptane.
The second compounds is 4-ethyl-3-methyl-4-propylheptane
The 3rd one is propanol.
The 4th one is 2-propanol.
The 5th one is methoxyethane
Answer:
The molarity of this solution is 14.82 mol/dm3 or 14.82 mol/L
Explanation:
- Molarity is the number of mole present in 1 Litre of solution. Molarity of a solution is a term referred to as concentration of a solution. The unit of Molarity is Mol/dm3 or Mol/L.
- let us make an assumption that the volume of the ammonia solution is 1L or 1dm3. Also, 1L = 1000 mL.
Step 1: calculate the mass of the solution
Density = 0.90g/ml (from the question)
Density = mass/ volume
Therefore Mass = density x volume
= 0.90g/ml x 1000ml
mass = 900 g
Step 2: calculate the mass of NH3 present in the solution
Since the concentrated aqueous of ammonia is 28%, It signifies that 1000ml of the solution contains 28% Ammonia
Recall from the above calculation that the mass of 1000 ml of solution is 900 g.
Therefore the mass of ammonia will be 28% of 900 g
mass of NH3 = 0.28 x 900 g
= 252 g
Step 3: calculate the number of mole of NH3
mole = mass/ molar mass
molar mass of NH3 = 17 g/mol
Therefore mole of NH3 = 252/17
= 14.82 mol
Step 4: Calculate Molarity
Molarity = number of moles/ volume of solution in Litre (L)
Molarity = 14.82 / 1
Molarity = 14.82 mol/L
