Correct Question: what is the oxidizing agent in the reaction.
2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Answer: MnO4-is the oxidizing agent
Explanation:
In the reaction 2MnO4–(aq) +10Cl–(aq) + 16H+(aq) --------> 5Cl2(g) + 2Mn2+(aq) +8H2O(l)
Oxidizing agent oxidizes other molecules while the themselves get reduced.
oxidizing agents give away Oxygen to other compounds.
MnO4-is the oxidizing agent because
On the reactants side
Oxidation number of Mn in 2MnO4- is +7
Oxidation number of Cl- is -1
On the products side
Oxidation number of Mn is +2
While oxidation number of Cl is zero
Therefore the oxidizing agent is 2MnO4 because is oxidizes Chlorine from -1 to 0 while itself got reduced from oxidation state of +7 to +2
Answer:
54 grams ammonium chloride and 40 grams sodium hydroxide
Explanation:
A buffer is a solution that contains either a weak acid and its salt or a weak base and its salt, the solution is resistant to changes in pH. This means that, a buffer is an aqueous solution of either a weak acid and its conjugate base or a weak base and its conjugate acid.
A Buffer is used to maintain a stable pH in a solution, buffers can neutralize small quantities of additional acid of base. For any buffer solution, there is always a working pH range and a set amount of acid or base that can be neutralized before the pH will change. The amount of acid or base that can be added to a buffer before changing its pH is called its buffer capacity.
A good buffer mixture is supposed to have about equal concentrations of its both components. It is a rule of thumb therefore, that a buffer solution has generally lost its usefulness when one component of the buffer pair is less than about 10% of the other component.
The implication of this is that the ammonium chloride and sodium hydroxide should be of approximately the same concentration. If the masses are dissolved as shown in the answer, then we will have 1molL-1 of each component of the buffer in accordance with the rule of thumb stated above.
3 mol H₂ → 2 mol NH₃
5 mol H₂ → x mol NH₃
x=2*5.0/3=3.3
n(NH₃)=3.3 mol
The molar concentration of the nitric acid solution was 0.6666 mol/L.
<em>Balanced equation</em>: KOH + HNO_3 → KNO_3 + H_2O
<em>Moles of KOH</em>: 32.33 mL KOH × (1.031 mmol KOH /1 mL KOH)
= 33.33 mmol KOH
<em>Moles of HNO_3</em>: 33.33 mmol KOH× (1 mmol HNO_3/1 mmol KOH)
= 33.33 mmol HNO_3
<em>Concentration of KOH</em>: <em>c </em>= "moles"/"litres" = 33.33 mmol/50.00 mL
= 0.6666 mol/L
