Answer:answers are in the explanation
Explanation:
(a). pH less than 7 between 1 - 3.5 are strong acid, and between 4.5-6.9 weak acid.
pH greater than 7; between 10-14 is a strong base, and between 7.1 - 9, it is weakly basic.
(b). Equation of reaction;
HBr + KOH ---------> KBr + H2O
One mole of HBr reacts with one mole of KOH to give one Mole of KBr and one mole of H2O
Calculating the mmol, we have;
mmol KOH = 28.0 ml × 0.50 M
mmol KOH= 14 mmol
mmol of HBr= 56 ml × 0.25M
mmol of HBr= 14 mmol
Both HBr and KOH are used up in the reaction, which leaves only the product,KBr and H2O.
The pH here is greater than 7
(C). [NH4^+] = 0.20 mol L^-1 × 50 ml. L^-1 ÷ 50 mL + 50mL
= 0.10 M
Ka=Kw/kb
10^-14/ 1.8× 10^-5
Ka= 5.56 ×10^-10
Therefore, ka= x^2 / 0.20
5.56e-10 = x^2/0.20
x= (0.20 × 5.56e-10)^2
x= 1.05 × 10^-5
pH = -log [H+]
pH= - log[1.05 × 10^-5]
pH = 4.98
Acidic(less than 7)
(c). 0.5 × 20/40
= 0.25 M
Ka= Kw/kb
kb= 10^-14/1.8× 10^-5
Kb = 5.56×10^-10
x= (5.56×10^-10 × 0.5)^2
x= 1.667×10^-5 M
pH will be basic
The balanced half-cell equation for the reaction occurring at the anode is H2 ---> 2H(+) + 2e(-)
E<u>xplanation:</u>
- The balanced half-cell equation taking place at the anode is explained below
- The product produced in the reaction in the fuel cell is water.
- H2 ---> 2H(+) + 2e(-)
- In the above reaction, the oxidation state of hydrogen switches from 0 to +1.
- It is becoming oxidized by delivering two electrons at the anode.
- In the fuel cell, hydrogen molecules get oxidized to hydronium ions.Thus half-reaction is the oxidation reaction.
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Answer:
0.01 moles of SrCO₃
Explanation:
In this excersise we need to propose the reaction:
K₂CO₃ + Sr(NO₃)₂ → 2KNO₃ + SrCO₃
As we only have data about the potassium carbonate we assume the strontium nitrite as the excess reactant.
1 mol of K₂CO₃ react to 1 mol of Sr(NO₃)₂ in order to produce 2 moles of potassium nitrite and 1 mol of strontium carbonate.
Ratio is 1:1. In conclussion,
0.01 mol of K₂CO₃ must produce 0.01 moles of SrCO₃
