When dissolved in water, acids donate hydrogen ions (H+). Hydrogen ions are hydrogen atoms that have lost an electron and now have just a proton, giving them a positive electrical charge. ... If a solution has a high concentration of H+ ions, then it is acidic.
Answer:
3.81 g Pb
Explanation:
When a lead acid car battery is recharged, the following half-reactions take place:
Cathode: PbSO₄(s) + H⁺ (aq) + 2e⁻ → Pb(s) + HSO₄⁻(aq)
Anode: PbSO₄(s) + 2 H₂O(l) → PbO₂(s) + HSO₄⁻(aq) + 3H⁺ (aq) + 2e⁻
We can establish the following relations:
- 1 A = 1 c/s
- 1 mole of Pb(s) is deposited when 2 moles of e⁻ circulate.
- The molar mass of Pb is 207.2 g/mol
- 1 mol of e⁻ has a charge of 96468 c (Faraday's constant)
Suppose a current of 96.0A is fed into a car battery for 37.0 seconds. The mass of lead deposited is:
Answer:
NH4+
Explanation:
NH4+ is the acid and NH3 is the base, so NH4+ is the stronger acid.
Answer:
66 g of CO₂
Solution:
The Balance Chemical Reaction is as follow,
C₂H₂ + 5/2 O₂ → 2 CO₂ + H₂O
Or,
2 C₂H₂ + 5 O₂ → 4 CO₂ + 2 H₂O ------- (1)
Step 1: Find out the limiting reagent as;
According to Equation 1,
56.1 g (2 mole) C₂H₂ reacts with = 160 g (5 moles) of O₂
So,
125 g of C₂H₂ will react with = X g of O₂
Solving for X,
X = (125 g × 160 g) ÷ 56.1 g
X = 356.5 g of O₂
It means for total combustion of Ethylene we require 356.5 g of O₂, but we are only provided with 60.0 g of O₂. Therefore, O₂ is the limiting reagent and will control the yield.
Step 2: Calculate Amount of CO₂ produced as;
According to Equation 1,
160 g (5 mole) O₂ produces = 176 g (4 moles) of CO₂
So,
60.0 g of O₂ will produce = X g of CO₂
Solving for X,
X = (60.0 g × 176 g) ÷ 160 g
X = 66 g of CO₂
