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:
79% of Mg-24
10% of Mg-25
11% of Mg-26
18.96 + 2.5 + 2.86 = 24.32u
Huh I do not understand this
Hey there!
Values Ka1 and Ka2 :
Ka1 => 8.0*10⁻⁵
Ka2 => 1.6*10⁻¹²
H2A + H2O -------> H3O⁺ + HA⁻
Ka2 is very less so I am not considering that dissociation.
Now Ka = 8.0*10⁻⁵ = [H3O⁺] [HA⁻] / [H2A]
lets concentration of H3O⁺ = X then above equation will be
8.0*10−5 = [x] [x] / [0.28 -x
8.0*10−5 = x² / [0.28 -x ]
x² + 8.0*10⁻⁵x - 2.24 * 10⁻⁵
solve the quardratic equation
X =0.004693 M
pH = -log[H⁺]
pH = - log [ 0.004693 ]
pH = 2.3285
Hope that helps!
Answer:
3.676 L.
Explanation:
- We can use the general law of ideal gas: PV = nRT.
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
- If n and P are constant, and have different values of V and T:
(V₁T₂) = (V₂T₁)
V₁ = 3.5 L, T₁ = 25°C + 273 = 298 K,
V₂ = ??? L, T₂ = 40°C + 273 = 313 K,
- Applying in the above equation
(V₁T₂) = (V₂T₁)
∴ V₂ = (V₁T₂)/(T₁) = (3.5 L)(313 K)/(298 K) = 3.676 L.
