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:
D) HCOOCH2CH3
Explanation:
An isomer of a compound is defined as a chemical substances with the same formula (That is, same atoms) but in different structures.
For propanoic acid, there are <em>3 atoms of C, 6 atoms of H and 2 atoms of oxygen.</em>
A) CH2CHCOOH
. Here you have 3 atoms of C but 4 atoms of H. That means this compound is not an isomer.
B) CH3CH2CH2COOH
. Here, there are 4 atoms of C. Thus, is not an isomer.
C) CH3CH(OH)CH2OH. This structure has 3 atoms of C, but 8 atoms of H. Thus, is not an isomer.
D) HCOOCH2CH3. Here, there are 3 atoms of C, 6 atoms of H and 2 atoms of O. Thus, this structure <em>is an isomer of propanoic acid.</em>
A: It prevents most reactions from occurring spontaneously.
The answer above is correct.
Answer: -2.373 x 10^-24J/K(particles
Explanation: Entropy is defined as the degree of randomness of a system which is a function of the state of a system and depends on the number of the random microstates present.
The entropy change for a particle in a system depends on the initial and final states of a system and is given by Boltzmann equation as
S = k ln(W) .
where S =Entropy
K IS Boltzmann constant ==1.38 x 10 ^-23J/K
W is the number of microstates available to the system.
The change in entropy is given as
S2 -S1 = kln W2 - klnW1
dS = k ln (W2/W1)
where w1 and w2 are initial and final microstates
from the question, W2(final) = 0.842 x W1(initial), so:
= 1.38*10-23 ln (0.842)
=1.38*10-23 x -0.1719
= -2.373 x 10^-24J/K(particles)