Balanced chemical equations show:

During a period of discharge of a lead-acid battery, 405 g of Pb from the anode is converted into PbSO4 (s).

Alexus [3.1K]

Answer:

The answers to the question are as follows

First part

The mass of PbO2 (s) reduced at the cathode during the period is = 467.55_g

Second part

The electrical charge are transferred from Pb to PbO2 is 377186.86_C or 3.909 F

Explanation:

To solve this, we write the equation for the discharge of the lead acid battery as

H₂SO₄ → H⁺ + HSO₄⁻

Pb (s) + HSO⁻₄ → PbSO₄ + H⁺ + 2e⁻

at the cathode we have

PbO₂ + 3H⁺ + HSO⁻₄ + 2e⁻ → PbSO₄ + 2H₂O

Summing the two equation or the total equation for discharge is

Pb (s) + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O

From the above one mole of lead and one mole of PbO₂ are consumed simultaneously hence

Number of moles of lead contained in 405 g of Pb with molar mass = 207.2 g/mole = (405 g)/ (207.2 g/mole) = 1.95 mole of Pb

Hence number of moles of PbO₂ reduced at the cathode = 1.95 mole

mass of PbO₂ reduced at the cathode = (number of moles)×(molar mass)

= 1.95 mole × 239.2 g/mol = 467.55 g of Lead (IV) Oxide is reduced at the cathode

Part B

Each mole of Pb transfers 2e⁻ or 2 electrons, therefore 1.95 moles of Pb will transfer 2 × 1.95 = 3.909 moles of electrons transferred

Each electron carries a charge equal to -1.602 × 10⁻¹⁹ C or one mole of electrons carry a charge equal to 96,485.33 coulombs

hence 3.909 moles carries a charge = 3.909 × 96,485.33 coulombs =377186.86 Coulombs of electrical charge

or transferred electrical charge = 377186.86 C or 3.909 Faraday

6 0

3 years ago

Are the 14 transition metals that follow lanthanum in the periodic table.

klio [65]

Answer: Lanthanide or Lanthanides

Explanation: Sorry that I don’t have an explanation. I just answered the question correctly myself. <33

3 0

3 years ago

3.

asambeis [7]

Answer:

zinc forms more than one ion

3 0

3 years ago

11. Assuming that the gases are ideal, calculate the amount of work done (joules) in each of the following reactions at 25 degre

gayaneshka [121]

Answer:

For a: Work done for the given reaction is 2477.572 J.

For b: Work done for the given reaction is 0 J

Explanation:

To calculate the work done for the reaction, we use the equation:

$W=-P\Delta V$

Ideal gas equation follows:

$PV=nRT$

Relating both the above equations, we get:

$W=-\Delta n_gRT$ ......(1)

where,

$\Delta n_g$ = difference in number of moles of products and reactants = $n_g_{(products)}-n_g_{(reactants)}$

R = Gas constant = 8.314 J/K.mol

T = temperature = $25^oC=[273+25]K=298K$

For a:

The chemical reaction follows:

$4HCl(g)+O_2(g)\rightarrow 2Cl_2(g)+2H_2O(g)$

$\Delta n_g=4-5=-1$

Putting values in equation 1, we get:

$W=-(-1mol)\times (8.314J/K.mol)\times 298K=2477.572J$

Hence, work done for the given reaction is 2477.572 J.

For b:

The chemical reaction follows:

$2NO(g)\rightarrow N_2(g)+O_2(g)$

$\Delta n_g=2-2=0$

Putting values in equation 1, we get:

$W=-(0mol)\times (8.314J/K.mol)\times 298K=0J$

Hence, work done for the given reaction is 0 J.

3 0

3 years ago

What is the density of motor oil in g/mL if 1.25 quart (qt) weighs 2.25 pounds (lb)?

Doss [256]

Answer:

0.86 g/mL

Explanation:

In order to solve the problem, you have to know the formula for "Density." Density refers to "mass per unit volume." It's formula is:

Density (ρ) = $\frac{mass}{Volume}$

Next, we have to convert the units of measurement in accordance to what is being asked in the problem. It is asking for "g/mL," which means we have to convert 1.25 quarts (qt.) to mL and 2.25 pounds (lbs.) to grams.

Let's solve.

1 quart = 946.35 mL

1.25 qt. x $\frac{946.35 mL}{1 qt.}$ = 1,182.94 mL

1 pound = 453.592 grams

2.25 pounds x $\frac{453.592 grams}{1 pound}$ = 1,020.58 grams

Now that we converted them to their appropriate units of measurement, we can now solve for density.

ρ = $\frac{mass}{Volume}$

ρ = $\frac{1,020.58 grams}{1,182.94 mL}$

ρ = 0.86 g/mL

The density of the motor oil is 0.86 g/mL

5 0

3 years ago