Consider the following intermediate chemical equations.

A student does an experiment to determine the molar solubility of lead(II) bromide. She constructs a voltaic cell at 298 K consi

attashe74 [19]

Answer:

The molar solubility of lead bromide at 298K is 0.010 mol/L.

Explanation:

In order to solve this problem, we need to use the Nernst Equaiton:

$E = E^{o} - \frac{0.0591}{n} log\frac{[ox]}{[red]}$

E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.

At equilibrium, E = 0, therefore:

$E^{o} = \frac{0.0591}{n} log \frac{[ox]}{[red]} \\\\log \frac{[ox]}{[red]} = \frac{nE^{o} }{0.0591} \\\\log[red] = log[ox] - \frac{nE^{o} }{0.0591}\\\\[red] = 10^{ log[ox] - \frac{nE^{o} }{0.0591}} \\\\[red] = 10^{ log0.733 - \frac{2x5.45x10^{-2} }{0.0591}}\\\\$

[red] = 0.010 M

The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.

7 0

3 years ago

I searched it up but it keeps saying chemical energy, but in the option doesn’t have it...I’m torn between B and C...helppppp!!!

jeyben [28]

Answer: The correct answer is in chemical bonds

Explanation:

When coal is burnt, these components burn and release energy. The energy released is by the chemical reaction between the constituents and oxygen

8 0

2 years ago

The other half is posted !!!!!!

koban [17]

Answer:

this isn't an answer just need these points

Explanation:

3 0

2 years ago

Read 2 more answers

Bicarbonato es metal o no?

givi [52]

Answer:

La palabra bicarbonato es un término químico para referirse a una sal ácida del ácido carbónico en combinación con un metal. ... Pero si sólo está sustituido un átomo de hidrógeno por el metal, entonces obtenemos un bicarbonato o carbonato ácido, por ejemplo, bicarbonato sódico: CO3H Na.

7 0

2 years ago

The molarity (M) of an aqueous solution containing 22.5 g of sucrose (C12H22O11) in 35.5 mL of solution is ________.

Nonamiya [84]

Answer:

1.86 M

Explanation:

From the question given above, the following data were obtained:

Mass of sucrose (C12H22O11) = 22.5 g

Volume of solution = 35.5 mL

Molarity of solution =?

Next, we shall determine the number of mole in 22.5 g of sucrose (C12H22O11). This can be obtained as follow:

Mass of sucrose (C12H22O11) = 22.5 g

Molar mass of C12H22O11 = (12×12) + (22×1) + (16×11)

= 144 + 22 + 176

= 342 g/mol

Mole of C12H22O11 =?

Mole = mass /Molar mass

Mole of C12H22O11 = 22.5 /342

Mole of sucrose (C12H22O11) = 0.066 mole

Next, we shall convert 35.5 mL to litres (L). This can be obtained as follow:

1000 mL = 1 L

Therefore,

35.5 mL = 35.5 mL × 1 L / 1000 mL

35.5 mL = 0.0355 L

Thus, 35.5 mL is equivalent to 0.0355 L.

Finally, we shall determine the molarity of the solution as follow:

Mole of sucrose (C12H22O11) = 0.066 mole

Volume of solution = 0.0355 L.

Molarity of solution =?

Molarity = mole /Volume

Molarity of solution = 0.066/0.0355

Molarity of solution = 1.86 M

Therefore, the molarity of the solution is 1.86 M.

8 0

3 years ago