CHEMISTRY HIGH SCHOOL

The standard cell potential for the aqueous reduction of Pb4+ to Pb2 + ions by the corresponding oxidation of Ce3+ to Ce4+ ions, Pb4 •(aq) + 2 Ce' •(aq) --'> Pb'•(aq) + 2 Ce4+(aq) is 0.06 V a

Answers

Answer 1

Answer:

Answer:the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.

Explanation:The given chemical reaction is:

Pb^4+(aq) + 2 Ce^3+(aq) -> Pb^2+(aq) + 2 Ce^4+(aq)

The standard cell potential (E°cell) for this electrochemical reaction is 0.06 V.

The standard cell potential for a galvanic cell can be calculated using the Nernst equation:

E°cell = E°cathode - E°anode

In this reaction, Pb^4+(aq) is being reduced to Pb^2+(aq), so it is the reduction half-reaction, and Ce^3+(aq) is being oxidized to Ce^4+(aq), so it is the oxidation half-reaction.

The standard reduction potentials (E°) for the half-reactions are as follows:

For the reduction half-reaction:

Pb^4+(aq) + 2 e^- -> Pb^2+(aq) E°red = x (we'll solve for x)

For the oxidation half-reaction:

2 Ce^3+(aq) -> 2 Ce^4+(aq) + 2 e^- E°red = 1.44 V (This value is usually given)

Now, plug these values into the Nernst equation:

E°cell = E°cathode - E°anode

0.06 V = x - 1.44 V

Now, solve for x:

x = 0.06 V + 1.44 V

x = 1.50 V

So, the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.

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What occurs when NaCl(s) is added to water?(1) The boiling point of the solution increases, and the freezing pointof the solution decreases.(2) The boiling point of the solution increases, and the freezing pointof the solution increases.(3) The boiling point of the solution decreases, and the freezing pointof the solution decreases.(4) The boiling point of the solution decreases, and the freezing pointof the solution increases.

Answers

The answer is (1) Boiling point increases and freezing point decreases. This is a colligative property of solutions. When you add a solvent to a solution, the effect is twofold: boiling point elevation and freezing point depression. In effect, you broaden the range of temperatures over which the solution remains liquid. This is why you might see people putting antifreeze in their cars during the summer. While antifreeze does depress the freezing point of the engine oil, it also raises its boiling point, so it doesn't boil as easily during the summer when the engine may become especially hot.

The vapour pressure of the solution is decreased by sodium chloride. With a drop in vapour pressure, the boiling point rises. When sodium chloride is added to boiling water, the resultant solution's boiling point rises relative to that of pure water. Here the correct option is 1.

NaCl solution is created when NaCl is added to water. The solution exerts less vapour pressure than pure water because there are comparatively fewer water molecules at the liquid's surface. A decrease in the freezing point of water is seen as a result of this drop in vapour pressure.

When you add a solvent to a solution, the effect is two-fold: boiling point elevation and freezing point depression. The Boiling point increases and the freezing point decreases. This is a colligative property of solutions.

Thus the correct option is 1.

To know more about colligative property, visit;

brainly.com/question/30799665

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Hydrogen sulfide burns form sulfur dioxide:2H2S(g) +3O2(g) → 2SO2(g) + 2H2O(g) ΔH= -1036 kJ

Calculate the enthalpy change when burning 26.7 g of hydrogen sulfide in kJ

pls help

Answers

Answer: 404.04 kJ.

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

moles of $H_2S$

$\text{Number of moles}=(26.7g)/(34.1g/mol)=0.78moles$

$2H_2S(g)+3O_2(g)\rightarrow 2SO_2(g)+2H_2O(g)$ $\Delta H=-1036kJ$

According to stoichiometry :

2 moles of $H_2S$ on burning produces = 1036 kJ

Thus 0.78 moles of $H_2S$ on burning produces = $(1036kJ)/(2)* 0.78=404.04$

Thus the enthalpy change when burning 26.7 g of hydrogen sulfide is 404.04 kJ.

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d) pressure

Answers

Answer:

Temperature, pressure, shape of container

Answer:

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Explanation:

1. An unknown compound was found to have a percent composition of: 47.0%potassium, 14.5% carbon, and 38.5 % oxygen. What is its empirical formula? If
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Answers

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Explanation:

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Answers

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Answer: chronic is long term and severe while acute lasts for a short amount of time.

What is the volume of the sample at stp ?

Answers

At stp the volume is 22.4 L .

hope this helps!

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