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3 years ago

For any galvanic cell to operate at standard conditions,

Chemistry

1 answer:

shepuryov [24]3 years ago

4 0

Answer:

ΔG° is negative and ΔE° is positive

Explanation:

A galvanic cell produces electrical energy from spontaneous electrochemical processes.

In a galvanic cell, ∆E° is always positive since the reaction is spontaneous at room temperature.

Generally in thermodynamics, a negative value of ∆G shows a spontaneous chemical process. Hence in a galvanic cell, ΔG° is negative and ΔE° is positive.

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If you have a pH of 3.35, what is the hydrogen ion concentration of the solution?

melamori03 [73]

Answer: 4.46 x 10^-4M

Explanation:

The pH of a solution is the concentration of hydrogen ion concentration in the solution. Mathematically, it is expressed as pH = -log(H+), where H+ is the concentration of hydrogen ion

On the pH scale, readings are from 1 to 14.

- pH values less than 7 are regarded as acidic. So, the solution with pH 3.35 is said to be acidic, and will produce hydrogen ions.

3.35 = -log(H+)

(H+) = Antilog (-3.35)

(H+) = 0.000446M

Place (H+) in standard form

(H+) = 4.46 x 10^-4M

Thus, the concentration of hydrogen ion in the solution with pH 3.35 is 4.46 x 10^-4M

3 0

4 years ago

The half-life of radon-222 is 3.8 days. How many grams of radon-222 remain

Ne4ueva [31]

I think it’s B but not 100% sure

3 0

3 years ago

What are clean sources of energy

solong [7]

solar energy

wind energy

hydro energy

tidal energy

geothermal energy

3 0

3 years ago

Read 2 more answers

Liquid octane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 63. g of octane is mi

Sever21 [200]

Answer:

52.1 g is the maximum mass of CO₂, that can be produced by this combustion

Explanation:

Mass of Octane: 63 g

Mass of O₂: 59.4 g

This is a combustion reaction where the products, are always water and CO₂. We define the equation:

2C₈H₁₈ (l) + 25O₂(g) → 16CO₂(g) + 18H₂O(g)

As we have, both mases of each reactant, we must define which is the limiting reagent. We convert the mass to moles:

63 g. 1mol / 114g = 0.552 moles

59.4 g . 1mol / 32g = 1.85 moles

Certainly, the limiting reagent is the oxygen:

2 moles of octane need 25 moles of O₂ to react

Therefore, 0.552 moles of octane must need (0.552 . 25) /2 = 6.9 moles of O₂ (I do not have enough moles of oxygen, I need 6.9 and I only got 1.85 moles)

When we know the limiting reagent we can do the calculations with the stoichiometry of the reaction:

25 moles of O₂ can produce 16 moles of CO₂

Therefore, 1.85 moles of O₂ may produce (1.85 . 16) /25 = 1.18 moles.

We convert the moles to mass to get the final answer:

1.18 mol . 44 g / 1mol = 52.1 g

5 0

3 years ago

Read 2 more answers

For the reaction below, Kp = 1.59 at 100°C. If 1.0 g of SrCO3 is placed in an empty 5.00 L reactor and allowed to reach equilibr

Sergio [31]

Answer:

$p_{CO_2}^{eq}=1.59atm$

Explanation:

Hello, in this case, one could consider the undergoing chemical reaction as:

$SrCO_3(s)\rightleftharpoons SrO(s)+CO_2(g)$

Thus, since 1.0 g of strontium carbonate is placed, the equilibrium equation takes the following form, excluding the solid-stated species and considering just the carbon dioxide as it is gaseous:

$Kp=p_{CO_2}^{eq}=1.59atm$

Hence, since at the beginning there is no carbon dioxide, its pressure at equilibrium equals Kp:

$Kp=p_{CO_2}^{eq}=1.59atm$

Which was clearly defined above.

Best regards.

8 0

3 years ago