Calculate the standard electrode potential of lead electrode, if the electrode potential is -0.18025 V at 301 K and a concentration of Pb2+ solution is 0.0096 M.

Explanation:

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The heat required to vaporize the water is 76.266 kJ or 76266.32 J.

Explanation:

Vaporization or commonly known as evaporation is the process where the liquid generates a vapor pressure above itself.

The heat of vaporization or the enthalpy of vaporization (ΔH vap) is the heat required in order to change the phase of the liquid.

The molecules of a liquid moves constantly and have a various range of kinetic energies. There will be a time that some of the liquid's molecules will have enough energy to escape from the surface and eventually will enter the gas phase (vapor phase).

For this problem, we can use the known enthalpy of vaporization of water.

The enthalpy of vaporization of water is 2256.4 J/g. This means that in order to convert 1 g of water at 100°C to steam at 100°C, a heat amounting to 2256.4 J is required.

Solution:

For our case, we have a mass of water amounting to 33.8 g at 100°C and we want to vaporize it to steam (also at 100°C).

Therefore, the heat required to vaporize the water is 76.266 kJ or 76266.32 J

Learn more about heat of vaporization here:

More example problems of vaporization/ evaporation here:

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Hello!

How many atoms do you have in 19 g of potassium ?  

A. 2.93x10^23 atoms  

B. 0.48 atoms  

C. 1.24x10^23 atoms  

D.2.05 atoms

Solving:

Molar mass of potassium ≈ 39 g/mol

Knowing that by Avogadro's Law for each mole of a substance we have 6.022 * 10²³ atoms or molecules. How many atoms do you have in 19 g of potassium?

39 g ---------------- 6.022*10²³ atoms

19 g ----------------  y atoms

multiply the means by the extremes

A. 2.93x10^23 atoms

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