How much heat is required to vaporize 33.8 g of water at 100 degrees celsius

Answer : Heat required will be 76.42 KJ

Explanation : Here we have the data as 33.8 g of water and temperature is 100°C

To calculate the heat required we need to use the formula as,

q= mol of water / heat of vaporization of water.

So we have convert the mass into moles,

33.8g / 18g = 1.877 moles of water.

and heat of vaporization is 40.7 KJ

Now, q = 1.877 X 40.7 KJ = 76.4 KJ

So the heat required will be 76.4 KJ

metals are good conductor of electricity. it has free electrons on their surface. according to electron pool theory, each atom present in a metal crystal losses all of its valence electron and as a result a pool of free is form. these pool of electrons are free to move and   conduct electricity easily   while making metal as good conductor.

the correct answer is a. the ability of electrons to flow throughout the metal.

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