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

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

we can conclude that the melting point of table salt is above room temperature (it's actually 801 deg c). the melting point of mercury must be below room temperature (- 39 deg c).