Liquid butane, C4H10, is stored in cylinders to be used
as a fuel. Suppose 35.5 g of butane gas is removed from a
cylinder. How much heat must be provided to vaporize this
much gas? The heat of vaporization of butane is 21.3 kJ/mol.

Explanation:

The heat to vaporize a liquid is equal to the amount of liquid in moles multiplied by the specific heat of vaporiztion per mole.

First, calculate the number of moles in 35.5g of butane.

Now, calculate the heat to vaporize that amount of liquid butane:

The answer must be reported with 3 significant figures.

explanation:

metals have excess of electrons whereas non-metals have deficiency of electrons. thus, in order to gain stability metals easily donate their electrons to the non-metals.

for example, electronic configuration of sodium is and electronic configuration of phosphorous is .

when a phosphorous atom and sodium atom combine together then sodium will donate is extra one electron to the phosphorous atom. as a result there will be formation of ionic bond between sodium and phosphorous.

the reaction will be as follows.

there, are three sodium sodium atoms and each of them will donate its one extra electron to phosphorous atom. thus, the sub-shell of phosphorous gets completely filled.  

thus, we can conclude that electrons are transferred from atoms of sodium to atoms of phosphorous. this transfer makes the sodium atoms donate one electron and the phosphorous atoms to gain one electron. as a result, the sodium and phosphorus atoms strongly attract each other.