How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and lower the temperature to 205.0 K? The following physical data may be useful. ΔHvap = 33.9 kJ/mol ΔHfus = 9.8 kJ/mol Cliq = 1.73 J/g°C Cgas = 1.06 J/g°C Csol = 1.51 J/g°C Tmelting = 279.0 K Tboiling = 353.0 K

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

The expression used will be:  

where,

= heat released for the reaction = ?

m = mass of benzene = 94.4 g

= specific heat of solid benzene =

= specific heat of liquid benzene =

= enthalpy change for fusion =

Now put all the given values in the above expression, we get:

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

mention the spill to the teacher clean the area and throw the broken pieces away

explanation:

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.