If the aluminum block is initially at 25 ∘c, what is the final temperature of the block after the evaporation of the alcohol? assume that the heat required for the vaporization of the alcohol comes only from the aluminum block and that the alcohol vaporizes at 25 ∘c. the heat of vaporization of the alcohol at 25 ∘c is 45.4 kj/mol, the specific heat of aluminum is 0.903 j/g⋅∘c

Explanation:

The given data is as follows.

       Heat of vaporization () = 45.4 kJ/mol

       Specific heat () = 0.903

Let us assume the alcohol given is and its mass is 1.12 g. Also, mass of aluminium block is 73.0 g.

First, calculate the moles of alcohol () as follows.

          No. of moles alcohol =

                                              =

                                              = 0.0186 mol

So, heat absorbed by the alcohol () = heat lost by aluminium ()

          =

              12.71 =

Thus, we can conclude that the final temperature of the block is 12.3^{o}C[/tex].

The final temperature of the block after the evaporation of alcohol is  12.3 .

The given data has:

The initial temperature of block =

Weight of block = 73 g

Weight of alcohol = 1.12 g.

The heat of vaporization of alcohol  (H)= 45.04 kJ/mol

Specific heat of aluminum = 0.903 J/g

Since heat energy can not be created nor be destroyed, the heat lost by the aluminum is equal to the heat absorbed by the alcohol.

Heat lost by aluminum = -mcT

Heat absorbed by alcohol = nH

moles of alcohol =

moles of alcohol =

moles of alcohol = 0.0186 moles

Heat absorbed by alcohol = 0.0186 45.04 kJ/mol

Heat absorbed by alcohol = 0.844 kJ

Heat absorbed by alcohol = Heat lost by aluminum

0.844 kJ =  -mcT

0.844 J = - 73 0.903 J/g (final temperature - )

0.844 J = -65.919 (final temperature - )

12 = - (final temperature - )

Final temperature = 12.3 .

The final temperature of the block after the evaporation of alcohol is  12.3 .

For more information about the heat of vaporization, refer to the link:

link

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.

it’s previous models   because when i picked that son of a gun’s answer it was wrong look for previous models because they might make it a different letter