Astudent enters the lab and conducts part a of the experiment. the student uses 32.00 ml of 2.202 m hcl, and adds naoh in excess as instructed. if the δh of the neutralization reaction is known to be -57,320 j/mol h2o, what is the total theoretical heat released (in joules)?

4012.4 Joules.

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

moles of

As is in excess , is the limiting reagent as will decide the amount of product formed and energy released.

Heat of neutralisation is the amount of heat released when 1 mole of an acid is completely neutralised by 1 mole of the base.

Given : = -57,320 J/mol

Thus if 1 mole of release = 57,320 J

0.070 moles release =

Thus the total theoretical heat released is 4012.4 Joules.

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.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3