Mrs Salge is very particular about her ice cream. Her
ice cream cone recipe is 1 cone, 1 scoop of blue ice
cream and 2 scoops of red ice cream. There is also a
cherry on top!
How many ice cream cones could Mrs Salge make if
she had 10 cones, 12 scoops of blue ice cream, 12
scoops of red ice cream and 10 cherries?

Mrs. Salge can make 6 ice cream cones

Explanation:

Mrs. Salge's recipe:

1 cone

1 scoop blue ice cream

2 scoops red ice cream

1 cherry

Now we will find what the limiting ingredient is:

We know she has:

10 cones.    →    10 ice cream cones

12 scoops blue ice cream.    →    12 ice cream cones

12 scoops red ice cream.    →    6 ice cream cones

10 cherries.    →    10 ice cream cones

The red ice cream is the limiting factor. Mrs. Salge can make 6 ice cream cones.

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.

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole