Question

Derine mudily as the number of moles liter of solution. of solute in one . What is the uncertainty in a 100 ml buret? The density of a solution is 1.41 g/mL. If we have 27.5 grams of this solution. calculate its volume. Show all work. DE V 1.417mL V = 21.59 . van V = 19.5 mlnom 4. If 13 out of every 180 students wear glasses, calculate the percent of students who wear glasses. Show all work. 5. Calculate the percent uncertainty in measuring out 8.50 mL with a 10 ml pipet. Show all work.

Answer 1

14- The impossible orbital is 1p

Because when we calculate the name and number of orbitals in a shell, we follow the rule-

l = n-1

For l = 0,1,2,3 ,4.... the name of the subsells are s,p,d,f,g....respectively

where l = azimuthal quantum number. This quantum number tells us the maximum numbers of sub-shell a shell can have

n = principal quantum number. This number tells us the name of the shell

Now for the given 1p orbital, the 1 = represents the 1st shell

For 1st shell, n = 1

Thus for 1st subshell, l = n-1 = 1+1 = 0  

Again for value of l = 0, the only possible subshell = s

Thus the possible orbital for 1st shell is 1s and not 1p

15-

The half filled p-orbitals are present on the Gr-15 in the periodic table.

Now if we go along down a group, then mass of the element goes on increasing. Thus the lightest element is always the lightest element of the group.

And the 1st element of Gr-15 is Nitrogen (N)

Hence the lightest atom containing half-filled p-subshell is Nitrogen (N)

b- lightest atom containing half-filled d-subshell is from Gr-6, starting from Chromium. Thus the lightest atom containing half-filled d-subshell is Chromium (Cr)

c- Heaviest element with no d-orbital = Argon (1s2 2s2 2p6 3s2 3p6)

d- Elements with 2 unpaired 2p electrons

Carbon (C) = 1s2 2s2 2p2

Oxygen (O) = 1s2 2s2 2p4

e- Configuration of Ne = 1s2 2s2 2p6

Dipositive means loss of 2 electron. Thus an dipositive atom to have configuration of Neon means having means having 10 electrons after loss of 2 electrons

Thus initial electrons = 10+2 = 12

So the dipositive ion is Mg+2. Or the atom is Magnesium (Mg)

f-

Configuration of Ne = 1s2 2s2 2p6

Similarly one negative means gain of 1 electron. Thus an element to have configuration of Neon means having means having 10 electrons after gain of 1 electron

Thus initial electrons = 10-1 = 9

So the element with 9 electron is Flourine (F)

16-

The electronic configurations of the following elements are-

a - V(23) = Ar 3d3 4s2 = unpaired = 2

b- Mn(25) = Ar 3d5 4s2 = unpaired = 5

c- Cr(24) = Ar 3d5 4s1 = unpaired = 6

d- Fe(26) = Ar 3d6 4s2 = unpaired = 4

e- Zn(30) = Ar 3d10 4s2 = unpaired = 0