I was homeless and busking, he said. I was living nowhere, I was
living in the streets. I met a lot of people but couldnt speak the
language so I had to just get on.
I realised that we are all equal and that its all about helping each
other and I learned

. Using the musim Tenors between W and Huh. m and ft. and K and FL the Etefan-Bdtzmam
mar: :55?- m-B W! rt? .Ic£1 isle be emreeaedintheEngim Lrit. Elu,‘ h_ﬂ? 514.
Analysis The mmﬁdﬂ'sfurw, rn, and K areg'vmincu‘umim labia-statue
1 W: 3.41214 Elufh
1m=123m

. We know hfa = ZWIS, so from Eq. {8.30) we may de'lemline the stream
velocny:
h = 20 = coil hfa ]= mil 2ﬂf15 :|, solve for U. II 12.9 E Ans.
a 15 2m!(U_,a) 2(2):m.15um} 5
Then —=—=l.ﬂ36, —=[l+2(1.036)] =1.75, 2L—53cm Ans.
Uﬂa 12.9{015} :1
Finally, Vm

. The souroe and sin]: are each
1'5=224mﬁ'ompointﬁ,sothesink
velocity is 1W224 = 4.47 this and the
source velocity is 132.24 = 5.37 mfg, as
shown. The vortex velocity is 912 = 4.5 mfg.
The net horizontal component is 6.44 11113.
The net vertical component

bl
Here are two examples of cyclic isomers of HOSCN with zero formal charges on all of the atoms. All non-
hyrdrogen atoms must have octets of electrons.
FE. = valence if — # of "dots” {if non—bonding e"s]l — # of "sticks" [1Q the if of bonded e"s}
i}

Focus on the most stable structure from part [a].
A group is deﬁned as an atom aridfor lone pair of electrons attached to an atom. The hybridization scheme
[and associated bond angles] of an atom based on the number of attached groups is:
2 groups —> sp h

iii}
To determine the number of unpaired electrons, allocate the electrons to the last subshell in the abbreviated
electron conﬁguration using Hund’s nJIe [maximizing parallel electron spins], then count the number of
unpaired e"s.
Mn: [Ad-1523f L; L; L B

H is not hybridized; it forms a bond mm a 13 orbital
CI has 4 groups attached; therefora it is spa hybridized
S has 4 groups anach-ad; merefore, it is sit:3 hyblidiz-ad
C has 2 groups atlabhed; therefore, it is 3p hybridized
N has 2 groups attached; there

The total number of 1«valence electrons in HOSE” is:
# of valenc e electrons
1XH=1
1x0=ﬁ
1XS=E
1xC=4
1XN=5
22electrons
3}
There are man-3,:r possible isomers of HDSCN and several are shown below. The most stable structure[s} must
have zero formal charges

CHEM 120 (Online) Hand-in Assignment #2 (Spring 2015) page 3 of 19
The minimum photon energy required to eject an electron from K metal is provided by the work function and is
3157,40"19 J. By setting the energy difference between the n = 5 to n = ? level

SCHEMATIC AND GWEN DATA:
T1=1DDEJ K 1
p1=8bar 1—1 bar
r5: =2kgfs
L _ I
T Sher
lﬂGﬂK
Iber
{a} As the expansion is isentrepie; E1 — it = (1', where
_ =yli1(§2 _’§1)‘NI "Pfennig: “Educ, +J’H10ﬁz _E1}H!ﬂ
and each 01" the y“: (52 — it)“: , You: ("3'1 — Er