College level general chemistry class notes - 2021

,
, significant
Date : 23.08.21
-




numbers
✗ ample si Explanation :




5 →
1 ,
so →
7,500 →
7,5000--7 . . .
-

In a number , the zero is not significant
50 Once

becomes
put decimal the
-




.
→
2 , 50.0 → 3
, 50.00 →
4 . .
we a , number


significant . In the case of 50 .

,



we round to the nearest decimal . The



accuracy is . . .
[ 49.5
,
50 .
5)
With
•




a decimal ,
the zeros after the non
-




Zero numbers a re
significant .




5 1 The [ 0.045,0 )
-

→
0.5
→
7 0.05 7 of 0.05 is OSS
→
accuracy
. - -
. . .
.

, ,




The 0 050 [ 5)
'




0.050 →
2 ,
0 .
0500 → 3 .
.
.




accuracy
of .
. . . 0 .

0495,0 .
050


and 0.0500 . . -
[ 0.04995 , 0.05005 )
contexts :


-


The temperature outside is

28°C .
→
2 Significant
numbers .




1am km/h There
- -




travelling at 100 are 3
Sig .

fig because the accuracy is .
. .




→
3 numbers [ 99.5
significant , 100.5 )


There
>




30 students There
-




a re in a re infinite Sig fig because the number
my
-
.




classroom .
is exact . So . . . 30.000 - - -




Order of oppe ration :



Additions and Subtractions : In this case , we round off to the least

accurate number .




EXAMPLE .

-15.7+2.593-6.46=11.833 ( on calculator)

* Since the least accurate number is 15.913 Sig figs -
, the awnser is 11.8




Multiplications and divisions : You round off
your awnser to the least Sig .




figs given . .




EXAMPLE 16.78×0.0048=-1.39--0.051909352
:



* Since the number with the least Sig -


figs . is 0.004812 Sig figs) .
, the awnser is 0.052


EXAMPLE :
23.499-4.39×1.29=23.499 -


92.8=15.99 →
16.011 point after decimal )

, Date
temperature
-



.
24.08.21 .




Formulas :
conversion
T( F)
•
= IT c) +32




1- ( F)
•

1,8T / c)
☐
=
-132
-




Examples :
Solving steps :



I cook chicken at 400°F , T( F) 1,817°C ) -132
☐
=
a
-




what will be the temperature 400 =
1,817°C) -132
in Celsius ? T= 204,4°C


volumetric
- density
Formula :
Solving steps :




A) ① v=¥ñr3 ② f= F-
7- ✓ =
¥7120.033 m=f
f- v
-




✓ = 33 510cm
>
m =
9.87 ¥3 -




33510dm
'




m : mass in
grams
m= 2.675×1058×1100*099=11
✓ : volume in cm3( solids) 2451b¥
Or volume in m2 ( liquids) m= 581.5 lbs



Examples :
B) ① ② V= Ab -
h
-




A 40,0cm diameter 18in ✗
74k¥ V= 45.92cm ✗ 60.96cm -




Cannonball has a den =
45.92cm =
2989h cms

sity of
7.87g / cm3 .




"% ¥°"



✓
"
" ^^ " " " " be +" ^^ " "*
× ✗
"


in pounds (V =
¥1T r 3) =
60.96cm .
=
0.0606µm
③ 7-
g=
Aluminum density ¥4s
>




has a of ✓ =

. 6×10 -3g
2-
70g / cm ? If I cut out a
piece 2787h --
2.9¥
with a mass of
45.6mg and h =
6.06×10
-
•
Clm ✗
t¥m
dimensions
with the
following :




aetbyieoincnes
is the foil in
µm
,



?
now thick
CONVERSION TRICK !
•
632hm in m .
.
.
•
99.5MHz in H2
-




EXTRA . .
.
nano is ✗ 10-9 Mega is ✗ 10°
°

SO -
- -
632×10-9m So - . .
99.5×10 HE
A has fuel
consumption
car a


42.85¥ C) ① of - If 600km ✗ 0.6>214 F¥m×t£¥mi 98¥ .¥6pt× ✗ ×



the distance between 1-7×65-1
Mtl and Toronto is 600km ,
=
53.85$
how much would it cost to

travel to Toronto ?