2. By the end of this chapter you should:
Know the 7 SI basic units and their prefixes.
Be able to convert from one unit to other.
Know to derive units from the 7 SI basic units.
Common units (L & mL)
Know the temperature Scales
Dr. LAILA AL-HARBI
3. Chemistry is the study of matter and the changes it
undergoes
There are three states of matter
solid
liquid
gas
Dr. LAILA AL-HARBI
5. 5
The SI unit of mass is
(a) The pound
(b) The gram
(c) The kilogram
(d) The mole
Examples
The Kg is the SI unit of
(a) length
(b) mass
(c) temperature
(d) current
9. Dr. LAILA AL-HARBI
I. Convert 134 pm to m?
II. How many meters are in 134 pm?
نقسم كبير الى صغير من للتحويل÷
1 pm = 1×10-12 m
134 pm = ?? m
االعداد خط باستخدام الحل
الحلوسطين في طرفين الجدول باستخدام
134 ×1×10-12 m = 1.34×10-10 m
10. The SI prefixes giga
and micro represent,
respectively:
A. 10-9 and 10-6.
B. 106 and 10-3.
C. 103 and 10-3.
D. 109 and 10-6.
The SI unit of mass is
(a). The pound
(b). The gram
(c). The kilogram
(d). The mole.
Dr. LAILA AL-HARBI
11. How many microseconds are in a second?
(a). 1 x 10-1
(b). 1 x 10-6
(c). 1 x 10-15
(d). 1 x 10 6
Explanation: Since the prefix micro means 1 x 10-6, there will be
1 x 10 + 6 microseconds in one second.
Dr. LAILA AL-HARBI
1 µs= 1 x 10 -6 s
x µs = 1 s
s µs
1 10 6x 1 x 10 6 µs
12. Which of the following is the smallest distance?
(a) 21 m → 21m
(b) 2.1 x 102 cm → 2.1m
(c) 21 mm → 0.021 m
(d) 2.1 x 104 pm → 2.1 x 10-8 m
Put all of them in the same unit
Explanation: Even though 2.1 x 104 is the largest
number in this question, the units of pm (picometers)
are the smallest units here, making it the smallest
distance.
Dr. LAILA AL-HARBI
13. A) 6.0 km is how many
micrometers?
Solution 1
1 km = 103 m
6 km = x = 6 × 103 m
1µm = 1× 10-6 m
x = 6 × 103 m
x = 6 × 109µm
Dr. LAILA AL-HARBI
Explanation: convert first to meter then from meter to micro
( two steps solution )
km µm
×6 109
6 × 109
14. Example
The diameter of an atom is
approximately 1 10-7 mm.
What is this diameter when
expressed in nanometers?
A. 1 10-18 nm
B. 1 10-15 nm
C. 1 10-9 nm
D. 1 10-1 nm
= 1× 10-7 × 1 × 106 =
1 × 10-1 nm = 0.1 nm
Example
Which of these quantities
represents the largest mass?
A. 2.0 102 mg
B. 0.0010 kg
C. 1.0 105 g
D. 2.0 102 cg
Put all of them in the same
unit
Dr. LAILA AL-HARBI
A) 0.2 g
B)1 g
C) 0.1 g
D) 2 g
15. are defined in terms of the seven base quantities via a
system of quantity equations.
The SI derived units for these derived quantities are
obtained from these equations and the seven SI base
units. For example
Area = width x length
Unit of width = m
Unit of length = m
Unit of Area = m× m = m2
Dr. LAILA AL-HARBI
Treat units like numbers
16. Volume –
Volume = width × length × heights = m × m × m = m3
SI derived unit for volume is cubic meter (m3)
Common unit of volume is liter (L) and milliliter (ml)
The relation ship between liter (L) and ml (1L= 1000mL)
The relation ship between liter (L) and metric system
1 L = 1 dm3
The relation ship between milliliter (ml) and metric system
1 mL = 1 cm3
1 cm3 = (1 x 10-2 m)3 = 1 x 10-6 m3
1 dm3 = (1 x 10-1 m)3 = 1 x 10-3 m3
Dr. LAILA AL-HARBI
17. Example (1)
Example (2)
How many liters are in 25
dm3 ?
Science 1L = 1 dm3
25 dm3 = 25L
How many milliliters are in
32 cm3 ?
Science 1mL = 1 cm3
32 cm3 = 32 mL
Dr. LAILA AL-HARBI
Example (3)
How many liters are in 250
cm3 ?
Science 1L = 1 dm3
and 1mL = 1 cm3
250 cm3 = 250 mL
L → mL
250/1000 = 0.25 L
18. How many cubic
centimeters are there in
exactly one cubic meter?
A. 1 10-6 cm3
B. 1 10-3 cm3
C. 1 10-2 cm3
D. 1 106 cm3
Solution
(1m )3 = (cm )3
1m3 = (1 102)3 cm 3
1m3 = 1 106 cm3
Dr. LAILA AL-HARBI
The diameter of an atom
is approximately 1 10-7
mm. What is this diameter
when expressed in
nanometers?
A. 1 10-18 nm
B. 1 10-15 nm
C. 1 10-9 nm
D. 1 10-1 nm
19. Mass is the measure of the amount of
matter in an object.
SI unit of mass is the kilogram (kg)
1 kg = 1000 g = 1 x 103 g
Weight is the measurement of the pull
of gravity on an object.
The Mass of an object doesn't change
when an object's location changes.
Weight, on the other hand does change
with location.
Chemist are interested primarily in mass
weight = c x mass
The weight of man on earth is 50 pounds.
is 8.25 pounds on moon
Dr. LAILA AL-HARBI
20. Density is defined as the amount of
matter in a given amount of space.
SI derived unit for density is kg/m3
common units of density are g/mL ,
g/L
Density decrease with temperature
(g/ml )g/cm3 for liquid and solids
g/L = 0.001g/ml for gases
Because density of gases are very
low The density of copper
is 8.94 g/cm3.
density =
mass
volume
Dr. LAILA AL-HARBI
21. 21
Given
m & V
? d
d = m/V
Given
m & d
? V
V = m /d
Given
V & d
? m
m = d x V
22. A piece of Gold metal
has a volume of 15.6
cm3, with a mass of
301 g What is its
density?
A piece of platinum
metal with a density of
21.5 g/cm3 has a
volume of 4.49 cm3.
What is its mass?
d =
m
V
m = d x V
= 21.5 g/cm3 x 4.49 cm3
= 96.5 g
d =
m
V
301 g/ 15.6 cm3
= 19.3 g/ cm3
Dr. LAILA AL-HARBI
23. The density of mercury
is 13.6 g/mL has a
volume of 5.50 mL.
What is its mass?
The density of sulfuric
acid is 1.41 g/mL has a
volume of 242 mL.
What is its mass?
d =
m
V
m = d x V
= 1.41 g/mL x 242 mL
= 341.22 g
d =
m
V
m = d x V
= 13.6 g/mL x 5.50 mL
= 74.8 g
Dr. LAILA AL-HARBI
24. Fahrenheit °F: °F = [ (9/5) × °C] + 32
Celsius °C: °C = (5/9) (°F - 32)
Kelvin K: K = °C + 273.15
Dr. LAILA AL-HARBI
25. 25
Temperature Units Conversion
1. Degrees Celsius 0C: Scale 0 → 100 Thus: 100
divisions or 100 degrees
2. Kelvin K: Scale 273 → 373
Thus: 100 divisions or 100 degrees
1K = 1 C
3. Degrees Fahrenheit 0F : Scale from 32→ 212
Thus: 180 divisions or 180 degrees
Thus: the size of degree in 0F scale is only 100/180 or
5/9 of a degree on the 0C scale 10F = (5/9) 10C
26. Convert 224 0C to degrees Fahrenheit?
°F = (9 0F /5 0C) × °C + 32
[°F = (9 0F /5 0C) × 224 °C] + 32 0C = 435 0F
Convert -452 0F to degrees Celsius.
°C = (5 0C /9 0F) (°F - 32 0F)
°C = (5 0C /9 0F) (-452 °F - 32 0F) = -269 0C
Convert -38.9 0C to degrees Kelvin..
° K = [-38.9 °C + 273.15 °C ] × 1 K/ 1 0C = 234.3 K
Dr. LAILA AL-HARBI
27. Ammonia boils at -33.4C. What temperature is
this in F?
A. -60.1F
B. -92.1F
C. -28.1F
D. +13.5F
F = (9 0F /5 0C) × °C + 32
[°F = (9 0F /5 0C) × -33.4 °C] + 32 0C = - 28.1 0F
Dr. LAILA AL-HARBI
Notas del editor
عزيزتي الطالبة تعلمي بناء بنك أسئلة الخاصة بك ... ومن الجدول السابق نجد انه يمكننا عمل 14 سؤال اما بالسؤال عن الوحده الاساسية كما في المثال الاول أو السؤال عن الكمية مثل السؤال الثاني
هذا الجدول يعطي الكمية بالمقارنه بالمتر مثلا الواحد كيلو يساوي 1000 متر و ممكن استخدام طريقة الطرفين بالوسطين لمعرفة الكمية المطلوبة مثل الامثلة التي سنراها في الامثلة التالية
عند استخدام طريقة خط الاعداد اضربي في معامل التحويل كما هو – بعض الطالبات يتلخبطوا بين هذه الطريقة و الجدول يمكنكم استنتاج الجدول من خط الاعداد ببساطة مثلا 1 تترا تكافئ = 1 اس 12 متر
بينما 1 بيكومتر = 1 اس -12 لماذ؟ لاننا حولنا من صغير الى كبير فقسمنا فتحول المعامل الى سالب
عزيزتي الطالبة لا تجعلي صيغة السؤال تغير من طريقة حلك بالرغم من اختلاف الصيغتين فان السؤال واحد و هو تحويل 134 بيكومتر الى متر لو استخدمنا طريقة خط الاعداد نظمي طريقة حلك حددي الوحده المعطاة ثم الوحدة المطلوبة حددي العملية ثم المعامل و توجدي الحل بسهولة
لا يمكننا مقارنة قيم عددية لوحدات مختلفة لذلك يجب تحويلها لنفس الوحده حتى تكون المقارنة سهله
لكي نختار الوحده أو الاكبر يجب وضعها بنفس الوحده حتى نقارن بينهم
نظرا لوجود عدد كبير من القوانين ... فانه لايمكن التعبير عنها فقط بالوحدات الاسياسية لذلك نقوم باشتقاق وحدات لهذه الكميات عن طريق وضع القانون للكمية المطلوبة ثم التعويض بوحدة كل كمية و معاملة الوحدات معاملة الارقام من حيث الضرب و القسمة
تذكري أنه لو كان لدي قانون فيه ثلاث متغيرات ممكن يأتي السؤال بأكثر من صيغة بأن يعطيك قيمتين و يطلب الثالث كما هو موضح في الشكل