# Guide n° 2 of solved exercises of units and notation

Guide n° 2 of solved exercises of units and notation

## Solve the following exercises

### problem #1

In one mole of molecules there are 602,000,000,000,000,000,000,000 molecules. Express this quantity as a power of ten with a single figure.

### problem #2

Make the following conversions:

a) 8h → s

b) 0.0200 mm → dm

c) 1 dl → µl

d) 8 cm → mm

e) 5 kg → mg

f) 9 m³ → l

g) 0.05 km → m

h) 2h 5m 15s → s

### Problem #3

How many significant figures does each of the following numbers have?

a) 8

b) 80

c) 8,000.00

d) 0.08

e) 0.080

f) 808

g) 3.14159

h) 3.1416

i) 3.14

i) 9.81

### Problem #4

Express in a single number:

a) 3.58 10 -2

b) 4.33 10³

c) 3.15 10 5

d) 5,303 10 -5

e) 6.94 10 -2

f) 0.003 10²

g) 6.02 10 23

h) 4.2 10³

i) 7.66 10 -4

j) 235 10 -5

### Problem #5

Carry out the following operations:

a) 4 10 5 2.56 10 4

b) 4.6 10 -5 – 6 10 -6

c) 5.4 10² + 3.2 10 -3

 d) 4.84 10 -5 2.42 10 -7
 and) 48.6 10² 0.524 10 -2 2.2 10³

### Problem #6

Express in scientific notation:

a) 4.59

b) 0.0035

c) 45,900,800

d) 0.0000597

e) 345,700,000

f) 0.03 10 5

### Problem #7

How many significant figures should appear in the results of the following accounts?

a) 5 0.006

b) 0.05 9.5 10²

c) 100 6

 d) 0.5 0.02
 and) 0.08 2 10 -2

# Solved problems of units and scientific notation

## Statement of exercise n° 1

In one mole of molecules there are 602,000,000,000,000,000,000,000 molecules. Express this quantity as a power of ten with a single figure.

### Solution

602,000,000,000,000,000,000,000 = 6.02 10 23

## Statement of exercise n° 2

Make the following conversions:

a) 8h → s

b) 0.0200 mm → dm

c) 1 dl → µl

d) 8 cm → mm

e) 5 kg → mg

f) 9 m³ → l

g) 0.05 km → m

h) 2h 5m 15s → s

### Solution

To convert units simply multiply and divide the “measure” (numerical value) by the unit expressed in both “magnitudes”.

a) We know that 1 = 60 minutes and that one minute is equal to 60 seconds, therefore:
8 h = 8 h (60 m/1 h) = 480 m = 480 m (60 s/1 m) = 28,800 s = 2.88 10 4 s

b) We multiply by 1 dm and divide by the equivalent 0.0000001 Mm:
0.0200 Mm = 0.0200 Mm (1 dm/0.0000001 Mm) = 200,000 dm = 2 10 5 dm

c) We multiply by 100,000 µl and divide by the equivalent 1 dl:
1 dl = 1 dl (100,000 µl/1 dl) = 100,000 µl = 1 10 5 µl

d) We multiply by 10 mm and divide by the equivalent 1 cm:
8 cm = 8 cm (10 mm/1 cm) = 80 m = 8 10¹ mm

e) We multiply by 1 kg and divide by the equivalent 1,000,000 mg:
5 kg = 5 kg (1,000,000 mg/1 kg) = 5,000,000 mg = 5 10 6 mg

f) We multiply by 1,000 l and divide by the equivalent 1 m 3 :
9 m 3 = 9 dm 3 ·(1,000 l/1 m 3 ) = 9,000 l = 9 10 3 l

g) We multiply by 1,000 m and divide by the equivalent of 1 km:
0.05 km = 0.05 km (1,000 m/1 km) = 50 m = 5 10¹ m

h) We know that 1 = 60 minutes and that one minute is equal to 60 seconds, in this case we proceed by parts and then add:

▫ 2h = 2h (60m/1h) = 120m = 120m (60s/1m) = 7200s

▫ 5m = 5m (60m/1h) = 300s

▫ 15s = 15s

▫ 2h 5m 15s = 7200s + 300s + 15s = 7515s = 7.515 10 3s

## Statement of exercise n° 3

How many significant figures does each of the following numbers have?

a) 8

b) 80

c) 8,000.00

d) 0.08

e) 0.080

f) 808

g) 3.14159

h) 3.1416

i) 3.14

i) 9.81

### Solution

 figures a) 8 b) 80 c) 8,000.00 d) 0.08 e) 0.080 f) 808 g) 3.14159 h) 3.1416 i) 3.14 i) 9.81 one one 6 one two 3 6 5 3 3

## Statement of exercise n° 4

Express in a single number:

a) 3.58 10 -2

b) 4.33 10³

c) 3.15 10 5

d) 5,303 10 -5

e) 6.94 10 -2

f) 0.003 10²

g) 6.02 10 23

h) 4.2 10³

i) 7.66 10 -4

j) 235 10 -5

### Solution

a) 3.58 10 -2 = 3.58/100 = 0.0358

b) 4.33 10³ = 4.33 1,000 = 4,330

c) 3.15 10 5 = 3.15 100,000 = 315,000

d) 5.303 10 -5 = 5.303/100,000 = 0.00005303

e) 6.94 10 -2 = 6.94/100 = 0.0694

f) 0.003 10² = 0.003 100 = 0.3

g) 6.02 10 23 = 6.02 100,000,000,000,000,000,000,000 = 602,000,000,000,000,000,000,000

h) 4.2 10³ = 4.2 1,000 = 4,200

i) 7.66 10 -4 = 7.66/10,000 = 0.000766

j) 235 10 -5 = 235/100,000 = 0.00235

## Statement of exercise n° 5

Carry out the following operations:

a) 4 10 5 2.56 10 4

b) 4.6 10 -5 – 6 10 -6

c) 5.4 10² + 3.2 10 -3

d) 4.84 10 -5 /2.42 10 -7

e) 48.6 10² 0.524 10 -2 /2.2 10³

### Solution

a) 4 10 5 2.56 10 4 = (4 2.56) 10 (5+4) = 10.24 10 9

b) 4.6 10 -5 – 6 10 -6 = 0.000046 – 0.00006 = -0.000014 = -1.4 10 -5

c) 5.4 10² + 3.2 10 -3 = 540 + 0.0032 = 540.0032 = 5.400032 10 2

d) 4.84 10 -5 /2.42 10 -7 = (4.84/2.42) 10 (-5-7) = 2 10 -12

e) 48.6 10² 0.524 10 -2 /2.2 10³ = (48.6 0.524/2.2) 10 (2-2-3) = 11.576 10 -3 = 1.1576 10 -2

## Statement of exercise n° 6

Express in scientific notation:

a) 4.59

b) 0.0035

c) 45,900,800

d) 0.0000597

e) 345,700,000

f) 0.03 10 5

### Solution

a) 4.59 = 4.59

b) 0.0035 = 3.5 10 -3

c) 45,900,800 = 4.5 10 7

d) 0.0000597 = 5.97 10 -5

e) 345,700,000 = 3.457 10 8

f) 0.03 10 5 = 3 10³

## Statement of exercise No. 7

How many significant figures should appear in the results of the following accounts?

a) 5 0.006

b) 0.05 9.5 10²

c) 100 6

d) 0.5/0.02

e) 0.08/2 10 -2

### Solution

a) 5 0.006 = 0.03 → 1 significant figures

b) 0.05 9.5 10² = 47.5 → 3 significant figures

c) 100 6 = 600 → 1 significant figures

d) 0.5/0.02 = 25 → 2 significant figures

e) 0.08/2 10 -2 = 4 → 1 significant figures

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