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
