1. Rekenen met grote en kleine getallen.

(1)

3 3

4 8

4 2 8 2

2 10 2,5 10

25 10 60 25 10 6 6 5 10

10 1 6 ) 10 5

( ⋅ ⁻ × ⋅ = ⋅ ⁻ × ⋅ = ⋅ = ⋅ = ⋅

Uitwerkingen hoofdstuk 1 versie 2014

1. Rekenen met grote en kleine getallen.

Opgave 1.1 Schrijf de volgende notatie zo kort mogelijk

a ₆ ⁶

2 10

3 2

2 5

2

10 10 2

10 2 10 )

10 (

10 2 ) 10

( × ⋅ = ⋅

⋅ =

×

b ₆ ⁶ ⁴

2 6

2 3

2

10 10 2

10 2 )

10 (

10 2 )

( ₋

⋅

⋅ =

= ×

⋅

× a b a a b

b

c ₆ ⁶ ⁶

3 3

3 2

3 3

2 10 1 10

2 1 )

10 ( 2

)

( ₋

⋅

−

⋅ =

×

⋅

= −

×

−a a a a a

d a²+3a² =1⋅a²+3⋅a² =4a²

e a² +a³ =1⋅a² +a⋅a² =(1+a)⋅a² of a²(a+1) is niet echt korter, dus laten staan is ook goed!

f (−3a²)³ =(−3)³⋅(a²)³ =−27a⁶

g ₃ ⁶ ³

6

3 3 2 3

2 ( )

)

( = = of a ⋅b⁻

b a b a b

a

Opgave 1.2 Schrijf als macht of getal in SCI-mode.

a ₂₃ ²⁷ ²⁷

4

3 10 10 1

6 2 10 6

10

2 ⁻ ₋ ₋

⋅

=

⋅

⋅ =

⋅

b ₈ ¹⁵ ⁵

3 15

4 2

3 3 5

) (

)

( x y

y y x y

y

x ⋅ = ⋅

⋅ =

−

c 1

1

1⁻³¹⁹ = ₃₁₉1 = d 23⁰ = 1

e (a+b)⁰ =1

f 5²×5⋅10² =5³⋅10²

g 1⁻³¹⁹ = 1 zelfde opgave als c) h (−1)³⁹⁹ =−1⋅(−1)³⁹⁸ =−1 i (−1)⁴⁰⁰ =1

j ² ² ² ⁴ ₂ ⁴ 10 ⁶ 4 10 ⁶

25 10 100 5 10 1 5 ) 10 5

( ⋅ ⁻ = ⁻ ⋅ ⁻ = ⋅ ⁻ = ⋅ ⁻ = ⋅ ⁻ = k (0,1)⁴ =(10⁻¹)⁴ =10⁻⁴

l

(2)

Opgave 1.3 Vul de ontbrekende exponent in.

a 3³×3⁴ =3⁷

b 12³×2³ =(2²×3)³×2³ =2⁶^⋅×3³×2³ =2⁹×3³

c ₃ ⁶ ² ³ ³ ⁹ ⁵

2 3

3 2

3 2 3 2 3 ) 2

3 2 (

) 3 2 ( 6

24 = × × × = ⋅

×

= × ₋

−

d ₃ ₂ ₆ ⁶ ⁶

0

3 ) 2

3 2 (

1 )

6 (

16 = ×

= × ₋

−

e ₂ ₂ ² ⁰

2

2 2

3 2 2

1 ) 8 2 (

8 16

8 = = ⋅

= × ⁻

f ₄ ² ⁴

2

4 2

5 ) 3

5 3 (

3 15

3 ₋ ₋

×

× =

=

g ₄ ² ²

2

4 2

5 5 2

) 5 2 ( 5

10 ₋

×

× =

=

h ₂ ⁰

2

10 10 =a

h ₀ = ⋅ ⁻¹⋅⁰ ⁻¹

⋅ J kg C

C kg

J

Opgave 1.4 Oefenen met vermenigvuldiging van machten.

Maak eerst een schatting en controleer vervolgens met je rekenmachine.

a (10²)³^,¹ ≈10⁶ met rekenmachine (10²)³^,¹ =1,58⋅10⁶!

b ₁₁ ¹⁷

6 3

11 3 2

10 10 2

4 10 2 10

4 ) 10 1 , 2

( ≈ ⋅

⋅

≈ ⋅

⋅

−

c ₃ ²¹

18 3 3 6

10 10 10 1 999

) 10 1 , 1

( ⁻ ⁻ ₋

⋅ ≈

⋅ ≈ d 1+10⁻⁵ ≈1

e ⁸ ⁸

6 3 2 3

10 8 12 10

100 12 10 12

) 10 ( 12

) 011 , 0

( ⁻ ⁻ ₋ ₋

⋅

≈

⋅

≈

f ₈ ¹⁴

3 3

3

8 3 3

10 10 4

4

10 2 10 2 10

2 , 4

10 2 ) 10 2

( ≈ ⋅

×

≈ ×

⋅

×

⋅

−

g ₆ ₆ ⁶ 10 ⁷ 2,5 10 ⁷

4 10 10 4 1 10 4

1 10

4 02 ,

1 ₋ ₋ ₋

⋅

≈

⋅

≈

⋅

⋅ ≈

h 0,25

4 1 10 4

) 110 , 2 (

0 0 2

≈

⋅ ≈

i 1,5 10 15

10 2

10 3 10 21 , 2

10 0 ,

3 ₁

3 4

=

⋅

⋅ =

≈ ⋅

⋅

j ₁ ³

4

1 4

10 10 2

2 10 4 10 21 , 2

10 16 ,

4 = ⋅

⋅

≈ ⋅

⋅

k ₁ ²

3 3

3

10 10 8

120 10 1000 1200

10 1000 1200

10 04 , 5

200 ≈ ⋅

⋅

= ⋅

≈ ⋅

⋅

×

(3)

l 2 10 20 210

2×2230 ≈ × ≈

m 2,5

12 30 10 12

10 3 300 400

300000

4 5

≈

⋅ ≈

≈ ⋅

×

Opgave 1.5 Welk getal is groter.

Maak eerst een schatting en controleer vervolgens met je rekenmachine.

a ³ ₃ 5 10 ³ 5 10 ³ 6 10 ³

200 1 6

6⁻ = 1 ≈ ≈ ⋅ ⁻ ⋅ ⁻ < ⋅ ⁻ b (2⋅10⁻²)² =4⋅10⁻⁴ 4⋅10⁻⁴ <0,004

c 225⋅10³ =2,25⋅10⁵ 2,25⋅10⁵ >2,25⋅10⁴

d ² ₂ ² ₂ ₂ ₂

2 1 4

1 2

) 1 2 4 (

) 1 4

( ⁻ = of ⁻ = <

e 0,25 4 0,25

2 ) 1 2 ( 25 4

, 0

1 ) 5 , 0 ( ) 1 5 , 0

( ⁻² = ₂ = = of ⁻² = ₂ = >

Opgave 1.6 Gebruik van voorvoegsels.

In de natuurkunde en techniek werkt men met grootheden, eenheden en voorvoegsels.

Zet de volgende eenheden om en geeft het antwoord in SCI-mode.

a 4,78 Mg =4,78⋅10⁶ g=4,78⋅10⁹ mg b 4,12⋅10^-3 m = 4,12⋅10⁻³×10³ mm=4,12mm

c 200 L = h200×0,01hL=2,00hL d 0,0032 mL =0,0032×10⁶ nL=3,2⋅10³ nL

e 23,2·10^-6 kg =23,2⋅10⁻⁶×10⁹μg =2,32⋅10⁴ μg

f 525000 N=5,25⋅10⁵×10⁻³ kN=5,25⋅10² kN g 2⋅10³ g =2⋅10³×10⁻³ kg=2kg

Opgave 1.7 Schrijf zonder voorvoegsel en bij voorkeur in SCI-mode.

a 2⋅10³ kg = 2⋅10³ × 10³ g = 2⋅10⁶ g

b 2⋅10^-3 mm = 2⋅10⁻³×10⁻³ m=2⋅10⁻⁶ m c 23,5 mL = 23,5×10⁻³ L=2,35⋅10⁻² L d 1,89⋅10²μL =1,89⋅10²×10⁻⁶ L=1,89⋅10⁻⁴ L e 2300 km =2300×10³ m=2,3⋅10⁶ m f 23,5 ms = 23,5×10⁻³s=2,35⋅10⁻² s g 2,1 MA =2 ⋅,1 10⁶A

h 700 nm =700×10⁻⁹ m=7×10⁻⁷ m i 23,5 GJ = 23,5×10⁹ J=2,35⋅10¹⁰ JJ j 2,1 ns =2,1⋅10⁻⁹ s

k 340 mm² =340×10⁻⁶ m² =3,4⋅10⁻⁴ m²

(4)

moleculen 10

34 , 3

mol deeltjes 10

022 , 6 mol 10 55 , 5 N

mol 10 5,55 015 mol

, 18 g 1 1

g 1 cm cm 1

1g ) water mL 1 (

22

23 2

A

2 -

3 3

⋅

=

→

⋅

×

⋅

=

→

⋅

=

⋅

=

×

=

→

⋅

=

−

N

n mol

aantal N

m V m ρ

moleculen 10

08 , 3

mol deeltjes 10

022 , 6 mol 10 107 , 5 N

mol 10 5,107 015 mol

, 18

920 , g 0 920 , 0

g 920 , 0 cm cm 1

920 g , 0 ) ijs mL 1 (

22

23 2

A

2 -

3 3

⋅

=

→

⋅

×

⋅

=

→

⋅

=

⋅

=

×

=

→

⋅

=

−

N

N mol

aantal N

m V m ρ

H atomen 10

6,9 (H) : afgerond

atomen 10

47 , 3 2 ) H ( moleculen 10

47 , 3 ) O H (

mol deeltjes 10

022 , 6 mol 10 77 , 5 N

mol 10 77 , 5 015 mol

, 18

10 4 , mg 10 4 , 10

g 18,105 mol

1

mg 4 , 10 ) lucht L 1 in waterdamp (

20

20 20

2

23 4

A

4 3

⋅

=

⋅

×

=

→

⋅

=

→

⋅

×

⋅

=

→

⋅

=

⋅

× =

=

−

N

N N

N mol

aantal N

m

l 6,3·10^-4 dm² = 6,3⋅10⁻⁴×10⁻² m² =6,3⋅10⁻⁶ m² m 2,1 mm³ =2,1⋅10⁻⁹ m³

Opgave 1.9 Eenheden converteren(omzetten) .

a 3,12·10^-3 m/min = 3,12⋅10⁻³×10³/60=5,20⋅10⁻²mm/s b 60 µm³ = 60×(10⁻⁶)³ =60×10⁻¹⁸ =6⋅10⁻¹⁷ m³

c 2,45 nm =2,45⋅10⁻⁹ m d 2,78·10^-6 L/min =

cm s 10 63 , 4 60 / 10 10 78 ,

2 ⋅ ⁻⁶× ³ = ⋅ ⁻⁵ ³

e 0,003 L = 0,003×(10⁸)³ =3⋅10²¹ nm³ f 3·10³ mm³ = 3⋅10³ μL

g 60 ms =60×10⁻³ s=6⋅10⁻² s i 340 GJ = 340×10⁹ =3,4⋅10¹¹ J

k 0,998·10³ kg·m^-3 = 0,998⋅10³×10³×10⁻⁶ =0,998g∙cm^-3 Opgave 1.10 Rekenen met mol 1 .

a

b

c

(5)

g 10 99 , 10 2 022 , 6

015 , ) 18 O H molecuul 1

(

g/mol 015 , 18 ) O H (

23 2 23

2

⋅ −

⋅ =

=

= m g

M

g 10 674 , 10 1 022 , 6

g 008 , ) 1 proton 1 (

g 008 , 1 ) protonen 10

022 , 6 (

24 23

23

⋅ −

⋅ =

=

⋅ m

m

g 10 369 , 2000 8

) 1

) ( elektron 1

( = m proton = ⋅ ⁻²⁸

m

g 1,008 10

674 , 1 10 022 , 6 ) H mol 1 (

g 10 1,674 g

10 8,369 g

10 674 , 1 ) atoom H

1 (

24 23

-24 -28

24

=

⋅

×

⋅

=

⋅

=

⋅ +

⋅

=

−

m m

μL 10 24 , 5 10 10 10 24 , 5

m 10 24 , 5 m ) 10 0 , 10 6 (

6 1 1

7 6

3 16

3 16 3

3 6 3

−

⋅

=

×

⋅

=

→

⋅

=

⋅

=

→

⋅

= V

V d

V π π

μg 10 1,31 μg 10 10 31 , 1

kg 10 31 , 1 m 10 24 , m 5 2500kg

3 - 9

12

12 3

16 3

⋅

=

×

⋅

=

→

⋅

=

⋅

×

=

→

⋅

=

−

m

m V m ρ

Pa p

A p p F^z

3 2

4 2 2

10 2 , m 2 10 N 22 , 0

cm 22N , cm 0 400

N 88

⋅

=

×

=

→

=

→

= d

Opgave 1.11 Rekenen met mol 2 . a

b

c

Opgave 1.12 Rekenen aan bolletjes a

b

Opgave 1.13 Parachutist.

a Een versnelling van 9,8 m/s² betekent dat de snelheid(v) toeneemt met 9,8 m/s per seconde of 0,98 m/s per 0,1 seconde.

v (na 1 seconde) ofwel v(1) = 9,8 m/s v (na 2 seconden) ofwel v(2) =

ms 6 , 19 8 , 9

2× = b F_z =m⋅g →F_z =80kg×9,81Nkg=785N

c F_w =F_z =785N want snelheidblijft hetzelfde d F_z =785N dezehangt alleen vandemassaaf e F_w =F_z =785N want snelheidblijft hetzelfde

f F_N = F_z =785N FN is de kracht van de grond op de parachutist Opgave 1.14 Rekenen aan kracht en druk

a G=F_z →G =9,0×9,81=88,3N afgerondG =88N b

(6)

Pa p

A p p F

F G

z z

3 2

4 2 2

10 8 , m 1 10 N 18 , 0

cm 176N , cm 0 500

N 88 N 88

⋅

=

×

=

→

=

→

=

mbar 1013 mbar 1000 1,013 bar

013 , 1

cm 13N , 10 10 / 10 013 , m 1 10 N 013 , 1

hPa 10 013 , 1 Pa 10 013 , 1

bar 013 , m 1 10 N 013 , 1

2 4

5 2

5

3 5

2 5

=

×

=

⋅

=

⋅

=

⋅

=

⋅

=

⋅

=

b b b b

p p p p

N 10 afgerond

N 13 , 10 m 10 m 1

101300N ₂× × ⁴ ² =

=

→

⋅

= p A F ⁻

F

m 764 , 0 Nkg 81 , m 9 10 kg 5 , 13

m 101300 N

3 3

2

=

×

⋅

=

⋅ →

=

→

⋅

= h

g h p h g

p ρ ρ

2 2

4

2 3

cm 147N , cm 0 10 N / 1472

m 1472N m 15 , kg 0 81N , m 9 1000kg

=

→

=

×

=

→

⋅

= p

p h g p ρ

2 2

4

2 3

cm 0981N , cm 0 10 N / 981

m 981N m 10 , kg 0 81N , m 9 1000kg

=

→

=

×

=

→

⋅

= p

p h g p ρ

Pa 10 96 , 1 :

m 1962N m

200 , kg 0 81N , m 9 1000kg

3

2 3

1 1

⋅

=

×

=

→

⋅

=

→

=

gas gas

gas h

gas

p afgerond

p

h g p

p

p ρ

c

d 20cm

cm 500

cm 10

2 3 4

=

→

=

→

⋅

= h

A h V h A V

Opgave 1.15 Rekenen aan luchtdruk.

a 600 10 m 6078N

m 10 N 013 ,

1 ⋅ ⁵ ₂× × ⁴ ² =

=

→

⋅

= p A F ⁻

F _b

b

Opgave 1.16 Kracht = druk × oppervlak a

b

Opgave 1.17 Rekenen aan vloeistofdruk in vat.

a

b

Opgave 1.18 Bepalen van de gasdruk a

(7)

Pa 10 80 , 9 :

Pa 98038 Pa

1962 Pa 10

10

4 5

2 5

2

⋅

=

−

=

→

⋅

−

=

→

−

=

gas gas

gas h

atm gas

p afgerond

p

h g Pa p

p p

p ρ

Pa 10 02 , 1 :

Pa 101962 Pa

1962 Pa 10

10

5 5

3 5

3

⋅

=

= +

=

→

⋅

⋅ +

=

→ +

=

gas gas

gas h

atm gas

p afgerond

p

h g Pa p

p p

p ρ

m3

640kg 1 , 0 1 64

, 0 0800 , 0 800

1 , 0 81 , 9 0800

, 0 81 , 9 800

=

→

×

=

×

→

×

=

×

→

⋅

=

⋅

→

=

benzine benzine

benzine

benzine benzine

olie olie

benzine

olie p g h g h

p

ρ ρ

ρ

ρ ρ

mol 10 04 , 2 mol JK 8,314

JK 17 , 0

JK 17 , K 0

) 273 20 (

m 10 m 500

10 N

2 3 6 2

5

−

⋅

=

⋅

=

→

=

+ =

⋅

×

=

⋅ →

=

R n n C

T C V C p

moleculen 10

2 , 1 :

moleculen 10

228 , 1 10 022 , 6 10 04 , 2 10 022 , 6

22

22 23

2 23

⋅

=

⋅

=

⋅

×

⋅

=

⋅

= ⁻

N afgerond

n N

g 90 , 0 ) ( :

g 8976 , mol 0 44g mol 10 04 , 2

2

=

×

⋅

=

→

⋅

= ⁻

CO m afgerond

m M n m

lucht m

gas mL ppm 10

10 = ₃

=

− waarde MAC

lucht m per mol 10 2 , 4 :

S H mol 10 158 , 4 mol JK 8,314

JK 10 457 , 3

JK 10 457 , K 3

) 273 20 (

m 10 m 10

101300 N

3 4

2 4

3

3 3

6 2

−

⋅

=

⋅

=

⋅

=

→

=

⋅ + =

⋅

×

=

⋅ →

=

n afgerond

R n n C

T C V C p b

Opgave 1.19 Bepalen van de dichtheid van een vloeistof met u-buis.

Opgave 1.20 Hoeveelheid gas meten.

a De massa van het gas wordt bepaald via de gasconstante.

b

c

Opgave 1.21 Maximaal aanvaardbare concentratie.

a

b

(8)

J 10 537 , 4 ) atoom per (

s m 10 kg

537 , 4 s) 10 m 9979 , 2 ( kg 10 048 , 5

kg 10 048 , 5 kg 10 6605 , 1 0304 , 0

u 0,0304 u

4,0026 - u 0330 , 4

12

2 12 2

2 8 29

29 27

2

⋅

=

→

⋅ ⋅

=

⋅

×

⋅

=

⋅

=

⋅

×

=

∆

→

=

∆

⋅

∆

=

−

E E

m m

c m E c N =

lucht m S H mmol 42 , S 0 H mol 10 158 ,

4 ⋅ ⁻⁴ ₂ = ² ₃

Opgave 1.22 Grote en kleine getallen in een stof.

1 gram water ( 1 mL) bevat 3,346·10²² moleculen H2O.

Water heeft bij 4 °C een dichtheid van 1000 kg/m³. IJs heeft bij 0 ⁰C een dichtheid 920 kg/m³ .

Waterdamp in de lucht heeft bij 20 ⁰C en een relatieve vochtigheid van 60% een dichtheid van 10,2 g/m³.

a Bereken het aantal watermoleculen in 1 gram ijs.

1 gram ijs = 1 gram water = 3,346·10²² moleculen H2O b Bereken het aantal watermoleculen in 1 gram waterdamp.

1 gram waterdamp = 1 gram water = 3,346·10²² moleculen H2O c 1 liter water = 1000 gram = 3,346∙10²⁵ moleculen H2O

d 1 liter ijs = 920 gram = 920 × 3,346∙10²² = 3,078∙10²⁵ moleculen H2O

e In 1 liter water zitten meer moleculen dan in 1 liter ijs.

In 1 liter water zit × 920

1000 meer moleculen dan in 1 liter ijs f In 1 liter lucht van 20⁰C en 60% vochtigheid zit 10,2 gram water.

23 22 3,41 10 10

346 , 3 2 , 10 g 2 ,

10 = × ⋅ = ⋅ moleculen H2O

g In 1 liter lucht van 20⁰C en 100% vochtigheid zit 10,2 17g 60

100× = water.

23 22 5,69 10 10

346 , 3 17 g

17 = × ⋅ = ⋅ moleculen H2O h De waarde bij g is = ×

3 5 60

100 groter dan die bij f.

i 3,346·10²² moleculen = 1 g

kg 10 99 , 10 2 346 , 3 molecuul 1

1 ₂₂ = ⋅ ⁻²³

= ⋅

j Een watermolecuul heeft een diameter van ongeveer 0,3 nm.

In 1 m passen ₉ 3,33 10⁹ 10

3 , 0

1 = ⋅

⋅ ⁻ moleculen

Opgave 1.23 Atoomkern splijten.

(9)

J 10 865 , 6 s) 10 m 9979 , 2 ( kg 10 638 , 7 ) atoom per (

kg 10 638 , 7 kg 10 6605 , 1 0046 , 0

u 0,0046 u)

0026 , 4 u (234.0436 -

u 0508 , 238

13 2

8 30

30 27

2

−

⋅

=

⋅

×

⋅

=

⋅

=

⋅

×

=

∆

→

= +

=

∆

⋅

∆

=

E m m

c m E

J 10 75 , 1 ) 1 ( :

J 10 794 , 1 J 10 865 , 6 10 613 , 2 ) 1 (

U atomen 10

613 , 2

U atomen 10

022 , 238 6 U 1

238 mol 1

9

9 13

21 238 21

238 23

238

⋅

=

⋅

=

⋅

×

⋅

=

⋅

=

→

⋅

×

=

→

=

−

gram per

E afgerond

gram per

E N

N n

2 2

2

μm 135 μm 2 μm 5 , 7 π ) μm 75 , 3 ( π 2

π π

2

=

×

× +

×

=

→

⋅

⋅ +

⋅

×

= A

h d r

A

s ery' 10 0 , 3 10 29,7 10

4 , 5 5 , 5 liter 5 , 5

s ery' 10 5,4 bloed L 1 s ery' 10 4 , 5 bloed mL 1

13 12

12

12 9

×

=

×

=

⋅

×

=

→

⋅

=

→

⋅

=

2 12

12 2 12

m 4000 10

135 10 7 , 29 ) s ery' alle (

m 10 135 ) ery 1 (

=

×

=

→

⋅

=

−

A A

L 2 , 5 ) bloed ( dan ) s ery' ( 2 ) bloed ( Als

L 6 , 2 m 10 6 , 2 10 88 10 7 , 29 ) s ery' alle (

m 10 88 ) ery 1 (

3 3 18

12 3 18

=

×

=

⋅

=

×

=

→

⋅

=

−

V V

V A V

mmolmL 10

1 , 1 :

mmol 10

076 , 1 moleculenmmol

10 022 , 6

moleculen 10

48 , bloed 6

mL 1

moleculen 10

48 , 6 10 2 , 1 10 4 , 5 bloed mL 1

s ery' 10 4 , 5 bloed mL 1

2

2 20

18

18 9

9 9

−

⋅

=

−

⋅

=

⋅

= ⋅

→

⋅

=

⋅

×

⋅

=

→

⋅

=

gehalte e

hemoglobin afgerond

e hemoglobin

mL 173mg

mmol 16114mg

mmolmL 10

076 ,

1 ⋅ ² × =

=

−gehalte ⁻

e hemoglobin Opgave 1.24 Radioactief verval.

a

b

Opgave 1.25 Rode bloedcellen (ery’s)

a V =π⋅r²⋅h→V =π×(3,75μm)²×2μm=88μm³ b

c

d

e

f s

s ery' 10 9 , 2 s 3600 24 120

s ery' 10 dagen 3

120 in s ery' 10 3

13 6

13 = ⋅

×

→ ⋅

⋅

g 5 10 ery's

m 10 2

m m 1

1 in aantal m

10

2 ⁶ _-₆ = ⋅ ⁵

= ⋅

→

⋅

= ⁻

d

Opgave 1.26 Hemoglobinegehalte in het bloed.

a

b

(10)

mL 4mg , mmol 2 56mg

mmolmL 10

304 , 4 ) Fe (

mmolmL 10

304 , 4 10 076 , 1 4 ) atomen Fe

(

2

2 2

=

×

⋅

=

→

⋅

=

⋅

×

=

−

m n

15 7

23 2

23 23

3 3

7 9

2

10 84 , 10 3 2,82

10 cm 1,08

1 van laagje per C atomen aantal

C atomen 10

08 , 1 10 022 , 6 18 , 0 ) 1 (

C mol 18 , 0 mol 12g

g 16 , ) 2 cm 1 (

g 16 , 2 ) 1 ( kg 16 , 2 ) dm 1 (

10 82 , 10 2 335 , 0

10 335

, 0 1 1

⋅

⋅ =

= ⋅

→

⋅

=

⋅

×

=

→

=

→

=

→

=

⋅

⋅ =

=

= ₋

−

cm N n

cm m m

m nm

cm cm in laagjes aantal

` c

Opgave 1.27 Grafeen, bijzonder materiaal.

a ₉ ⁶

3

10 82 , 10 2 335 , 0

10 335

, 0

1 1 = ⋅

= ⋅

= ₋

− m

nm mm mm

in laagjes aantal

b

c Zie uitwerking bij b)