Question

ex 11 sau 12 va rog!!! ​

Answer 1

Răspuns:

Explicație:

11.

a.

1mol NH3----------6,022.10²³molecule NH3----------4 . 6,022.10²³atomi total

1,5 moli NH3 ------------------x-------------------------------------------y

x= 1,5 . 6,022. 10²³= 9,033.10²³molecule NH3

y= 1,5  .  4  . 6,022.10²³ =  36,132.10²³atomi

b.

nCO= 14g : 28g/moli= 0,5 moli CO

1mol CO--------6,022.10 ²³molecule CO---------2  . 6,022.10²³atomi

0,5 moli CO-------------------x--------------------------------y

x= 0,5 . 6,022.10²³molecule CO = 3,011.10²³molecule CO

y= 0,5  . 2  . 6,022.10²³atomi = 6,022.10²³atomi

c.

19,6 kg=19600 g H2SO4

M= 2 + 32 + 4.16=98-------> 98g/moli

n= 19600g : 98g/moli= 200moli H2SO4

1mol H2SO4 -------6,022.10²³molecule -------7  . 6,022.10²³atomi

200moli ------------------------x--------------------------------y

x= 200. 6,022.10²³= 1204,4 .10²³molecule H2SO4

y= 200 .  7  . 6,022.10²³=8430,8 .10²³atomi

d.

MSO3= 32 + 3.16=80-------> 80g/moli

80kg= 80000g SO3

n= 80000g : 80g/moli= 1000moli SO3

1mol SO3---------6,022.10²³molecule---------4  .  6,022.10²³atomi

1000 moli-----------------x-----------------------------------y

x= 1000 . 6,022.10²³ =6022.10²³molecule

y= 1000 . 4  . 6,022.10²³atomi =

e.

2kmoli=2000 moli He

1mol He--------6,022 .10²³molecule He-----------1.6,022.10²³atomi

2000 moli ----------x--------------------------------------------------y

x= 2000 .  6,022.10²³molecule H2 =

y= 2000 . 6,023.10²³atomi=

Answer 2

Răspuns:

12.

n(niu)= 2x4 mol O

se stie   1mol atomi..........6,023x10²³atomi

               8mol O................x

x= 48,18x10²³ atomi O

CO2

n(niu)= 11X2molO

1mol atomi.........6,023x10²³atomi\

22moli...................x

x=132,5x10²³ atomi O

Explicație: