Question

raspundeti si aici va rog ka ce puteti.​

Answer 1

Răspuns:

Bună!

2.

T=273+27=300K

Folosim ecuația de stare a gazului ideal:

p×V=n×R×T

4×V=4×0,082×300

V=(4×0,082×300)/4= 24,6L

3.

1 mol.........22,4L

x moli........448L

x=448/22,4= 20 moli

4.

1mol de CO2........6,022×10²³ molecule

3 moli de CO2.......x molecule

x=3×6,022×10²³=18,066×10²³ molecule CO2

1mol CO2......6,022×10²³ atomi C......2×6,022×10²³ atomi O2

3 moli Co2.....y atomi C.........z atomi O2

y=3×6,022×10²³= 18,066×10²³ atomi C

z=3×2×6,022×10²³= 36,132×10²³ atomi O2

5.

1mol CO2......22,4L

x moli..........2,24L

=> x=2,24/22,4=0,1 moli CO2

MCO2 = 12+2×16= 44g/mol

n=m/M => m=n×M

m=0,1×44= 4,4g CO2

Sper că te-am ajutta!