Question

852kg dintr-o substanta gazoasa elementara ocupa 340,48 m cubi la 364 K si 800 mm Hg. Sa se afle:
a) numarul de atomi si molecule continute in volumul respectiv;
b) masa moleculara a substantei gazoase;
c) care este substanta gazoasa;
d) densitatea si densitatea fata de aer a substantei date.

Answer 1

1atm..........760 mmHg
xatm..........800 mmHg
x=1,05 atm
pV=nRT
b)n=pV/RT=1,05(atm)*340480(L)/0,082(atm*L/mol*K)*364K≈12000moli=12kmoli gaz
n=m/M => M=m/n=852kg/12kmoli=71kg/kmol
c)substanta noastra se gaseste sub forma de molecula => 2A=71 => A=35,5kg/kmol=35,5g/mol => Elementul: Cl si substanta: Cl₂
a) eu lucrez cu 6,0225*10²³, tu lucrezi cu ce iti cere cartea
n=N/NA => N=n*NA=12000*6,0225*10²³ molecule Cl₂=72270*10²³ molecule Cl₂
1molecula=2atomi => avem 144540*10²³ atomi Cl
d)pV=mRT/M => pM=mRT/V=dRT => d=pM/RT=1,05(atm)*71(g/mol)/0,082*364=2,49g/L
daer=MCl₂/Maer=71/28,9=2,45