Question

CS2, XeO4, CCl4, PCl5, Mn2O7.​

Answer 1

Треба знайти масову частку?
Дано: W=n*Ar.
CS2. ————-. *100%
————- Mr
W(C)? Mr(CS2)=C+S*2=12+ 32*3=
W(S)? =76
W(C)= 1*12
—- *100% = 16%
76
W(S)=2*32
—————- *100%=84%
76



Дано:
XeO4. W=n*Ar
————- ————- *100%
W(Xe)-? Mr
W(O)-? Mr(XeO4)= Xe+O*4=131+16*
*4=195
W(Xe)=1*131
——————. *100%. =
195
= 67,18%
W(O)=4*16
—— *100% = 32,82
195



Дано:
CCl4. W=n*Ar
————. ————— *100%
W(C)-? Mr
W(Cl)-? Mr(CCl4)= C+Cl*4=12+35*4=
=152
W(C)=1*12
—— *100%= 0,08%
152
W(Cl)= 4*35
——- *100% =0,92%
152

Answer 2

The compounds you've listed are:

1. CS2 - Carbon disulfide
2. XeO4 - Xenon tetroxide
3. CCl4 - Carbon tetrachloride
4. PCl5 - Phosphorus pentachloride
5. Mn2O7 - Manganese heptoxide