8.5L H2O (1 mol H2O/22.4L)=.37946 mol H2O
.3796 mol H2O(1 mol CH4/2 mol H2O)=.18973 mol CH4
.18973 mol CH4(22.4L/1mol CH4)=4.25L CH4 gas
Answer:
Final volume 30.513 L.
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 17 L
Initial pressure = 2.3 atm
Initial temperature = 299 K
Final temperature = 350 K
Final volume = ?
Final pressure = 1.5 atm
Solution:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 2.3 atm × 17 L × 350 K / 299 K × 1.5 atm
V₂ = 13685 atm .L. K / 448.5 K . atm
V₂ = 30.513 L
Answer:
No, i will not use a water pipe consisting of the two metals
Explanation:
Looking at the reduction potential of the both metals, it is clear that an electrochemical cell is set up with iron as the anode and copper as the cathode.
This will make the iron to quickly corrode and eventually destroy the water pipe. It is better to have a set up in which another metal that is higher than iron in the electrochemical series is combined with it.
Use M x V = M' x V'
0.300 x V = 0.100 x 250
V = .......... ml
Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=
106.16×1000
17.12
=0.00016moles
Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO
2
=
44.01×1000
56.77
=0.0013
Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H
2
O==
18.02×1000
14.53
=0.0008
Moles ratios
\frac{0.0013}{0.0008}=1.625
0.0008
0.0013
=1.625
\frac{0.0008}{0.0008}=1
0.0008
0.0008
=1
Hence molecular fomula
The empirical formula is C 4H 5.
The molecular formula C8H10
