Methane is the compound CH4, and burning it uses the reaction:
CH4 + O2 -> CO2 + H2O, which is rather exothermic. To find the heat released by burning a certain amount of the substance, you should look at the bond enthalpy of each compound, and then compare the values before and after the reaction. In methane, there are 4 C-H bonds, which have bond energy of 416 kj/mol, resulting in a total bond energy of 1664 kj/mol. O2 is 494 kj/mol. Therefore we have a total of 2080 kj/mol on the left side. On the right side we have CO2, which has 2 C=O bonds, each at 799 kj/mol each, resulting in 1598 kj/mol, and H2O has 2 O-H bonds, at 459kj/mol each, resulting in a total of 2516 kj/mol on the right hand side. Now, this may be confusing because the left hand side seems to have less heat than the right, but you just need to remember: making minus breaking, which results in a total change of 436kj/mol heat evolved.
Now it is a simple matter of find the mols of CH4 reacted, using n=m/mr.
n = 9.5/16.042 = 0.592195 mol
Therefore, if we reacted 0.592195 mol, and we produced 436 kj for one mol, the total amount of energy evolved was 436*<span>0.592195 kj, or 258.197 kj.</span>
Answer:
molar composition for liquid
xb= 0.24
xt=0.76
molar composition for vapor
yb=0.51
yt=0.49
Explanation:
For an ideal solution we can use the Raoult law.
Raoult law: in an ideal liquid solution, the vapor pressure for every component in the solution (partial pressure) is equal to the vapor pressure of every pure component multiple by its molar fraction.
For toluene and benzene would be:
Where:
is partial pressure for benzene in the liquid
is benzene molar fraction in the liquid
vapor pressure for pure benzene.
The total pressure in the solution is:
And
Working on the equation for total pressure we have:
Since 
We know P and both vapor pressures so we can clear from the equation.
So
To get the mole fraction for the vapor we know that in the equilibrium:
So
Something that we can see in these compositions is that the liquid is richer in the less volatile compound (toluene) and the vapor in the more volatile compound (benzene). If we take away this vapor from the solution, the solution is going to reach a new state of equilibrium, where more vapor will be produced. This vapor will have a higher molar fraction of the more volatile compound. If we do this a lot of times, we can get a vapor that is almost pure in the more volatile compound. This is principle used in the fractional distillation.
Answer:
D. fields moves in longitudinal waves
Explanation:
Answer:
Explanation:
Explanation:
All you have to do here is use the ideal gas law equation, which looks like this
P
V
=
n
R
T
−−−−−−−−−−
Here
P
is the pressure of the gas
V
is the volume it occupies
n
is the number of moles of gas present in the sample
R
is the universal gas constant, equal to
0.0821
atm L
mol K
T
is the absolute temperature of the gas
Rearrange the equation to solve for
T
P
V
=
n
R
T
⇒
T
=
P
V
n
R
Before plugging in your values, make sure that the units given to you match those used in the expression of the universal gas constant.
In this case, the volume is given in liters and the pressure in atmospheres, so you're good to go.
Plug in your values to find
T
=
3.10
atm
⋅
64.51
L
9.69
moles
⋅
0.0821
atm
⋅
L
mol
⋅
K
T
=
251 K
−−−−−−−−−
The answer is rounded to three
A or D I think...sorry if it’s wrong
