Answer:
a)
b)
![[PCl_5]=0.0375M](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D0.0375M)
Explanation:
Hello!
a) In this case, since we can see that the second reaction is equal to the half of the first reaction, we can relate the equilibrium constants as shown below:
Thus, by plugging in the the equilibrium constant of the first reaction we obtain:
b) In this case, for the described reaction we can write:
Thus, the corresponding equilibrium expression is:
![K=\frac{[PCl_3][Cl_2]}{[PCl_5]}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7B%5BPCl_5%5D%7D)
In such a way, since we know the equilibrium constant and the concentrations of PCl3 and Cl2 at equilibrium, we can compute the concentration of PCl5 at equilibrium as follows:
![[PCl_5]=\frac{[PCl_3][Cl_2]}{K}\\](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D%5Cfrac%7B%5BPCl_3%5D%5BCl_2%5D%7D%7BK%7D%5C%5C)
![[PCl_5]=\frac{\frac{0.20mol}{4L} *\frac{0.12mol}{4L} }{0.04}](https://tex.z-dn.net/?f=%5BPCl_5%5D%3D%5Cfrac%7B%5Cfrac%7B0.20mol%7D%7B4L%7D%20%2A%5Cfrac%7B0.12mol%7D%7B4L%7D%20%7D%7B0.04%7D)
Best regards!
Hello.
Energy is absorbed by the atom.
Emission lines occur only in the case of transitions from higher to lower energy levels in which energy is emitted.
Have a nice day
<span>the reaction is endothermic
</span><span>the δh is positive</span>
If we were to make room for errors, there should really be no limiting reagent because practically all of both Nitrogen and Hydrogen is used up during this reaction. If this values were actually exact, then Nitrogen would be the limiting reagent, but a very very little amount of Nitogen is needed for all the Hydrogen to react.
We solve this problem by first writing the equation
N2 + 3H2 = 2NH3
N2 = 14g*2 = 28g, 3H2 = 3(1*2) = 6g
so 28g of Nitrogen needs 6g of Hydrogen for this reaction. Thus if we had 10.67g of Hydrogen in the reaction, 6g*49.84g/28g of hydrogen is needed to react = 10.68g of Hydrogen, but since we have 10.7g of it thus it is excess and thus the limiting reagent has to be Nitrogen, but notice that 10.68g and 10.7g are practically the same, so there might actually not be a limiting reagent. Using the other value(10.7), the amount of Nitrogen required would be 10.7g*28g/6g = 49.93, and since this is slightly more than the 49.84g we have, this confirms that Nitrogen is the limiting reagent. But note still that since this values are really close, there is a possibility that there is neither a limiting nor an excess reagent
Answer:
1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.
2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.
Explanation:
The molecular formula of boron trifluoride is 
.
So, one mole of boron trifluoride has one mole of boron atoms.
1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.
The number of atoms in 2.20 moles of boron is:
One mole of boron has ---- 
atoms.
Then, 2.20 moles of boron has
-
2. Calculate the number of moles of BF3 in 5.35*1022 molecules.
One mole of boron trifluoride has three moles of fluorine atoms.
Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.
=0.266mol of fluorine atoms.
