Answer:
P = 14.1 atm
Explanation:
Given data:
Mass of methane = 64 g
pressure exerted by water vapors = ?
Volume of engine = 24.0 L
Temperature = 515 K
Solution:
Chemical equation:
CH₄ + 2O₂ → CO₂ + 2H₂O + energy
Number of moles of methane:
Number of moles = mass / molar mass
number of moles = 64 g/ 16 g/mol
Number of moles = 4 mol
Now we will compare the moles of water vapors and methane.
CH₄ : H₂O
1 : 2
4 : 2/1×4 = 8 mol
Pressure of water vapors:
PV = nRT
R = general gas constant = 0.0821 atm.L/mol.K
P = 8 mol × 0.0821 atm.L/mol.K× 515 K / 24.0 L
P = 338.25 atm.L/ / 24.0 L
P = 14.1 atm
The question is partially incorrect, because nitration of <span> methyl benzoate results in generation of methyl 3-nitrobenzoate, and not methyl 2-benzoate.
This a because of the present of ester group, which deactivated benzene ring at ortho and para position. Due to this, the electrophile (NO2+) attackes on meta position.
The detailed mechanism is attached below.</span>
This problem has two parts; the first one asking for the concentration of NaBr given both its mass and volume and the second one asking for its volume given both mass and concentration. The answers turn out to be 0.158 M and 211 mL.
<h3>Molarity</h3>
In chemistry, the use of units of concentration depends on both the substances to analyze and their amounts. In such a way, for molarity, one needs the following relationship between the moles of solute and volume of solution:

Thus, for the first part of the problem we first calculate the moles in 2.60 g of NaBr via its molar mass:

Next, we convert the 160. mL to L by dividing by 1000 in order to obtain 0.160 L to subsequently calculate the molarity:

Next, since the moles remain the same and for the second part we are asked for the volume given the concentration, one can solve for the volume so as to obtain:

That in milliliters turns out to be:

Learn more about molarity: brainly.com/question/10053901
Answer: The equilibrium constant for the overall reaction is 
Explanation:
Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios.
a) 
![K_a=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D)
b) 
![K_b=\frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
For overall reaction on adding a and b we get c
c) 
![K_c=\frac{[PCl_5]}{[Cl_2]^\frac{5}{2}}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5E%5Cfrac%7B5%7D%7B2%7D%7D)
![K_c=K_a\times K_b=\frac{[PCl_3]}{[Cl_2]^{\frac{3}{2}}}\times \frac{[PCl_5]}{[Cl_2]\times [PCl_3]}](https://tex.z-dn.net/?f=K_c%3DK_a%5Ctimes%20K_b%3D%5Cfrac%7B%5BPCl_3%5D%7D%7B%5BCl_2%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%7D%5Ctimes%20%5Cfrac%7B%5BPCl_5%5D%7D%7B%5BCl_2%5D%5Ctimes%20%5BPCl_3%5D%7D)
The equilibrium constant for the overall reaction is 
the god ran down the street