Basically team B would win since it is exerting a force of 900N unlike team A ( you can tell by doing 4900N minus 4000N ). It is very unbalanced.
Answer:
B = CHCl2 + Cl2 --> CHCl3 + Cl
Explanation:
Free radical halogenation is a chlorination reaction on Alkane hydrocarbons. This involves the splitting of molecules into radicals/ unstable molecules in the presence of sunlight/ U.V light which ensures bonding of the molecules.
Free radical chlorination is divided into 3 steps which are:
The initiation step
The propagation step
The termination step
So in reference to the question, propagation step involves two steps.
The first step is where the molecule in this case the methylene chloride(CH2Cl2) loses a hydrogen atom and then bond with a chlorine atom radical to give a nethylwnw chloride radical and HCl.
The second step involves the reaction of this methylene chloride got in the first step with chlorine molecule to form trichloride methane and a chlorine radical.
You would find in the attachment the 2 step mechanism.
1L = 1000ml
1ml = 1cm^3
2.1L = 2100mL = 2100cm^3
According to this formula:
K= A*(e^(-Ea/RT) when we have K =1.35X10^2 & T= 25+273= 298K &R=0.0821
Ea= 85.6 KJ/mol So by subsitution we can get A:
1.35x10^2 = A*(e^(-85.6/0.0821*298))
1.35x10^2 = A * 0.03
A= 4333
by substitution with the new value of T(75+273) = 348K & A to get the new K
∴K= 4333*(e^(-85.6/0.0821*348)
= 2.16 x10^2
If we write the equation of the reaction that will take place, it is:
2HNO₃ + Na₂CO₃ → 2NaNO₃ + H₂CO3
The molar ratio of 2HNO₃ : Na₂CO₃ = 1 : 2
Therefore, we can set up the equation:
M₁V₁ = 2M₂V₂
Where the left side of the equation has the molarity and volume of HNO₃ and the right side has the molarity and concentration of Na₂CO₃. Substituting:
M₁ = (2 x 0.108 x 35.7) / 25
M₁ = 0.308 M