What mass of nickel (Ni) is in a 2.4 Kg sample of propanol if the concentration is 20 ppb ? (atomic mass of Ni = 58.69)
1 answer:
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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.
The new volume at standard pressure of 1 atm is 21294 liters.
Explanation:
Data given:
Initial volume of the gas V1 = 338 liters
initial pressure on the gas P1 = 63 atm
standard pressure as P2 = 1 atm
Final volume at standard pressure V2 =?
The data given shows that Boyle's law equation is to used:
P1V1 = P2V2
rearranging the equation to calculate V2,
V2 =
Putting the values in the equation:
V2 =
= 21294 L
as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.
Answer:
0.3023 M
Explanation:
Let Picric acid =
So, +
⇄
+
The ICE table can be given as:
+
⇄
+
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant () = 0.42
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
=
=
= OR
= OR
= OR
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [ ] = [
] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).