Answer:
CH3Cl + Cl2 -> CH2ClCH2Cl + HCl
Explanation:
Chlorination is an addition reaction that involves addition of Chlorine to a chemical compound. Methane undergo addition reaction with chlorine to produce chloromethane, which undergo further further chlorination to produce dichloromethane.
For the chlorination of methane, one of the hydrogen atoms of methane is replaced by a chlorine atom of the chlorine gas (molecule)
CH4 + Cl2 -> CH3Cl + HCl
methane chlorine chloromethane Hydrochloric acid
Using their structure:
H H
I I
H - C - H + Cl - Cl -> H - C - Cl + HCl
I I
H H
methane chloromethane
In further chlorination of chloromethane, one of the hydrogen atom of chloromethane is replaced by the chlorine atom in the chlorine gas (molecule)
Chlorination of Chloromethane to dichloromethane is represented below:
CH3Cl + Cl2 -> CH2ClCH2Cl + HCl
Chloromethane Chlorine dichloromethane Hydrochloric acid
Using their structure:
Cl Cl
I I
H - C - H + Cl - Cl -> H - C - Cl + HCl
I I
H H
Chloromethane chlorine dichloromethane
Organic compound Inorganic compound
From the chemical equation above, an organic compound (Chloromethane) reacts with an inorganic compound (Chlorine) .
Answer:
because it could self drive
Explanation:
idrk but thats my answer
Answer:
The concentration is [-1 + sqrt(1+0.11t)]/0.1542 M
Explanation:
Let the concentration of CH3CHO after selected reaction times be y
Rate = Ky^2 = change in concentration of CH3CHO/time
K = 0.0771 M^-1 s^-1
Change in concentration of CH3CHO = 0.358 - y
0.0771y^2 = 0.358-y/t
0.0771ty^2 = 0.358 - y
0.0771ty^2 + y - 0.358 = 0
The value of y must be positive and is obtained in terms of t using the quadratic formula
y = [-1 + sqrt(1^2 -4(0.0771t)(-0.358)]/2(0.0771) = [-1 + sqrt(1+0.11t)]/0.1542 M
A 40.3 ml sample of a 0.587 m aqueous hydrocyanic acid solution is titrated with a 0.363 m aqueous sodium hydroxide solution. The pH after 19.9 ml of base have been added is 12
After 19.9 mL of NaOH have been added we have gone way past the equivalence point, so there is a lot of excess NaOH,
⇒ 40.3 - 19.9 = 20.4 mL excess
⇒ 0.0204 L (0.363 M) = 0.0074 moles NaOH excess,
⇒ 0.0074 moles / (0.0199 + 0.0403)
⇒ 0.122 M NaOH at this point,
[OH]⁻ = [NaOH]
pOH = - log (0.122) = 2.103
pH = 14 - pOH = 11.897 = 12
Hence, pH = 12, After 19.9 mL base have been added.
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