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MrRa [10]
3 years ago
5

Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine because A.the bromine radical i

s more reactive and therefore more selective. B.the chlorine radical is more reactive and therefore more selective. C.the chlorine radical is less reactive and therefore more selective. D.the bromine radical is less reactive and therefore more selective. E.none of these choices.
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine because the bromine radical is less reactive and therefore more selective.

Explanation: Halogenation occurs when a halogen replaces  one or more hydrogen atoms in an organic compound ie  chlorine or  bromine with the reactivity of the halogens decreasing  in the order of  F2 > Cl2 > Br2 > I2

Since fluorine reacts  explosively making it is difficult to control, and iodine is unreactive. Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine  with Chlorination  ie chlorine radical being more reactive and not selective and the Bromination of alkanes ie bromine radical  occurring similarly but slower and less reactive but  more selective which is due to the fact that a bromine atom is less reactive in  the hydrogen abstraction  than a chlorine atom evidence in the  higher bond energy of H-Cl than H-Br.

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3 years ago
How many grams of magnesium metal will react completely with 5.2 liters of 4.0 M HCl?
neonofarm [45]
1) Find the number of mols of HCl in 5.2 liters of 4.0M solution:

n = M*V(L) = 4.0 mol/L * 5.2 L = 20.8 mol

2) Find the number of mols of Mg that will react with 20.8 mol of HCl, using the coefficients of the balanced equation

[1mol Mg / 2 mol HCl] * 20.8 mol HCl = 10.4 mol Mg

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2 years ago
What is the correct formula
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7 0
3 years ago
Argon (Ar) and helium (He) are initially in separate compartments of a container at 25°C. The
love history [14]

Answer:

(a) V_B=11.68L

(b) x_{He}=0.533

Explanation:

Hello,

In this case, since the both gases behave ideally, with the given information we can compute the moles of He in A:

n_A=\frac{0.082\frac{atm*L}{mol*K}*298K}{1.974 atm*6.00L}=2.063mol

Thus, since the final pressure is 3.60 bar, we can write:

P=x_{Ar}P_A+x_{He}P_B\\\\P=\frac{n_{Ar}}{n_{Ar}+n_{He}} P_A+\frac{n_{He}}{n_{Ar}+n_{He}} P_B\\\\3.60bar=\frac{2.063mol}{2.063mol+n_{He}} *2.00bar+\frac{n_{He}}{2.063mol+n_{He}} *5.00bar

The moles of helium could be computed via solver as:

n_{He}=2.358mol

Or algebraically:

3.60bar=\frac{1}{2.063mol+n_{He}} *(4.0126+5.00*n_{He})\\\\7.314+3.60n_{He}=4.013+5.00*n_{He}\\\\7.314-4.013=5.00*n_{He}-3.60n_{He}\\\\n_{He}=\frac{3.3}{1.4}=2.358mol

In such a way, the volume of the compartment B is:

V_B=\frac{n_{He}RT}{P_B}=\frac{2.358mol*0.082\frac{atm*L}{mol*K}*298.15K}{4.935atm}\\  \\V_B=11.68L

Finally, he mole fraction of He is:

x_{He}=\frac{2.358}{2.358+2.063}\\ \\x_{He}=0.533

Regards.

8 0
2 years ago
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