3. The Kp for the reaction below is 1.49 × 108 at 100.0 °C:CO (g) + Cl2 (g) → COCl2 (g)In an equilibrium mixture of the three ga
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Answer:- 1467 K
Solution:- It asks to calculate the kelvin temperature of the light bulb. Looking at the given info, it is based on ideal gas law equation, PV=nRT.
Given:
V = 75.0 mL = 0.0750 L
P = 116.8 kPa
We know that, 101.325 kPa = 1 atm
So,
= 1.15 atm
R is universal gas constant and it's value is .
T = ?
Let's plug in the values in the equation and solve it for T.
0.08625 = 0.00005878(T)
T = 1467 K
So, the temperature of the light bulb would be 1467 K.
Answer:
Explanation:
In the chlorination of alkanes, the condition necessary is UV light so free radical substitution can take place. For alkanes like pentane, the primary, secondary and tertiary Hydrogen atoms (Hydrogen atoms bonded to their respective carbon) þare taken into consideration and this is because the tertiary Hydrogen is the most reactive (due to bond dissociation energy) hence the easiest to be substituted. The trend is as follows in the order of their reactivity;
1° < 2° < 3°
So, the products of the chlorination of pentane, the principal monochloride constituted is 3 - chloropentane while the other two monomers are:
2- chloropentane
1- chloropentane
Below is the attachment showing the structural formula of the three monochloride constituted pentane.
