ΔHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
Bond enthalpies,
N ≡ N ⇒ 945 kJ mol⁻¹
N - Cl ⇒ 192 kJ mol⁻¹
Cl - Cl⇒ 242 kJ mol⁻¹
According to the balanced equation,
ΣδΗ(bond breaking) = N ≡ N x 1 + Cl - Cl x 3
= 945 + 3(242)
= 1671 kJ mol⁻¹
ΣδΗ(bond making) = N - Cl x 3 x 2
= 192 x 6
= 1152 kJ mol⁻¹
δHrxn = ΣδΗ(bond breaking) - ΣδΗ(bond making)
= 1671 kJ mol⁻¹ - 1152 kJ mol⁻¹
= 519 kJ mol⁻¹
Answer:
See explanation
Explanation:
Atomic size increases down the group due to the addition of more shells.
As more shells are added and repulsion of inner electrons become more significant, atomic size increases down the group. However, across the period, atomic size decreases due to increase in effective nuclear charge without any increase in the number of shells. This causes increased attraction between the nucleus and the outermost shell thereby decreasing the size of the atom.
Ionization energy decreases down the group because the outermost electron is more shielded by inner electrons making it easier for this outermost electron to be lost. Across the period, ionization energy increases due to increase in effective nuclear charge which makes it more difficult to remove the outermost electron due to increased nuclear attraction.
Answer:
A. 4-ethyl-hex-3,5-dien-2-ol.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol.
Explanation:
Hello there!
In this case, according to the given problems, it is possible to apply the IUPAC rules to obtain the following names:
A. 4-ethyl-hex-3,5-dien-2-ol because we have an ethyl radical at the fourth carbon and the beginning of the parent chain is on the Me (CH3) because it is closest to first OH.
B. 2-chloro-3-methyl-5-<em>tert</em>-butylphenol: because we start at the alcohol and have a chlorine atom on the second carbon, a methyl radical on the third carbon, a <em>tert</em>-butyl on the fifth carbon and the parent chain is benzene which is phenol as an alcohol.
Regards!
Answer:
Statements Y and Z.
Explanation:
The Van der Waals equation is the next one:
(1)
The ideal gas law is the following:
(2)
<em>where n: is the moles of the gas, R: is the gas constant, T: is the temperature, P: is the measured pressure, V: is the volume of the container, and a and b: are measured constants for a specific gas. </em>
As we can see from equation (1), the Van der Waals equation introduces two terms that correct the P and the V of the ideal gas equation (2),<u> by the incorporation of the intermolecular interaction between the gases and the gases volume</u>. The term an²/V² corrects the P of the ideal gas equation since the measured pressure is decreased by the attraction forces between the gases. The term nb corrects the V of the ideal gas equation, <u>taking into account the volume occuppied by the gas in the total volume, which implies</u> a reduction of the total space available for the gas molecules.
So, the correct statements are the Y and Z: the non-zero volumes of the gas particles effectively decrease the amount of "empty space" between them and the molecular attractions between gas particles decrease the pressure exerted by the gas.
Have a nice day!
