The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
The bond energies data is given as follows:
BE for C≡O = 1072 kJ/mol
BE for Cl-Cl = 242 kJ/mol
BE for C-Cl = 328 kJ/mol
BE for C=O = 766 kJ/mol
The enthalpy change for the reaction is given as :
ΔHr×n = ∑H reactant bond - ∑H product bond
ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )
ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )
ΔHr×n = 1314 - 1422
ΔHr×n = - 108 kJ
Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.
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Answer:
4.00 is the pH of the mixture
Explanation:
The ethyl amine reacts with HNO3 as follows:
C2H5NH2 + HNO3 → C2H5NH3⁺ + NO3⁻
To solve this question we need to find the moles of ethyl amine and the moles of HNO3:
<em>Moles C2H5NH2:</em>
0.0500L * (0.100mol/L) = 0.00500 moles ethyl amine
<em>Moles HNO3:</em>
0.201L * (0.025mol/L) = 0.005025 moles HNO3
That means HNO3 is in excess. The moles in excess are:
0.005025 moles HNO3 - 0.00500 moles ethyl amine =
2.5x10⁻⁵ moles HNO₃
In 50 + 201mL = 251mL = 0.251L:
2.5x10⁻⁵ moles HNO₃ / 0.251L = 9.96x10⁻⁵M = [H+]
As pH = -log [H+]
pH = -log 9.96x10⁻⁵M
pH = 4.00 is the pH of the mixture
Answer:
- Option A): <em>Due to the constraints upton the angular momentum quantum number, the subshell </em><u><em>2d</em></u><em> does not exist.</em>
Explanation:
The <em>angular momentum quantum number</em>, identified with the letter l (lowercase L), number is the second quantum number.
This number identifies the shape of the orbital or <em>kind of subshell</em>.
The possible values of the angular momentum quantum number, l, are constrained by the value of the principal quantum number n: l can take values from 0 to n - 1.
So, you can use this guide:
Principal quantum Angular momentum Shape of the orbital
number, n quantum number, l
1 0 s
2 0, 1 s, p
3 0, 1, 2 s, p, d
Hence,
- <u>the subshell 2d (n = 2, l = 2) is not feasible</u>.
- 2s (option B) is possible: n = 2, l = 0
- 2p (option C) is possible: n = 2, l = 1
In the first shell there is 2 electrons and on the second shell there is 4 elections.
