Answer: Option (c) is the correct answer.
Explanation:
Atomic number of sodium is 11 and its electronic configuration is 
. When sodium loses one electron then it will attain +1 charge and its electronic configuration will be as follows.
: 
Atomic number of fluorine is 9 and its electronic configuration is 
. When fluorine gains an electron then it acquires -1 charge and its electronic configuration is as follows.
:
Atomic number of aluminium is 13 and its electronic configuration is 
. When aluminium loses its valence electrons then it acquires +3 charge and its electronic configuration is as follows.
:
Thus, we can conclude that the listing for aluminum is correct.
1) V(CH₄) = 0,376 L.T(CH₄) = 304 K.p(CH₄) = 1,5 atm 101325 Pa/atm = 151987,5 Pa = 151,9875 kPa.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(CH₄) = p · V ÷ R · T.n(CH₄) = 151,9875 kPa · 0,376 L ÷ 8,314 J/K· mol · 304 K.n(CH₄) = 0,0226 mol.V(CH₄) = n(CH₄) · Vm.V(CH₄) = 0,0226 mol · 22,4 dm³/mol.V(CH₄) = 0,506 dm³ = 0,506 L.
2) V(SO₂) = 5,2 L.p(SO₂) = 45,2 atm = 45,2 atm · 101,325 kPa/atm = 4579,89 kPa.T(SO₂) = 293 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(SO₂) = p · V ÷ R · T.n(SO₂) = 4579,89 kPa · 5,2 L ÷ 8,314 J/K· mol · 293 K.n(CH₄) = 9,77 mol.There is not enogh SO₂, 225 mol - 9,77 mol = 215,23 mol is needed.
3) p(He) = 3,50 atm · 101,325 kPa/atm = 354,63 kPa.V(He) = 4,00 L.n(He) = 0,410 mol.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.T = p · V ÷ R · n.T(He) = 354,63 kPa · 4,00 L ÷ 8,314 J/K· mol · 0,410 mol.T(He) = 416,14 K.n - amount of substance.
4) p(Ar) = 1,00 atm · 101,325 kPa/atm = 101,325 kPa.V(Ar) = 3,4 L.T(Ar) = 263 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(Ar) = p · V ÷ R · T.n(Ar) = 101,325 kPa · 3,4 L ÷ 8,314 J/K· mol · 263 K.n(Ar) = 0,157 mol.n(Ar) = 0,157 mol + 2,5 mol = 2,657 mol.p(Ar) = 2,657 mol · 8,314 J/K· mol · 263 K ÷ 3,4 L.p(Ar) = 1708,74 kPa.
Answer:
CH3CH=NH2+>CH3CH2NH3+
Explanation:
If we look at the both species under review, we will realize that they are both amines hence they possess the polar N-H bond.
Electrons are ordinarily attracted towards the nitrogen atom hence making both compounds acidic. It is worthy of note that certain features of a compound may make it more acidic than another of close structural proximity. 'More acidic' simply means that the proton is more easily lost.
CH3CH=NH2+ contains an sp2 hybridized carbon atom which is highly electronegative and further withdraws electron density from the N-H bond thereby leading to a greater acidity of CH3CH=NH2+ compared to CH3CH2NH3+
Answer:
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
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(a) 3
(b) 6
(c) 7
Explanation:
We can state the ground-state electron configuration for each element following Aufbau's principle.
(a) ₁₉K: 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹
(b) ₁₀Ne: 1s² 2s² 2p⁶
Second part
(a) Al belongs to Group 13 in the Periodic Table. It has 13-10=3 electrons in the valence shell.
(b) O belongs to Group 16 in the Periodic Table. It has 16-10=6 electrons in the valence shell.
(c) F belongs to Group 17 in the Periodic Table. It has 17-10=7 electrons in the valence shell.
