For C, they would be the same. At the equivalence, you have equal moles of both the acid and the base.
Answer:
E. Gain of electrons
Explanation:
A reduction reaction is one part of the two concurrent reactions that take place in a redox (reduction-oxidation) reaction.
During reduction, an atom gains electrons from a donor atom, and it's oxidation number becomes smaller.
Option A is wrong because reduction does not increase oxidation state nor are neutrons involved
Option B is wrong because reduction is not a nuclear reaction (does not involve the nucleons)
Option C is wrong because reduction leads to reduction in oxidation state
Option D is wrong leads to a reduction in oxidation state when electrons are gained
Option E is correct because reduction involves gain of electrons
I can help you! What is your question?
<u>Answer:</u> The ion formed by lithium element is 
<u>Explanation:</u>
Ions are formed when an atom looses or gains electrons.
If an atom gains electrons, it leads to the formation of negative ions known as anions. For Example: Fluorine gains 1 electron to form
ions.
If an atom looses electrons, it leads to the formation of positive ions known as cations. For Example: Sodium looses 1 electron to form
ions.
Lithium is the 3rd element of the periodic table having 3 electrons.
The electronic configuration of lithium element = 
This element will loose 1 electron to attain stable electronic configuration and will lead to the formation of
ion.
Hence, the ion formed by lithium element is 
Answer:
ΔU = 103.54 KJ
Explanation:
∴ ΔU = Q + W
ideal gas:
∴ PV = nRT
∴ R = 8.314 L.KPa/K.mol
∴ n = 3000 g N2 * ( mol/28.0134g N2) = 107.143 mol N2
∴ m N2 = 3.0 Kg
∴ T1 = 300 K
∴ P1 = 100 KPa
∴ V1 = nRT1/P1 = 2672.36 L = 2.67 m³
⇒ V2 = 0.9*V1 = 2405.12 L = 2.41 m³
∴ P2 = 140 KPa
⇒ T2 = P2.V2/n.R = 377.99 ≅ 378 K
⇒ W = P1V1 - P2V2
⇒ W = ((100KPa)*(2.67m³)) - ((140KPa)*(2.41m³))
⇒ W = - 70.164 KJ
∴ Q = nCpΔT
∴ Cp = (5/2)*R = 20.785 J/mol.K ....ideal gas
⇒ Q = (107.143mol)*(20.785 J/mol.K)*(378 - 300)
⇒ Q = 173703.446 J = 173.703 KJ
⇒ ΔU = 173.703 KJ - 70.164 KJ
⇒ ΔU = 103.54 KJ