Answer:
you can correct it by adding an acid to make it neutral or less basic
Answer: 1) only a salt and water
An acid and base reacts together to undergo neutralization to form salt and water.
where HCl is an acid and NaOH is a base to form salt (NaCl) and water 
.
Answer: 2) The formula of the hydrogen ion is often written as 
.
All acids dissociate in water to give ions.
Answer: 3) Arrhenius acids are substances that ionizes to yield protons in aqueous solution.
Arrhenius bases are substance that ionizes to yield hydroxide ions in aqueous solution.
Answer: 4) A conjugate acid base pair transfers hydrogen ion.
Here 
accepts a proton and thus act as a base and the corresponding 
is its conjugate acid.
Answer: 5) Bronsted Lowry acid is a substance that donates protons. Bronsted Lowry base is a substance that accepts protons.
and 
is a acid which accepts proton and thus acts as base to form conjugate acid .
Answer:
Elements in the same period have same number of electronic shell and electron is increased by one in every coming element with in same electronic shell.
Explanation:
Consider the second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.
Electronic configuration of lithium:
Li₃ = [He] 2s¹
Electronic configuration of beryllium:
Be₄ = [He] 2s²
Electronic configuration of boron:
B₅ = [He] 2s² 2p¹
Electronic configuration of carbon:
C₆ = [He] 2s² 2p²
Electronic configuration of nitrogen:
N₇ = [He] 2s² 2p³
Electronic configuration of oxygen:
O₈ = [He] 2s² 2p⁴
Electronic configuration of fluorine:
F₉ = [He] 2s² 2p⁵
Electronic configuration of neon:
Ne₁₀ = [He] 2s² 2p⁶
All these elements present in same period having same electronic shell and number of electron increased by 1.
Answer:
A) E° = 4.40 V
B) ΔG° = -8.49 × 10⁵ J
Explanation:
Let's consider the following redox reaction.
2 Li(s) +Cl₂(g) → 2 Li⁺(aq) + 2 Cl⁻(aq)
We can write the corresponding half-reactions.
Cathode (reduction): Cl₂(g) + 2 e⁻ → 2 Cl⁻(aq) E°red = 1.36 V
Anode (oxidation): 2 Li(s) → 2 Li⁺(aq) + 2 e⁻ E°red = -3.04
<em>A) Calculate the cell potential of this reaction under standard reaction conditions.</em>
The standard cell potential (E°) is the difference between the reduction potential of the cathode and the reduction potential of the anode.
E° = E°red, cat - E°red, an = 1.36 V - (-3.04 V) 4.40 V
<em>B) Calculate the free energy ΔG° of the reaction.</em>
We can calculate Gibbs free energy (ΔG°) using the following expression.
ΔG° = -n.F.E°
where,
n are the moles of electrons transferred
F is Faraday's constant
ΔG° = - 2 mol × (96468 J/V.mol) × 4.40 V = -8.49 × 10⁵ J
