Answer:
the stabilization of the negative charge in orbitals with higher s character
Explanation:
Acetylide anion is a carbon anion compound or popularly called carbanion. Now Acetylide anion is sp hybridized. However acetylide anion tends to be more acidic as we move from sp³ to sp, hence acidicity increases, which makes sp to have the highest acidity and become the most stable.
So, we can conclude that the acetylide anion is more acidic due to the stabilization of the negative charge in orbitals with higher s character and as the s character increases, acidic nature of acetylide anion also increases.
We can skip option B and D because NaCl is salt and H₂SO₄ is a strong acid.
Neutralization reactions are those reactions in which acid and base react to form salt and water.
As water being amphoteric in nature can react with HCl as follow,
HCl + H₂O ⇆ H₃O⁺ + OH⁻
In this case no salt is formed, so we can skip this option.
Ammonia being a weak base can abstract proton from HCl as follow,
HCl + NH₃ → NH₄Cl
Ammonium Chloride is a salt. So, among all four options, Option-C is the correct answer.
Answer:
HCl for acid, NaOH for base.
Explanation:
Acids start with H.
Bases end with OH.
137 K
The volume is constant, so you can use <em>Gay-Lussac’s Pressure-Temperature Law </em>to calculate the new temperature (you don’t have to use the number of moles).
P1/T1 = P2/T2
Solve for T2: T2= T1 x P2/P1
P1 = 1.83 atm; T1 = 122 K
P2 = 2.05 atm; T2 = ?
∴ T2 = 122 K x (2.05 atm)/(1.83 atm) = 137 K
This result makes sense. Temperature is directly proportional to pressure. You increased the pressure by about 10 %, so the temperature increased by about 10 %.
