The correct answer would be the first option. Material A having a smaller latent heat of fusion would mean that it will take only less energy to phase change into the liquid phase. Latent of heat of fusion is the amount of energy needed of a substance to phase change from solid to liquid or liquid to solid.
PH is calculated using <span>Handerson- Hasselbalch equation,
pH = pKa + log [conjugate base] / [acid]
Conjugate Base = Acetate (CH</span>₃COO⁻)
Acid = Acetic acid (CH₃COOH)
So,
pH = pKa + log [acetate] / [acetic acid]
We are having conc. of acid and acetate but missing with pKa,
pKa is calculated as,
pKa = -log Ka
Putting value of Ka,
pKa = -log 1.76 × 10⁻⁵
pKa = 4.75
Now,
Putting all values in eq. 1,
pH = 4.75 + log [0.172] / [0.818]
pH = 4.072
<span>2 KClO3(s) → 3 O2(g) + 2 KCl(s)
</span><span>Note: MnO2 (Manganese Dioxide) is not part of the reaction. A catalyst lowers the activation energy and increases both forward and reverse reactions at equal rates.
</span>
molar mass of KClO3 = 122.5
Moles of KClO3 = 3.45 / 122.55 = 0.028
Moles of O2 produce =

= 0.042 moles
molar mass of O2 = 32
so, mass of O2 = 32 x 0.042 = 1.35 g
Answer
- continuous removal of PH3
- adding more of P into the system
Explanation:
In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);
- The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:
heat+A⇌BΔH=+
- Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
- According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.
Answer:
41.3 minutes
Explanation:
Since the reaction is a first order reaction, therefore, half life is independent of the initial concentration, or in this case, pressure.

So, fraction of original pressure = 
n here is number of half life
therefore, 
⇒ n= 3
it took 124 minutes to drop pressure to 1/8 of original value, half life = 124/3= 41.3 minutes.