A white insoluble solid would appeaer
Answer:
It is always twice the value of the actual yield.
Explanation:
Answer:
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Explanation:
The pH of a buffer solution is calculated using following relation

Thus the pH of buffer solution will be near to the pKa of the acid used in making the buffer solution.
The pKa value of HC₃H₅O₃ acid is more closer to required pH = 4 than CH₃NH₃⁺ acid.
pKa = -log [Ka]
For HC₃H₅O₃
pKa = 3.1
For CH₃NH₃⁺
pKa = 10.64
pKb = 14-10.64 = 3.36 [Thus the pKb of this acid is also near to required pH value)
A mixture of 100. mL of 0.1 M HC3H5O3 and 50. mL of NaOH
Half of the acid will get neutralized by the given base and thus will result in equal concentration of both the weak acid and the salt making the pH just equal to the pKa value.
The volume of NH₃ produced at STP : 0.237 L
<h3>Further explanation</h3>
Reaction
N₂ + 3H₂ → 2NH₃
1 mol = 6.02 x 10²³ particles
9.6 X 10²¹ molecules of Hydrogen, mol :

mol H₂ : mol NH₃ = 3 : 2
mol NH₃ :

Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). <em>At STP, Vm is 22.4 liters/mol.</em>
The volume of NH₃ :

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