Answer:
60.55%
Explanation:
nO3=155/48
nO2 used: 3/2nO3=4.84375
percent yield: 4.84375/8=60.55%
Answer:
pH = 2.69
Explanation:
The complete question is:<em> An analytical chemist is titrating 182.2 mL of a 1.200 M solution of nitrous acid (HNO2) with a solution of 0.8400 M KOH. The pKa of nitrous acid is 3.35. Calculate the pH of the acid solution after the chemist has added 46.44 mL of the KOH solution to it.</em>
<em />
The reaction of HNO₂ with KOH is:
HNO₂ + KOH → NO₂⁻ + H₂O + K⁺
Moles of HNO₂ and KOH that react are:
HNO₂ = 0.1822L × (1.200mol / L) = <em>0.21864 moles HNO₂</em>
KOH = 0.04644L × (0.8400mol / L) = <em>0.0390 moles KOH</em>
That means after the reaction, moles of HNO₂ and NO₂⁻ after the reaction are:
NO₂⁻ = 0.03900 moles KOH = moles NO₂⁻
HNO₂ = 0.21864 moles HNO₂ - 0.03900 moles = 0.17964 moles HNO₂
It is possible to find the pH of this buffer (<em>Mixture of a weak acid, HNO₂ with the conjugate base, NO₂⁻), </em>using H-H equation for this system:
pH = pKa + log₁₀ [NO₂⁻] / [HNO₂]
pH = 3.35 + log₁₀ [0.03900mol] / [0.17964mol]
<h3>pH = 2.69</h3>
Explanation:
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Option (a) is correct.
A reducing agent is the one which loses electrons to other substance.
Here, Zn has oxidation number 0 in the L.H.S of the equation, but on R.H.S its oxidation number is +2 i.e. it Zn has donated two of its electrons to

.
Hence, Zn is the reducing agent here.
Given:
No of atoms present= 8.022 x 10^23 atoms
Now we know that 1 mole= 6.022 x 10^23 atoms
Hence number of moles present in 8.022 x 10^23 atoms is calculated as below.
Number of moles
= 8.022 x 10^23/6.022x 10^23
=1.3 moles.
Hence we have 1.3 moles present.