No. of atoms = Atomicity × mole ×NA (A) No. of atoms =6×106212×NA=12 NA (B) No. of atoms =3×62248×NA=12 NA (C) No. of atoms =3×40240×NA=18 NA (D) No. of atoms =2×212×NA=12 NA (E) No. of atoms =3×44220×NA=15NA
Among the following choose the ones with equal number of atoms. A. 212 g of Na2CO3( s) [molar mass =106 g ] B. 248 g of Na2O (s) [molar mass =62 g ] C. 240 g of NaOH(s) [molar mass =40 g] D. 12 g of H2( g) [molar mass =2 g ] E. 220 q of CO2(q) [molar mass =44 g] Choose the correct answer from the options given below:
Held on 30 Apr 2025 · Verified 9 Jul 2026.
A, B and C only
A, B and D only
B, C and D only
B, D and E only
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For the reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = -198 kJ. Which condition favours forward reaction?
The unit of rate constant for a first-order reaction is:
For the reaction $\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g})$, the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500 , at 1000 K . [Given : $\mathrm{R}=0.0831 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$ ] $\mathrm{K}_{\mathrm{p}}$ for the reaction at 1000 K is
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$\begin{aligned} &\text { Consider the following compounds: }\\ &\mathrm{\underline{K}O}_2, \mathrm{H}_2 \mathrm{\underline{O}}_2 \text { and } \mathrm{H}_2 \mathrm{\underline{S}O}_4 \text {. } \end{aligned}$ The oxidation states of the underlined elements in them are, respectively,
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