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NX is produced by the following step of reactions M + X2 → M X2 MX2 + X2 → M3X8 M3X8 + N₂CO₃ → N X + CO₂ + M3O₄ How much M (metal) is consumed to produce 206 gm of NX. (Take At. wt of M = 56, N=23, X = 80)
Assertion : 1mole of H₂SO₄ is neutralised by 2 moles of NaOH but 1 equivalent of H₂SO₄ is neutralised by 1 equivalent of NaOH. Reason : Equivalent weight of H₂SO₄ is half of its molecular weight while e…
Assertion :For a binary solution of two liquids, A and B, with the knowledge of density of solution, molarity can be converted into molality. Reason: Molarity is defined in terms of volume and molality …
Assertion : If 30 mL of H₂ and 20 mL of O₂ react to form water, 5 mL of H₂ is left at the end of the reaction Reason :H₂ is the limiting reagent.
Assertion :Number of molecules present in SO₂ is twice the number of molecules present in O₂. Reason : Molecular mass of SO₂ is double to that of O₂.
Assertion : Both 138 g of K₂CO₃ and 12 g of carbon have same number of carbon atoms. Reason : Both contains 1 g atom of carbon which contains 6.022 × 1023 carbon atoms.
5 mL of N-HCl, 20 mL of N/2 H₂SO₄ milli equivalent and 30 mL of N/3 HNO₃ are mixed together and the volume is made to 1L. The normality of the resulting solution is
N₂ + 3H₂ → 2NH₃ Molecular weight of NH₃ and N₂ are x1 and x2, respectively. Their equivalent weights are y1 and y2, respectively. Then (y1 – y2) is (a) 2x x
Given the reaction : Na O(s) H O( ) 2NaOH(aq) → l What is the molarity of the solution formed if 1.35g of Na₂O is mixed with H₂O such that the final volume is 100 mL?
A solution containing 12.0% NaOH by mass has a density of 1.131 g/mL. What volume of this solution contains 5.00 mol of NaOH ?
1 mole of oxalic acid is treated with conc. H₂SO₄. The resultant gaseous mixture is passed through a solution of KOH. The mass of KOH consumed will be (where KOH absorbs CO₂.) (COOH)₂ CO + CO₂ + H₂O 2 KOH + CO₂…
In a compound C, H and N atoms are present as 9%, 1%, 3.5% respectively. Molecular weight of compound is 108. Molecular formula of compound is (a) C H N (b) C H N (c) C H N (d) C H N
When a hydrate of Na₂CO₃ is heated until all the water is removed, it loses 54.3 per cent of its mass. The formula of the hydrate is
A 27.0 g sample of an unknown hydrocarbon was burned in excess O₂ to form 88g of CO₂ and 27g of H₂O. What is possible molecular formula of hydrocarbon ?
Number of atoms in 560g of Fe (atomic mass 56 g mol–1) is
If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will
P and Q are two elements which forms P₂Q3 and PQ2. If 0.15 mole of P₂Q3 weights 15.9g and 0.15 mole of PQ2 weights 9.3g, the atomic weight of P and Q is (respectively) :
100 mL of mixture of NaOH and Na₂SO₄ is neutralised by 10 mL of 0.5 M H₂SO₄. Hence, NaOH in 100 mL solution is
NX is produced by the following step of reactions M + X2 → M X2 MX2 + X2 → M3X8 M3X8 + N₂CO₃ → N X + CO₂ + M3O₄ How much M (metal) is consumed to produce 206 gm of NX. (Take At. wt of M = 56, N=23, X = 80)
Assertion : 1mole of H₂SO₄ is neutralised by 2 moles of NaOH but 1 equivalent of H₂SO₄ is neutralised by 1 equivalent of NaOH. Reason : Equivalent weight of H₂SO₄ is half of its molecular weight while e…
Assertion :For a binary solution of two liquids, A and B, with the knowledge of density of solution, molarity can be converted into molality. Reason: Molarity is defined in terms of volume and molality …
Assertion : If 30 mL of H₂ and 20 mL of O₂ react to form water, 5 mL of H₂ is left at the end of the reaction Reason :H₂ is the limiting reagent.
Assertion :Number of molecules present in SO₂ is twice the number of molecules present in O₂. Reason : Molecular mass of SO₂ is double to that of O₂.
Assertion : Both 138 g of K₂CO₃ and 12 g of carbon have same number of carbon atoms. Reason : Both contains 1 g atom of carbon which contains 6.022 × 1023 carbon atoms.
5 mL of N-HCl, 20 mL of N/2 H₂SO₄ milli equivalent and 30 mL of N/3 HNO₃ are mixed together and the volume is made to 1L. The normality of the resulting solution is