JEE Main 2019 (Online) 10th April Evening Slot | Some Basic Concepts of Chemistry Question 120 | Chemistry

JEE Main 2019 (Online) 10th April Evening Slot

MCQ (Single Correct Answer)

-1

The minimum amount of O₂(g) consumed per gram of reactant is for the reaction :
(Given atomic mass : Fe = 56, O = 16, Mg = 24, P = 31, C = 12, H = 1)

4Fe(s) + 3O₂(g) $$ \to $$ 2Fe₂O₃(s)

P₄(s) + 5O₂(g) $$ \to $$ P₄O₁₀(s)

C₃H₈(g) + 5O₂(g) $$ \to $$ 3CO₂(g) + 4H₂O(l)

2Mg(s) + O₂(g) $$ \to $$ 2MgO(s)

JEE Main 2019 (Online) 10th April Morning Slot

MCQ (Single Correct Answer)

-1

At 300 K and 1 atmospheric pressure, 10 mL of a hydrocarbon required 55 mL of O₂ for complete combustion, and 40 mL of CO₂ is formed. The formula of the hydrocarbon is :

C₄H₇Cl

C₄H₆

C₄H₈

C₄H₁₀

JEE Main 2019 (Online) 9th April Evening Slot

MCQ (Single Correct Answer)

-1

What would be the molality of 20% (mass/ mass) aqueous solution of KI?
(molar mass of KI = 166 g mol^–1)

1.51

1.35

1.08

1.48

JEE Main 2019 (Online) 9th April Morning Slot

MCQ (Single Correct Answer)

-1

The correct order of the oxidation states of nitrogen in NO, N₂O, NO₂ and N₂O₃ is :

NO₂ < NO < N₂O₃ < N₂O

NO₂ < N₂O₃ < NO < N₂O

N₂O < NO < N₂O₃ < NO₂

N₂O < N₂O₃ < NO < NO₂

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