1
JEE Main 2019 (Online) 10th April Evening Slot
MCQ (Single Correct Answer)
+4
-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)
A
4Fe(s) + 3O2(g) $$ \to $$ 2Fe2O3(s)
B
P4(s) + 5O2(g) $$ \to $$ P4O10(s)
C
C3H8(g) + 5O2(g) $$ \to $$ 3CO2(g) + 4H2O(l)
D
2Mg(s) + O2(g) $$ \to $$ 2MgO(s)
2
JEE Main 2019 (Online) 10th April Morning Slot
MCQ (Single Correct Answer)
+4
-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 :
A
C4H7Cl
B
C4H6
C
C4H8
D
C4H10
3
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
What would be the molality of 20% (mass/ mass) aqueous solution of KI?
(molar mass of KI = 166 g molā€“1)
A
1.51
B
1.35
C
1.08
D
1.48
4
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is :
A
NO2 < NO < N2O3 < N2O
B
NO2 < N2O3 < NO < N2O
C
N2O < NO < N2O3 < NO2
D
N2O < N2O3 < NO < NO2
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