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Q.

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a

a-q,b-r;c-s;d-p

b

a-r;b-q;c-p;d-s

c

a-s;b-r;c-p;d-q

d

a-p;b-q-c-r;d-s

answer is A.

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Detailed Solution

(a→q) 2molNH4⊕≡1mol of NH42CO3=96g 0.4molNH4⊕=96×0.42=19.2g (b→r) 8molH atoms ≡1molNH42CO3=96g 1molH atoms 6.02×1023H atoms  =968=12.0g  (c→s) NH42CO31mol+2HCl⟶2NH4Cl+CO21mol+H2O 1mol of CO2≡1 mol of (NH4)2CO3=96 g 3 mol of CO2≡96 × 3 = 288.0 g (d→p) 100mL×0.2M=20mmolNH42CO3 =20×10−3×93=1.92g
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