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Answers
Explanation:
1)Atomicity is defined as the total number of atoms present in a molecule. For example, each molecule of oxygen (O2) is composed of two oxygen atoms.
2)Molecular formula =XY
Molecular formula =XY 2
Molecular formula =XY 2
Molecular formula =XY 2 Z
Molecular formula =XY 2 Z 3
Molecular formula =XY 2 Z 3
Molecular formula =XY 2 Z 3
Molecular formula =XY 2 Z 3 10
Molecular formula =XY 2 Z 3 10 23
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z .
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound =
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 3
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01 ∴ Molecular mass =
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01 ∴ Molecular mass = 0.01
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01 ∴ Molecular mass = 0.014.40
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01 ∴ Molecular mass = 0.014.40
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01 ∴ Molecular mass = 0.014.40 =440
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01 ∴ Molecular mass = 0.014.40 =440 ⇒300+2y=440
Molecular formula =XY 2 Z 3 10 23 , atomic weight of Y=y Molecular mass =60+240+2y =300+2y 1.15×10 23 atoms of Y≃2 mol of Y ∴ Limiting agent is Z . ∴ No. of moles of compound = 30.03 =0.01 ∴ Molecular mass = 0.014.40 =440 ⇒300+2y=440 ⇒y=70