calculate the number of atoms of H present in 5.6dm3
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The sample D. (ii) Volume of A will get doubled. (iii) Gay Lussac’s Law.
(iv)
Gases A D
Volumes 1 : 4
5.6 dm3 4 × 5.6 dm3 at S. T. P.
= 22.4 dm3 (Molar volume)
= 6 × 1023 molecules.
(v) 6 × 1023 molecules is Avogadro’s number of molecules contained in 1 gram mole of the substance. If gas d is N2O then ,
1 gram mole of N2O = 2 × 14 + 16 = 44g.
(iv)
Gases A D
Volumes 1 : 4
5.6 dm3 4 × 5.6 dm3 at S. T. P.
= 22.4 dm3 (Molar volume)
= 6 × 1023 molecules.
(v) 6 × 1023 molecules is Avogadro’s number of molecules contained in 1 gram mole of the substance. If gas d is N2O then ,
1 gram mole of N2O = 2 × 14 + 16 = 44g.
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