Complete combustion of 3 g of ethane gives x*10²² molecules of water. The value of x is _____. (Round off to the nearest integer).
[ Use : NA = 6.023 *10²³; Atomic masses in u: C:12.0,O:16.0; H:1.0]
Complete combustion of 3 g of ethane gives x*10²² molecules of water. The value of x is _____. (Round off to the nearest integer).
[ Use : NA = 6.023 *10²³; Atomic masses in u: C:12.0,O:16.0; H:1.0]
Combustion of ethane: C? H? + 7/2 O? → 2CO? + 3H? O.
Moles of C? H? = 3g / 30 g/mol = 0.1 mol.
Moles of H? O produced = 0.3 mol.
Number of H? O molecules = 0.3 * 6.023 * 10²³ ≈ 18.06 * 10²².
So, x = 18.
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CH3COOH + NaOH → CH3COONa + H2O
ΔH = –50.6 kJ/mol
NaOH + SA [HCl] → NaCl + H2O
ΔH = –55.9 kJ/mol
the value of ΔH for ionisation of CH3COOH
⇒ ΔH = +55.9 – 50.6
5.3 kJ/mol
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Chemistry NCERT Exemplar Solutions Class 11th Chapter Three 2025
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