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how-is-solution-72-5gr-of-the-C-3-H-6-O-how-many-Molecule-of-H-exist-




Question Number 188775 by mustafazaheen last updated on 07/Mar/23
how is solution  72.5gr   of the [C_3 H_6 O] how many Molecule of [H] exist?
howissolution72.5grofthe[C3H6O]howmanyMoleculeof[H]exist?
Answered by witcher3 last updated on 06/Mar/23
M(C_3 H_6 O)=3.12+6.1+16=58g/Mol  n=(m/M)=((72.5)/(58))  N=n.N_A =((72.5)/(58)).6,02.10^(23) ..Molecule of [C_3 H_6 O]  atom of H=6.N
M(C3H6O)=3.12+6.1+16=58g/Moln=mM=72.558N=n.NA=72.558.6,02.1023..Moleculeof[C3H6O]atomofH=6.N

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