Question

Olá poderiam me ajudar nesta questão de Química?
Para uma amostra de 1980g de aspirina (C9H8O4) calculae

Answer 1

Resposta:

1 mol (C9H8O4)---------------------180g

x mols (C9H804)--------------------1980g

180x = 1980

x = 1980 ÷ 180

x = 11 mols de (C9H804)

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1 mol (C9H804)----------180g-------------6,02×$10^{23}$ moléculas

                                  1980g----------------  X moléculas tenho

180x = 6,02×$10^{23}$ × 1980

180x = 11919.6

×$10^{23}$

x = 11919.6

×$10^{23}$ ÷ 180

x = 6,622×$10^{24}$ moléculas de C9H804

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1 mol(C9H804)----180g---6,02×$10^{23}$---9×6,02×$10^{23}$

                            1980g-------------x átomos de C

180x =  1980 × 9×6,02×$10^{23}$

180x = 107276.4×$10^{23}$

x = 107276.4×$10^{23}$ ÷ 180

x = 5,9598×$10^{25}$ átomos de Carbono.

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1mol(C9H804)----180g----6,02×$10^{23}$----4×6,02×$10^{23}$

                          1980g--------x átomos de O

180x = 47678.4$10^{23}$

x = 47678.4$10^{23}$ ÷ 180

x = 2,6488×$10^{25}$ átomos de O

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1mol(C9H804)----180g----6,02×$10^{23}$----126.42$10^{23}$

                           1980g-------- x átomos totais

180x = 250311.6$10^{23}$

x = 250311.6$10^{23}$ ÷ 180

x = 1,39062×$10^{26}$ átomos totais

Explicação: