Question

determina conpozitia procentuala masica a unui amestec format din 18,066 la puterea 23 atomi de Fe si 5 moli de atomi de C​

Answer 1

Răspuns:

$18.066 \times 10 ^{23} at \: \: fe..........x \: \: \: moli \\ \\ 6.022 \times 10^{23} at \: \: fe.........1 \: \: mol$

$x = \frac{6.022 \times {10}^{23} }{18.066 \times 10^{23} } = \frac{1}{3} \: \: moli \: \: fier$

16/3 moli amestec...1/3 moli Fier...5moli Cupru

100.............................x % Fier .........y % Cupru

$x = \frac{100 \times \frac{1}{3} }{ \frac{16}{3} } = \\ \\ = \frac{100 \times 3 \times \frac{1}{3} }{16} = \\ \\ = \frac{100}{16} = \\ \\ = 6.25\%$

$y = \frac{100 \times 5}{ \frac{16}{3} } = \\ \\ = \frac{100 \times 5 \times 3}{16} = \\ \\ = \frac{1500}{16} = \\ \\ = 93.75\%$

Daca adunam cele doua procente, ne da fix 100%.

Succes! ♡︎♥︎♡︎