1) ответ:
n=N/Na
6.02 * 10^23 * 3 = 18.06 * 10^23
n = 18.06 * 10^23 /(6.02 * 10^23) = 3 моль
Объяснение:
В воде H2O есть всего 3 атома - 2 атома водорода и 1 атом кислорода
Значит, в воде будет в 3 раза больше атомов, чем молекул
тогда умножаем данное кол-во молекул на 3
6.02 * 10^23 * 3 = 18.06 * 10^23
теперь через формулу n=N/Na находим кол-во вещества(моли), где Nа - постоянна Авогадро(6.02 * 10^23)
2) ответ:
N=Nα×n отсюда N=6,02*10²³×3=18,06*10²³ атомов
3) ответ:
n(CO2) = N/Na = 3.01*10^23/6.02*10^23 = 0.5 моль
4) ответ:
N(O)=0,5*4*NA=05*4*6,02*10^23=12,04*10^23 атомов.
5) ответ:
n (H2O) = 0,9 / 18 = 0,05 (моль).
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pH = -lg[H⁺] = -lg10⁻⁵ = 5
pOH = 14 - pH = 14 - 5 = 9
pOH = - lg[OH⁻]
[OH⁻] = 10^-pOH = 10⁻⁹ моль/л
Реакция раствора муравьиной кислоты с данным значением рН будет слабокислой