1
[H+] * [OH-]= <span><span /></span><span><span><span><span /></span></span></span><span><span><span><span /></span></span></span><span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span></span>10^ -14
[H+] = <span><span><span /></span></span><span><span><span><span><span /></span></span></span></span><span><span><span><span><span /></span></span></span></span><span><span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span></span></span><span>10^ -14 / </span><span>10^ -9 =</span>10^ -5 <span>моль/л СРЕДА СЛАБОКИСЛАЯ</span>
2/
[OH-] = <span><span /></span><span><span><span><span /></span></span></span><span><span><span><span /></span></span></span><span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span></span>10^ -14 / 10^ -2 =10^ -12 моль/л СРЕДА <span>КИСЛАЯ
метилоранж красный
</span>
3.
[H+] = <span /><span><span><span /></span></span><span><span><span /></span></span><span><span><span><span><span><span><span><span><span><span /></span></span></span></span></span></span></span></span></span>10^ -14 / 10^ -5 =10^ -9 моль/л
рН = -lg[H+] = 9 среда щелочная
лакмус - синий
... неполном сгорании угля
2C + O2 => 2CO
вариант 3, так как по периоду радиус уменьшается
S(+6) +8ē = S(-2) окислитель
Al(0) - 3ē = Al(+3) восстановитель
15H2SO4 + 8Al = 3H2S + 4Al2(SO4)3 + 12H2O