6h 8min
3tx + 6t + 3hx +6h = 3(tx + 2t +hx +2h) = 3[t(x + 2) + h(x + 2)] = 3(x + 2)(t + h)
In the reaction (2H^+ + SO_4^{2-} + Ca^{2+} + 2I^- \rightarrow CaSO_4 + 2H^+ + 2I^-), the spectator ions are those that do not change during the reaction. Here, the ( H^+ ) ions and ( I^- ) ions are present on both sides of the equation and do not participate in the formation of the precipitate ( CaSO_4 ). Therefore, the spectator ions are ( H^+ ) and ( I^- ).
It is: 11h
Carbonic acid, H2CO3
2h + 2h + 2h = 6h
- 3h - 2h + 6h + 9 = h + 9
6h 8min
22h plus 3
6h+4g
3tx + 6t + 3hx +6h = 3(tx + 2t +hx +2h) = 3[t(x + 2) + h(x + 2)] = 3(x + 2)(t + h)
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "equals".
h + 6h = 7h
6h 10m - 3h 45m = 2h 25m
If you mean: 6h+7 then 6*5 plus 7 = 37
In the reaction (2H^+ + SO_4^{2-} + Ca^{2+} + 2I^- \rightarrow CaSO_4 + 2H^+ + 2I^-), the spectator ions are those that do not change during the reaction. Here, the ( H^+ ) ions and ( I^- ) ions are present on both sides of the equation and do not participate in the formation of the precipitate ( CaSO_4 ). Therefore, the spectator ions are ( H^+ ) and ( I^- ).
It is: 11h