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^- ).
The conditional constant= 1.8*1010
Barium.
The question is poorly specified, since the given triangle can be right angled at A or C. If it is right angled at A, then bc2 = ab2 + ac2 so that ab2 = 100 - 16 = 84 and ab = sqrt(84) = 9.165 Or it is right angled at C, and ab2 = bc2 + ca2 = 100 + 16 = 116 so that ab = sqrt(116) = 10.770
The net ionic equation is SO42- + Ca2+ CaSO4.
2H+ + SO42- + Ca2+ + 21 > CaSO4 + 2H+ + 21-
The net ionic equation is SO42- + Ca2+ CaSO4.
2H+ + SO42- + Ca2+ + 2I- CaSO4 + 2H+ + 2I-
2H+ + SO42- + Ca2+ + 2I- CaSO4 + 2H+ + 2I
Ca2+(aq) + SO42-(aq)- --> CaSO4(s) is the net ionic equation.
CaSO4 is an ionic compound because it contains two ions namely CA2+ and SO42-.
Complete Ionic Equation. 2K+(aq) + SO42-(aq) + Ca2+(aq) + 2I-(aq) --> 2K+(aq) + 2I-(aq) + CaSO4(s) Net Ionic Equation Minus the Spectator Ions K+ and I-. Ca2+(aq) + SO42-(aq) --> CaSO4(s)
The net ionic equation for the reaction between Na2SO4 and CaCl2 is: Ca2+ + SO4^2- → CaSO4 This reaction involves the formation of calcium sulfate (CaSO4) precipitate.
Ca2+ + 2OH- + 2H+ + SO42- -------> 2H2O + Ca2+ + So42- 2OH- + 2H+ -------> 2H2O
2Na+(aq) + SO42-(aq) + Ca2+(aq) + 2Cl-(aq) CaSO4(s) + 2Na+(aq) + 2Cl-(aq)
The balanced molecular equation is CaCl2 + Na2S -> CaS + 2NaCl. To write the ionic equation, we need to break down the reactants and products into their respective ions. This results in the ionic equation: Ca2+ + 2Cl- + 2Na+ + S2- -> CaS + 2Na+ + 2Cl-. Cross out spectator ions that appear on both sides of the equation to obtain the net ionic equation: Ca2+ + S2- -> CaS.