Question

.22

Answer 1

Answer:its b

Explanation:

Answer 2

Answer:

(a)  

First, write the balanced equation as follows:

Na2S(aq) + ZnCl2(aq) -> 2 NaCl(aq) + ZnS  (s)

Then, rewrite the equation so that all dissolved compounds are separated into their constituted ions. This is called, total ionic equation:

$2 Na^+(aq) + S^{2-}(aq) + Zn^{2+}(aq) + 2 Cl^-(aq) -> ZnS (s) + 2 Na^+(aq) + 2 Cl^-(aq)$

Finally, cancel the ions that appear both in reactants and products. This is called, net ionic equation:

$S^{2-}(aq) + Zn^{2+}(aq) -> ZnS (s)$

(b)

Balanced equation:

2 K3PO4(aq) + 3Sr(NO3)2(aq) -> Sr3(PO4)2 (s) + 6 KNO3 (aq)

Total ionic equation:

$6 K^+(aq) + 2 PO_4^{3-} (aq) + 3 Sr^{2-} (aq) + 6 NO_3^- (aq) -> Sr3(PO4)2 (s) + 6 K^+(aq) + 6 NO_3^- (aq)$

Net ionic equation:

$2 PO_4^{3-} (aq) + 3 Sr^{2-} (aq) -> Sr3(PO4)2 (s)$

(c)

Balanced equation:

Mg(NO3)2(aq) + 2NaOH(aq) -> Mg(OH)2 (s) + 2 NaNO3 (aq)

Total ionic equation:

$Mg^{2+} (aq) + 2 NO_3^- (aq) + 2 Na^+ (aq) + 2 OH^- (aq) -> Mg(OH)_2 (s) + 2 NO_3^- (aq) + 2 Na^+ (aq)$

Net ionic equation:

$Mg^{2+} (aq) + 2 OH^- (aq) -> Mg(OH)_2 (s)$