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noname [10]
3 years ago
6

When aqueous solutions of manganese(II) iodide and sodium phosphate are combined, solid manganese(II) phosphate and a solution o

f sodium iodide are formed.
Give the net ionic equation for this reaction.
Chemistry
1 answer:
Feliz [49]3 years ago
8 0

Answer:

The net ionic equation for the given reaction :

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

Explanation:

3MnI_2(aq)+2Na_3PO_4_2(aq)\rightarrow Mn_3(PO_4)_2(s)+6NaI(aq)...[1]

MnI_2(aq)\rightarrow Mn^{2+}(aq)+2I^-(aq)..[2]

Na_3PO_4(aq)\rightarrow 3Na^{+}(aq)+PO_4^{3-}(aq)...[3]

NaI(aq)\rightarrow Na^+(aq)+I^-(aq)

Replacing MnI_2(aq) , NaI and Na_3PO_4(aq) in [1] by usig [2] [3] and [4]

3Mn^{2+}(aq)+6I^-(aq)+6Na^{+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)+6Na^+(aq)+6I^-(aq)

Removing the common ions present ion both the sides, we get the net ionic equation for the given reaction [1]:

3Mn^{2+}(aq)+2PO_4^{3-}(aq)\rightarrow Mn_3(PO_4)_2(s)

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Answer:

ΔS=0.148  KJ/K

Explanation:

Given that

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T₂=5 + 273 = 278 K

Reservoir 1 is rejecting heat that is why it taken as negative while the reservoir 2 is gaining the heat that is why it is taken as positive.

So the total change in entropy given as

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ΔS=  - 100/473 + 100/278  KJ/K

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4 0
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Thus we can conclude that Fe₂P causes decrease in yield

Learn more about production of phosphorus at brainly.com/question/13337198

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6 0
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