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mamaluj [8]
3 years ago
11

A solution contains one or more of the following ions: Ag + , Ca 2 + , and Co 2 + . Ag+, Ca2+, and Co2+. Lithium bromide is adde

d to the solution and no precipitate forms. An excess of lithium sulfate is then added to the solution and a precipitate forms. The precipitate is filtered off and lithium phosphate is added to the remaining solution, producing a precipitate. A.Which ions are present in the original solution? Co 2 + Co2+ Ca 2 + Ca2+ Ag + Ag+ B.Write a net ionic equation for the formation of the precipitate observed after the addition of lithium sulfate. Include physical states. net ionic equation: C.Write a net ionic equation for the formation of the precipitate observed after the addition of lithium phosphate. Include physical states. net ionic equation:
Chemistry
1 answer:
Eva8 [605]3 years ago
8 0

Answer:

Since with LiBr no precipitation takes place. So, Ag+ is absent

When we add Li2SO4 to it, precipitation takes place.

Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate

Thus, Ca2+ is present.

When Li3PO4 is added, again precipitation takes place.Reaction is:

Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate

A. Ca2+ and Co2+ are present in solution

B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)

C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)

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<u>Answer:</u>

<u>For a:</u> The edge length of the crystal is 533.5 pm

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<u>Explanation:</u>

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To calculate the lattice parameter or edge length of the crystal, we use the equation:

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To calculate the edge length, we use the relation between the radius and edge length for BCC lattice:

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