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ella [17]
2 years ago
5

Write the formulas for the following ionic compounds: (a) copper bromide (containing the Cu+ ion), (b) manganese oxide (containi

ng the Mn3+ ion), (c) mercury iodide (containing the (Hg2)+2 ion), (d) magnesium phosphate (containing the (PO4)^-3 ion).

Chemistry
1 answer:
ivann1987 [24]2 years ago
4 0

Answer:The formulas of ionic compounds are:

a)CuBr

b)Mn_2O_3

c)Hg_2I_2

d)Mg_3(PO_4)_2

Explanation:

Formulas for the an ionic compounds is determine by:

Criss-cross method, the oxidation state of the ions gets exchanged and they form the subscripts of the other ions. This results in the formation of a neutral compound.

(a) Copper bromide :Given that it contains Cu^+ ion.

Cu^++Br^-\rightarrow CuBr

(b) Manganese oxide : Given that it contains Mn^{3+} ion.

Mn^{3+}+O^{2-}\rightarrow Mn_2O_3

(c)Mercury iodide :Given that it contains Hg_2^{2+}

Hg_2^{2+}+I^-\rightarrow Hg_2I_2

(d) Magnesium phosphate :Given that it contains PO_4^{3-}

Mg^{2+}+PO_4^{3-}\rightarrow Mg_3(PO_4)_2

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the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

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molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield  x 100%

81.0  % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

7 0
2 years ago
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Explanation:

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2 years ago
A 0.56 M solution of AlCl₃ is determined to have a concentration of particles of 1.79 M. What is the van't Hoff factor for AlCl₃
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Answer:

Van't Hoff factor for AlCl₃ = 3 (Approx)

Explanation:

Given:

Number of observed particular = 1.79 M

Number of theoretical particular = 0.56 M

Find:

Van't Hoff factor for AlCl₃

Computation:

Van't Hoff factor for AlCl₃ = Number of observed particular / Number of theoretical particular

Van't Hoff factor for AlCl₃ = 1.79 M / 0.56 M

Van't Hoff factor for AlCl₃ = 3.19

Van't Hoff factor for AlCl₃ = 3 (Approx)

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