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2 years ago

a solution of silver (i) nitrate and a solution of calcium bromate react and produce a splice precipitate of silver bromate

Chemistry

1 answer:

kykrilka [37]2 years ago

7 0

Answer:

2 AgNO₃(aq) + Ca(BrO₃)₂(aq) ⇒ Ca(NO₃)₂(aq) + 2 AgBrO₃(s)

2 Ag⁺(aq) + 2 NO₃⁻(aq) + Ca²⁺(aq) + 2 BrO₃⁻(aq) ⇒ Ca²⁺(aq) + 2 NO₃⁻(aq) + 2 AgBrO₃(s)

2 Ag⁺(aq) + 2 BrO₃⁻(aq) ⇒ 2 AgBrO₃(s)

Explanation:

The question is missing but I think it must be about the chemical equations.

Let's consider the molecular equation that occurs when a solution of silver nitrate and a solution of calcium bromate react.

2 AgNO₃(aq) + Ca(BrO₃)₂(aq) ⇒ Ca(NO₃)₂(aq) + 2 AgBrO₃(s)

The complete ionic equation includes all the ions and the insoluble species.

2 Ag⁺(aq) + 2 NO₃⁻(aq) + Ca²⁺(aq) + 2 BrO₃⁻(aq) ⇒ Ca²⁺(aq) + 2 NO₃⁻(aq) + 2 AgBrO₃(s)

The net ionic equation includes only the ions that participate in the reaction and the insoluble species.

2 Ag⁺(aq) + 2 BrO₃⁻(aq) ⇒ 2 AgBrO₃(s)

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A solution with pH of 9 has [OH-] concentration of

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3 years ago

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2 years ago

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An iron ore sample weighing 0.5562 g is dissolved HCl (aq), and the iron is obtained as Fe2 in solution. This solution is then t

erik [133]

Answer:

$\% Fe^{+2}=70%$

Explanation:

Hello,

In this case, we could considering this as a redox titration:

$Fe^{+2}+(Cr_2O_7)^{-2}\rightarrow Fe^{+3}+Cr^{+3}$

Thus, the balance turns out (by adding both hydrogen ions and water):

$Fe^{+2}\rightarrow Fe^{+3}+1e^-\\(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow 2Cr^{+3}+7H_2O\\\\6Fe^{+2}+(Cr_2^{+6}O_7)^{-2}+14H^++6e^-\rightarrow Cr^{+3}+7H_2O+6Fe^{+3}+6e^-$

Thus, by stoichiometry, the grams of Fe+2 ions result:

$m_{Fe^{+2}}=0.04021\frac{molK_2Cr_2O_7}{L}*0.02872L*\frac{6molFe^{+2}}{1molK_2Cr_2O_7}*\frac{56gFe^{+2}}{1molFe^{+2}}=0.388gFe^{+2}$

Finally, the mass percent is:

$\% Fe^{+2}=\frac{0.388g}{0.5562g}*100\%\\ \% Fe^{+2}=70%$

Best regards.

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2 years ago