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Chapter 1: Problem 1

If solutions of iron (III) nitrate and sodium carbonate are mixed, what would be the formula of the precipitate? (A) \(\mathrm{Fe} 3 \mathrm{CO}_{3}\) (B) \(\mathrm{Fe}_{2}\left(\mathrm{CO}_{3}\right)_{3}\) (C) \(\mathrm{NaNO}_{3}\) (D) No precipitate would form.

Short Answer

Expert verified
The formula of the precipitate that forms when iron(III) nitrate and sodium carbonate are mixed is \(\mathrm{Fe({CO_{3})_{3}}\).

Step by step solution

01

Write Down the Chemical Reaction

First write down the reaction: \(\mathrm{Fe(NO_{3})_{3}}\) + \(\mathrm{3Na_{2}CO_{3}}\) -> \(\mathrm{2Fe({CO_{3})_{3}}\) + \(\mathrm{6NaNO_{3}}\) Here, iron(III) nitrate \(\mathrm{Fe(NO_{3})_{3}}\) and sodium carbonate \(\mathrm{Na_{2}CO_{3}}\) are the reactants. A possible product is iron carbonate \(\mathrm{Fe({CO_{3})_{3}}\) and sodium nitrate \(\mathrm{NaNO_{3}}\).
02

Check the Solubility

The next step is to check the solubility of the products. Sodium nitrate, \(\mathrm{NaNO_{3}}\), is soluble in water, so it stays in solution. The solubility rules say that most compounds with carbonate, \(\mathrm{CO_{3}^{2-}}\), are insoluble. This means, that iron(III) carbonate, \(\mathrm{Fe({CO_{3})_{3}}\), would precipitate, as this compound is insoluble in water.
03

Write the Formula of the Precipitate

Since iron(III) carbonate is the compound that would precipitate, its formula \(\mathrm{Fe({CO_{3})_{3}}\) is the answer.

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Most popular questions from this chapter

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . If some of the solution evaporates overnight, which of the following will occur? (A) The mass of the solid and the concentration of the ions will stay the same. (B) The mass of the solid and the concentration of the ions will increase. (C) The mass of the solid will decrease, and the concentration of the ions will stay the same. (D) The mass of the solid will increase, and the concentration of the ions will stay the same.

A gaseous mixture at \(25^{\circ} \mathrm{C}\) contained 1 mole of \(\mathrm{CH}_{4}\) and 2 moles of \(\mathrm{O}_{2}\) and the pressure was measured at 2 \(\mathrm{atm}\) . The gases then underwent the reaction shown below. \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) What was the pressure in the container after the reaction had gone to completion and the temperature was allowed to return to \(25^{\circ} \mathrm{C} ?\) (A) 1 atm (B) 2 \(\mathrm{atm}\) (C) 3 \(\mathrm{atm}\) (D) 4 \(\mathrm{atm}\)

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) Based on the information given in the table below, what is \(\Delta H^{\circ}\) for the above reaction? \(\begin{array}{cc}{\text { Bond }} & {\text { Average bond energy }(\mathrm{kJ} / \mathrm{mol})} \\ {\mathrm{H}-\mathrm{H}} & {500} \\\ {\mathrm{O}=\mathrm{O}} & {500} \\ {\mathrm{O}-\mathrm{H}} & {500}\end{array}\) (A) \(-2,000 \mathrm{kJ}\) (B) \(-500 \mathrm{kJ}\) (C) \(+1,000 \mathrm{kJ}\) (D) \(+2,000 \mathrm{kJ}\)

$$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)+\mathrm{Cl}_{2}(g) \mapsto 2 \mathrm{NOCl}(g) \Delta G^{\circ}=132.6 \mathrm{kJ} / \mathrm{mol}$$ For the equilibrium above, what would happen to the value of \(\Delta G^{\circ}\) if the concentration of \(\mathrm{N}_{2}\) were to increase and why? (A) It would increase because the reaction would become more themodynamically favored. (B) It would increase because the reaction would shift right and create more products. (C) It would decrease because there are more reactants present. (D) It would stay the same because the value of \(K_{e q}\) would not change.
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