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RideAnS [48]
3 years ago
5

Consider the reaction: 3Co2+(aq) + 6NO3¯(aq) + 6Na+(aq) + 2PO43¯(aq) â Co3(PO4)2(s) + 6Na+(aq) + 6NO3¯(aq) Identify the net i

onic equation for this reaction.A. 3CO2(aq) + 6NO3(aq) + 6Na+(aq) + 2PO43-(aq) â Co3(PO4)2(s) + 6NaNO3(aq). B. Na+(aq) + NO3-(aq) â NaNO3(aq).C. 3CO2(aq) + NO3(aq) + Na*(aq) + 2PO43-(aq) â Co3(PO4)2(s) + NaNO3(aq).D. 3C02(aq) + 2PO43-(ag) â CO3(PO4)2(s).E. None of these choices are correct.
Chemistry
1 answer:
irina [24]3 years ago
3 0

Answer:

D. 3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

Explanation:

The complete ionic equation includes all the ions and the insoluble species. Let's consider the following complete ionic equation.

3 Co²⁺(aq) + 6 NO₃⁻(aq) + 6 Na⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s) + 6 Na⁺(aq) + 6 NO₃⁻(aq)

The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the insoluble species. The corresponding net ionic equation is:

3 Co²⁺(aq) + 2 PO₄³⁻(aq) ⇒ Co₃(PO₄)₂(s)

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This will happen when we add the reactant to a chemical reaction. According to Le-Chatelier's principle, by increasing the reactant, the equilibrium will shift in the direction where this effect is minimal. Hence, forward reaction is favored.

2) The rate of the revere reaction increases.

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3) The concentration of product increases.

This will happen when we add reactants to a chemical reaction. According to  Le-Chatelier's principle, when we increase the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the forward direction which means that the concentration of product will increase.

4) The concentration of products decreases.

This will happen when we remove reactants from a chemical reaction. According to  Le-Chatelier's principle, when we decrease the concentration of reactants, the equilibrium will shift in the direction where this effect is minimal. Hence, the reaction will be in the reverse direction which means that the concentration of reactants will increase or concentration of products will decrease.

3 0
3 years ago
A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What
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Answer:

Approximately 1.854\; \rm mol\cdot L^{-1}.

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of \rm Sr, \rm O, and \rm H on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

  • \rm Sr: 87.62.
  • \rm O: 15.999.
  • \rm H: 1.008.

Calculate the formula mass of \rm Sr(OH)_2:

M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}.

<h3>Number of moles of strontium hydroxide in the solution</h3>

M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1} means that each mole of \rm Sr(OH)_2 formula units have a mass of 121.634\; \rm g.

The question states that there are 10.60\; \rm g of \rm Sr(OH)_2 in this solution.

How many moles of \rm Sr(OH)_2 formula units would that be?

\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}.

<h3>Molarity of this strontium hydroxide solution</h3>

There are 8.71467\times 10^{-2}\; \rm mol of \rm Sr(OH)_2 formula units in this 47\; \rm mL solution. Convert the unit of volume to liter:

V = 47\; \rm mL = 0.047\; \rm L.

The molarity of a solution measures its molar concentration. For this solution:

\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}.

(Rounded to four significant figures.)

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