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3 years ago

Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?

Chemistry

1 answer:

Rus_ich [418]3 years ago

4 0

Answer:

$Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^{+2}(aq)+2(PO_4)^{-3}(aq)$

$Ksp=[Sr^{+2}]^3[(PO_4)^{-3}]^2$

Explanation:

Hello,

In this case, for strontium phosphate, we find an ionic equation for its dissociation as shown below:

$Sr_3(PO_4)_2(s)\rightleftharpoons 3Sr^{+2}(aq)+2(PO_4)^{-3}(aq)$

Next, the solubility product is found by applying the law of mass action, considering that the solid salt is not considered but just the aqueous species due to heterogeneous equilibrium:

$Ksp=[Sr^{+2}]^3[(PO_4)^{-3}]^2$

Best regards.

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What is the density of an object that has a volume of 34.2 cm^3 and a mass of 19.6 g? (a) 0.573 g/cm^3 (b) 1.74 g/cm^3 (c) 670 g

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Read 2 more answers

Consider the reaction of magnesium metal with hydrochloric acid to produce magnesium chloride and hydrogen gas. If 3.56 mol of m

pishuonlain [190]

<h3>Answer:</h3>

43.27 g Mg

<h3>Explanation:</h3>

The balanced equation for the reaction between magnesium metal and hydrochloric acid is;

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

From the equation;

1 mole of magnesium reacts with 2 moles of HCl

We are given;

3.56 moles of Mg and 3.56 moles of HCl

Using the mole ratio;

3.56 moles of Mg would react with 7.12 moles of HCl, and

3.56 moles of HCl would react with 1.78 moles of Mg

Therefore;

The amount of magnesium was in excess;

Moles of Mg left = 3.56 moles - 1.78 moles

= 1.78 moles

But; 1 mole of Mg = 24.305 g/mol

Therefore;

Mass of magnesium left = 1.78 moles × 24.305 g/mol

= 43.2629 g

= 43.27 g

Thus, the mass of magnesium that remained after the reaction is 43.27 g

4 0

3 years ago

Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?​

Ionic equation for Sr3(PO4)2 and its solubility product (Ksp)?