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3 years ago

Why would Magnesium Phosphate (Mg3(PO4)2) not make an aqueous solution? Please help!

Chemistry

2 answers:

VARVARA [1.3K]3 years ago

7 0

Answer:

$\boxed {\boxed {\sf Almost \ all \ phosphates \ are \ insoluble}}$

Explanation:

For magnesium phosphate to make an aqueous solution, it must be soluble in water.

Let's check the solubility rules. There are many different lists and versions, but it should mention a rule about phosphates.

All phosphates are insoluble except Na₃PO4 (sodium phosphate), K₃PO4 (potassium phosphate), and H₁₂N₃PO₄ (ammonium phosphate).

Magnesium phosphate is included in "all phosphates" so it is insoluble and can't become an aqueous solution.

netineya [11]3 years ago

5 0

Answer:

Basically, all phosphates except Sodium phosphates, Potassium phosphates and Ammonium phosphates are insoluble in water. That, of course, includes Magnesium phosphate.

Explanation:

Hope this helped!

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Complete one rotation.

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3 years ago

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How many moles of ba(oh)2 are present in 275 ml of 0.200 m ba(oh)2?

DENIUS [597]

You have to put your attention to the unit of concentration. It is expressed in terms of molarity, which is represented in M. It is the number of moles solute per liter solution. So, you simply have to multiply the molarity with the volume in liters.

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3 years ago

What metalloid has commonly been used as an insecticide due<br> to its effectiveness as a poison.

Taya2010 [7]

Answer:

Arsenic.

Explanation:

Hello there!

In this case, since insecticides are substances that act as poisons to get rid of insects in order to prevent their presence and/or reproduction in houses, companies, crops and others, a substance that has been widely used is the metalloid arsenic due to its direct affection of the insect's body (movement, performance, cellular functions).

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3 years ago

Identify the name for the following compound: Al2O3

Vera_Pavlovna [14]

Answer:

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3 years ago

Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1

Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

$[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M$

$[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

$[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M$

Explanation:

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

$SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)$

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of $SbCl_5$ is increasing .So, the equilibrium will shift in the right direction.