When an aqueous solution of sodium phosphate (Na3PO4) and an aqueous solution of magnesium sulfate (MgSO4) are mixed, a precipitate forms. Write the chemical formula of the precipitate.

Answers

Answer 1

Answer: The chemical formula of this precipitate is [tex]Mg_{3}(PO_{4})_{2}[/tex].

Explanation:

An equation which depicts the chemical reaction of substances in the form of chemical formulas is called a chemical equation.

For example, chemical equation for an aqueous solution of sodium phosphate [tex](Na_{3}PO_{4})[/tex] and an aqueous solution of magnesium sulfate [tex](MgSO_{4})[/tex] are mixed, a precipitate forms is as follows.

[tex]3MgSO_{4}(aq) + 2Na_{3}PO_{4}(aq) \rightarrow Mg_{3}(PO_{4})_{2}(s) + 3Na_{2}SO_{4}(aq)[/tex]

Here, trimagnesium phosphate is the precipitate. The chemical formula of this precipitate is [tex]Mg_{3}(PO_{4})_{2}[/tex].

Thus, we can conclude that the chemical formula of this precipitate is [tex]Mg_{3}(PO_{4})_{2}[/tex].

Related Questions

THIS NOT MY WORK. its for my sister. HELP HER

Answers

Answer:

Your answer will be b(molten material from the outer core makes its way to the surface of earth)

Explanation:

Answer:

Explanation:

I'm pretty sure molten material comes from the inner core. I think you can search up this though, try to find where molten material comes from. Goodluck!

what is the pH of a 2.58 x 10^-6 M solution of an acid?

Answers

Answer:

5.59

Explanation:

The pH of a solution can be found by using the formula

[tex]pH = - log [ {H}^{+} ][/tex]

From the question we have

[tex]ph = - log(2.58 \times {10}^{ - 6} ) \\ = 5.58838[/tex]

We have the final answer as

5.59

Hope this helps you

Determine the spec gravity of 500 mL alcohol if the density is 0.79g/ml

Answers

Answer:

S.G = 0.79.

Explanation:

Hello there!

In this case, according to the given information in the problem, it turns out possible for us to calculate the specific gravity of this alcohol by simply dividing its density, 0.79 g/mL by that of the water, 1 g/mL just as a reference for us to work with:

[tex]S.G=\frac{\rho _{alc}}{\rho _w}[/tex]

Thus, we plug in the densities to obtain:

[tex]S.G=\frac{0.79g/mL}{1g/mL}\\\\S.G=0.79[/tex]

Which is dimensionless as g/mL is cancelled out due to its presence on both top and bottom of the previous formula.

Regards!

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