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2 years ago

What is the correct formula for calcium phosphate?

Chemistry

1 answer:

coldgirl [10]2 years ago

6 0

<span>your answer is Ca3</span>(PO4)2<span>, </span>

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Write the balanced electrochemical reaction for when zinc reacts with a copper solution. Label what is being oxidized and what i

Fed [463]

The balanced equation would be $Zn + Cu^{2+} ---- > Cu + Zn^{2+}$

<h3>Electrochemical equations</h3>

Zn reacts with Cu solution according to the following equation:

$Zn + Cu^{2+} ---- > Cu + Zn^{2+}$

In the reaction, $Cu^{2+}$ is reduced according to the following: $Cu^{2+} + 2 e^- -- > Cu$

While Zn is oxidized according to the following: $Zn - 2e^- --- > Zn^{2+}$

Thus, giving the overall equation of; $Zn + Cu^{2+} ---- > Cu + Zn^{2+}$

More oxidation-reduction equations can be found here: brainly.com/question/13699873

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1 year ago

Explain how Earth is heated through radiation.

sergiy2304 [10]

With this temperature the Earth Radiation will be centered on a Wavelength!

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2 years ago

Which of the following is considered a form of technology?

anygoal [31]

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3 years ago

Read 2 more answers

How many electrons do nonmetals atoms tend to gain when forming ions

amm1812

Since nonmetals have five, six, or seven electrons in their valence shells, it takes less energy to gain the necessary electrons, and therefore form anions.

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3 years ago

Hydrogen fluoride is used in the manufacture of Freons (which destroy ozone in the stratosphere) and in the production of alumin

Lerok [7]

<u>Answer:</u> The percentage yield of HF is 73.36 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ....(1)

For calcium fluoride:

Given mass of calcium fluoride = 6.25 kg = 6250 g (Conversion factor: 1 kg = 1000 g)

Molar mass of calcium fluoride = 78.07 g/mol

Putting values in above equation, we get:

$\text{Moles of calcium fluoride}=\frac{6250g}{78.07g/mol}=80.05mol$

For the given chemical reaction:

$CaF_2+H_2SO_4\rightarrow CaSO_4+2HF$

By Stoichiometry of the reaction:

1 mole of calcium fluoride produces 2 moles of hydrofluoric acid

So, 80.05 moles of calcium fluoride will produce = $\frac{2}{1}\times 80.05=160.1mol$ of hydrofluoric acid

Now, calculating the theoretical yield of hydrofluoric acid using equation 1, we get:

Moles of of hydrofluoric acid = 160.1 moles

Molar mass of hydrofluoric acid = 20.01 g/mol

Putting values in equation 1, we get:

$160.1mol=\frac{\text{Theoretical yield of hydrofluoric acid}}{20.01g/mol}=3203.6g=3.20kg$

To calculate the percentage yield of hydrofluoric acid, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield of hydrofluoric acid = 2.35 kg

Theoretical yield of hydrofluoric acid = 3.20 kg

Putting values in above equation, we get:

$\%\text{ yield of hydrofluoric acid}=\frac{2.35g}{3.20g}\times 100\\\\\% \text{yield of hydrofluoric acid}=73.36\%$

Hence, the percentage yield of HF is 73.36 %

4 0

2 years ago