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3 years ago

Why does the polyatomic anion OH need to gain

1 answer:

5 0

Answer:Well-known examples are sodium hydroxide (NaOH) with OH- as the polyatomic anion, calcium carbonate (CaCO3), and ammonium nitrate (NH4NO3), which contains two polyatomic ions: NH+ and NO3-. ... The properties of compounds containing polyatomic ions are very similar to those of binary ionic compounds.

Explanation:

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2 years ago

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3 years ago

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3 years ago

What is the percent error if the measured value is 30.0 g and<br> the accepted value is 32.0 g?

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3 years ago

what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield

earnstyle [38]

Answer : The percent yield of titanium (II) oxide is, 142.5 % and the impurities could have caused the percent yield to be so high.

Explanation : Given,

Mass of titanium(II) sulfide = 20.0 g

Molar mass of titanium(II) sulfide = 79.9 g/mole

Molar mass of titanium(II) oxide = 63.9 g/mole

First we have to calculate the moles of titanium(II) sulfide.

$\text{ Moles of titanium(II) sulfide}=\frac{\text{ Mass of titanium(II) sulfide}}{\text{ Molar mass of titanium(II) sulfide}}=\frac{20.0g}{79.9g/mole}=0.2503moles$

Now we have to calculate the moles of titanium(II) oxide.

The balanced chemical reaction is,

$TiS+H_2O\rightarrow TiO+H_2S$

From the reaction, we conclude that

As, 1 mole of titanium(II) sulfide react to give 1 mole of titanium(II) oxide

So, 0.2503 mole of titanium(II) sulfide react to give 0.2503 mole of titanium(II) oxide

Now we have to calculate the mass of titanium(II) oxide.

$\text{ Mass of titanium(II) oxide}=\text{ Moles of titanium(II) oxide}\times \text{ Molar mass of titanium(II) oxide}$

$\text{ Mass of titanium(II) oxide}=(0.2503moles)\times (63.9g/mole)=15.99g$

To calculate the percentage yield of titanium (II) oxide, we use the equation:

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

Experimental yield of titanium (II) oxide = 22.8 g

Theoretical yield of titanium (II) oxide = 15.99 g

Putting values in above equation, we get:

$\%\text{ yield of titanium (II) oxide}=\frac{22.8g}{15.99g}\times 100\\\\\% \text{yield of titanium (II) oxide}=142.5\%$

Hence, the percent yield of titanium (II) oxide is, 142.5 %

If the percent yields is greater than 100% that means the product of the reaction contains impurities which cause its mass to be greater than it actually.

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3 years ago