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3 years ago

Well over _______ different organic compounds are known to exist.

Chemistry

1 answer:

stich3 [128]3 years ago

6 0

The answer is 4 million, there are about 9-10 million unique organic compounds that are known to exist

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At what time of the day is most photosynthesis happens

Elodia [21]

I assume that it would be whenever the sun is at it's highest point because it is normally the hottest and covers the most surface area in a place. I'd say like 2pm maybe.

8 0

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.

5 0

3 years ago

Gaseous neon has a density of 0.900 g/L at standard conditions. How many neon atoms are in 1.00 L of neon gas at standard condit

Anvisha [2.4K]

Answer: 1.348 ×10^23 atoms

Explanation:

Given that volume = 1.00L

At standard condition, the volume of a gas is 22.4L/mol (at S.T.P)

Volume = mole /volume at STP

1 = mole/22.4

Mole= 22.4mol.

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