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jasenka [17]
3 years ago
7

Please help me with this.And show all work as well ASAP!!​

Chemistry
1 answer:
qaws [65]3 years ago
6 0

Answer: The partial pressure of oxygen is 187 torr.

Explanation:

According to Raoult's law, the partial pressure of a component at a given temperature is equal to the mole fraction of that component multiplied by the total pressure.

p_1=x_1p_{total}  

where, x = mole fraction  

p_{total} = total pressure  

x_{oxygen}=\frac{\text {moles of oxygen}}{\text {total moles}},  

x_{oxygen}=\frac{3.0}{12.33}=0.243,  

p_{oxygen}=0.243\times 770torr=187torr  

Thus the partial pressure of oxygen is 187 torr.

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What is the product of SO2+H2O. ​
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A gas at 42. 0°c occupies a volume of 1. 32 l. if the volume increases to 2. 24 l, what is the new temperature in kelvin?
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what is the percent yield of titanium (II) oxide if 20.0 grams of titanium (II) sulfide is reacted with water? The actual yield
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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

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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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