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oksian1 [2.3K]
3 years ago
14

For a particular isomer of C8H18, the following reaction produces 5093.7 KJ of heat per mole of C8H18(g) consumed, under stander

ed conditions. What is the standard enthalpy of formation of this isomer of C8H18(g)?
Chemistry
1 answer:
lapo4ka [179]3 years ago
3 0
Delta Hf = dHf 
<span>dHf rxn = (n*dHf products) - (n*dHf reactants). </span>
<span>Substitute 5113.3 kJ for dHf rxn and solve for the only unknown in the equation which is dHf C8H18. 
</span>-5133.3-[(8x-393.5)+9(-241.8)] = 190.9 

<span>Since it is an enthalpy of formation, the value has to be negative, </span>
<span>thus the correct answer is -190.9</span>
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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

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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A 50.00 mL sample of groundwater is titrated with 0.0300 M EDTA . Of 12.40 mL of EDTA is required to titrate the 50.00 mL sample
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Answer:

7.44x10⁻³ mol/L and 744 ppm

Explanation:

Let's assume that the hardness of the water is totally from Ca⁺² ions only(the hardness is the measure of Ca⁺² and Mg⁺² ions). The titration with EDTA will form a complex. The EDTA is always in 1:1 proportion, so the number of moles of it will be the number of moles of Ca⁺², which will be the number of moles of CaCO₃.

n = 0.0124 L * 0.0300 mol/L

n = 3.72x10⁻⁴ mol

The molarity is the number of moles divided by the volume (0.05 L)

M = 3.72x10⁻⁴/0.05

M = 7.44x10⁻³ mol/L

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How many atoms of oxygen are on the reactants side of this balanced equation?
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<h3>Answer:</h3>

6

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

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Order of Operations: BPEMDAS

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<u>Step 2: Identify</u>

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