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Mrac [35]
3 years ago
5

Heptane (C7H16) and octane (C8H18) are constitutents of gasoline. At 80 degree celsius, the vapor pressure of heptane is 428mmHg

and the vapor pressure of octane is 175mmHg. What Xheptane in a mixture of heptane and octane that has a vapor pressure of 305mmHg and 80 degree celsius?
Chemistry
2 answers:
harkovskaia [24]3 years ago
7 0

<u>Answer:</u> The mole fraction of heptane at 80°C is 0.506

<u>Explanation:</u>

We are given:

Vapor pressure of heptane at 80°C = 428 mmHg

Vapor pressure of octane at 80°C = 175 mmHg

Total pressure at 80°C = (428 + 175) = 603 mmHg

To calculate the mole fraction of heptane at 80°C, we use the equation given by Raoult's law, which is:

p_{heptane}=p_T\times \chi_{heptane}

where,

p_A = partial pressure of heptane = 305 mmHg

p_T = total pressure  = 603 mmHg

\chi_A = mole fraction of heptane = ?

Putting values in above equation, we get:

305mmHg=603mmHg\times \chi_{heptane}\\\\\chi_{heptane}=0.506

Hence, the mole fraction of heptane at 80°C is 0.506

ss7ja [257]3 years ago
4 0
Looking at this equation P= (pa*pb)/ (pa+(pb-pa)) ya  where pa=vap press a and ya= vap composition a and P= total pressure,it relates vapor pressure mixture to vapor composition. This is derived using the combination of Dalton's and Raoult's laws.
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Metals don't form covalent bonds because of the low ionization energes of the metal atoms. It is easier for them to release electrons rather than sharing it. But this is not always the case, there are some metals that can form covalent bonds.
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Good evening! Does anyone know this? Earth Science
riadik2000 [5.3K]

The fraction of Earth's radius (6371 km) relative to the thickness of the oceanic (7.5 km) and continental crust (35 km) is 0.12 and 0.55, respectively.    

What we know:

  • The average radius of Earth (E) = 6371 km
  • The average thickness of oceanic crust (O) = 7.5 km
  • The average thickness of continental crust (C) = 35 km

We need to convert all the above units from kilometers to miles:

E = \frac{0.6214 mi}{1 km}*6371 km = 3958.9 mi

O = \frac{0.6214 mi}{1 km}*7.5 km = 4.7 mi

C = \frac{0.6214 mi}{1 km}*35 km = 21.7 mi

Now, we can calculate the fraction of Earth's radius relative to each type of crust, with the given equation:

X = \frac{avg. \: thickness}{avg. \: radius} \times 100

  • <u>For the oceanic crust (O)</u>:

X = \frac{4.7 mi}{3958.9 mi}\times 100 = 0.12

  • <u>For the continental crust (C)</u>:

X = \frac{21.7 mi}{3958.9 mi}\times 100 = 0.55

Therefore, the fraction of Earth's radius relative to the oceanic and continental crust is 0.12 and 0.55, respectively.

You can see another example of calculation of fractions of Earth's radius here: brainly.com/question/4675868?referrer=searchResults

I hope it helps you!

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2 years ago
What changes could be made to optimize the manufacturing process
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3 years ago
When the equation Sn + HNO₃ → SnO₂ + NO₂ + H₂O is balanced in acidic solution, what is the smallest whole-number coefficient for
balu736 [363]

Answer:

1.

Explanation:

Hello,

In this case, for the given reaction we first assign the oxidation state for each species:

Sn^0 + H^+N^{5+}O^{-2}_3 \rightarrow Sn^{4+}O_2 + N^{4+}O^{2-}_2 + H^+_2O^-

Whereas the half reactions are:

Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O

Next, we exchange the transferred electrons:

1\times(Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-)\\\\4\times (H^++H^+N^{5+}O^{-2}_3 +1e^-\rightarrow  N^{4+}O^{2-}_2+H_2O)\\\\\\Sn^0+2H_2O \rightarrow Sn^{4+}O_2 +4H^++4e^-\\\\4H^++4H^+N^{5+}O^{-2}_3 +4e^-\rightarrow  4N^{4+}O^{2-}_2+4H_2O

Afterwards, we add them to obtain:

Sn^0+2H_2O+4H^++4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4H^++4N^{4+}O^{2-}_2+4H_2O

By adding and subtracting common terms we obtain:

Sn^0+4H^+N^{5+}O^{-2}_3  \rightarrow Sn^{4+}O_2 +4N^{4+}O^{2-}_2+2H_2O

Finally, by removing the oxidation states we have:

Sn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O

Therefore, the smallest whole-number coefficient for Sn is 1.

Regards.

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A. o. I know gamma can I Hope this helps


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