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3 years ago

Help me answer this please

Chemistry

1 answer:

Vlada [557]3 years ago

8 0

<span>1=H, 2=B, 3=F, 4=A,5=C,6=E, 7=D, 8=G
</span>9: 69Ga=60.12% and 71Ga=39.88%; total=69.797amu
10: 27 27.977 92.23; 28 28.976 4.67; 29 29.974 3.10; abundance =28.07 Silicon

I hope this helps!

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If the vapor pressure of pure benzene is 96.1 mm Hg, what must the vapor pressure of pure toluene be in order for the 50/50 % mi

mylen [45]

Answer:

$P_{tol}=30.34mmHg$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the vapor pressure of pure toluene by using the Raoult's law as shown below:

$y_{tol}P_{mix}=x_{tol}P_{tol}\\\\y_{ben}P_{mix}=x_{ben}P_{ben}$

Thus, we solve for the mole fraction of benzene in the vapor phase first:

$y_{ben}=\frac{x_{ben}P_{ben}}{P_{mix}} =\frac{0.5*96.1mmHg}{63.2mmHg}=0.76$

Which means that the mole fraction of toluene in the vapor phase is 0.24, and therefore, the vapor pressure of pure toluene turns out to be:

$P_{tol}=\frac{y_{tol}P_{mix}}{x_{tol}} =\frac{0.24*63.2mmHg}{0.5}=30.34mmHg$

Regards!

7 0

3 years ago

Which option correctly describes the pattern of valence electrons on the periodic table? (1 point)

uysha [10]

Answer:

The number of valence electrons increases from left to right.

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3 years ago

Calculate the change in entropy when 10.0 g of CO2 isothermally expands from a volume of 6.15 L to 11.5 L. Assume that the gas b

USPshnik [31]

Answer:

The change in entropy of the carbon dioxide is $1.183\times 10^{-3}$ kilojoules per Kelvin.

Explanation:

By assuming that carbon dioxide behaves ideally, the change in entropy ( $\Delta S$ ), measured in kilojoules per Kelvin, is defined by the following expression:

$\Delta S = m\cdot \bar c_{v}\cdot \ln \frac{T_{f}}{T_{o}}+m\cdot \frac{R_{u}}{M}\cdot \ln \frac{V_{f}}{V_{o}}$ (1)

Where:

$m$ - Mass of the gas, measured in kilograms.

$\bar c_{v}$ - Isochoric specific heat of the gas, measured in kilojoules per kilogram-Kelvin.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the gas, measured in Kelvin.

$V_{o}$ , $V_{f}$ - Initial and final volumes of the gas, measured in liters.

$R_{u}$ - Ideal gas constant, measured in kilopascal-cubic meter per kilomole-Kelvin.

$M$ - Molar mass, measured in kilograms per kilomole.

If we know that $T_{o} = T_{f}$ , $m = 0.010\,kg$ , $R_{u} = 8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$ , $M = 44.010\,\frac{kg}{kmol}$ , $V_{o} = 6.15\,L$ and $V_{f} = 11.5\,L$ , then the change in entropy of the carbon dioxide is:

$\Delta S = \left[\frac{ (0.010\,kg)\cdot \left(8.315\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)}{44.010\,\frac{kg}{kmol} } \right]\cdot \ln \left(\frac{11.5\,L}{6.15\,L}\right)$