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Xelga [282]
2 years ago
13

What do you think happens to Difluoroethane at –24°C? Provide evidence to support your claim.

Chemistry
1 answer:
balandron [24]2 years ago
6 0

Answer:

The following subsections explain the explanation according to the particular circumstance.

Explanation:

  • The boiling point seems to be the temperature beyond which the working fluid as well as the boiling phase would be at a predetermined pressure or voltage at equilibrium among one another and.  
  • The vapor or boiling temperature of 1,1 difluoroethane seems to be -25oC at 1 atm, although as a gas it can remain at a higher temperature around -24oC.
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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma
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To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

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Here, R is radius and is related to a as follows:

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A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}

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V=Area\times c

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n=\frac{\rho N_{A}V_{c}}{A}

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n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18

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Converting them into cm:

1 pm=10^{-10}cm

Thus,

r_{Cr^{3+}}=6.2\times 10^{-9}cm

and,

r_{O^{2-}}=1.4\times 10^{-8}cm

Volume of sphere will be sum of volume of total number of cations and anions thus,

V_{S}=V_{Cr^{3+}}+V_{O^{2-}}

Since, volume of sphere is V=\frac{4}{3}\pi r^{3},

V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )

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V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}

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