Answer:
Its molecules are made up of 60 carbon atoms joined together by strong covalent bonds. Molecules of C 60 are spherical. There are weak intermolecular forces between molecules of buckminsterfullerene. These need little energy to overcome, so buckminsterfullerene is slippery and has a low melting point.
Explanation:
The question is incomplete, here is the complete question:
The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?
<u>Answer:</u> The rate constant at 324°C is 
<u>Explanation:</u>
To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:
![\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7BK_%7B244%5EoC%7D%7D%29%3D%5Cfrac%7BE_a%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%5Cfrac%7B1%7D%7BT_2%7D%5D)
where,
= equilibrium constant at 244°C = 
= equilibrium constant at 324°C = ?
= Activation energy = 71.0 kJ/mol = 71000 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
= initial temperature = ![244^oC=[273+244]K=517K](https://tex.z-dn.net/?f=244%5EoC%3D%5B273%2B244%5DK%3D517K)
= final temperature = ![324^oC=[273+324]K=597K](https://tex.z-dn.net/?f=324%5EoC%3D%5B273%2B324%5DK%3D597K)
Putting values in above equation, we get:
![\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}](https://tex.z-dn.net/?f=%5Cln%28%5Cfrac%7BK_%7B324%5EoC%7D%7D%7B6.7%7D%29%3D%5Cfrac%7B71000J%7D%7B8.314J%2Fmol.K%7D%5B%5Cfrac%7B1%7D%7B517%7D-%5Cfrac%7B1%7D%7B597%7D%5D%5C%5C%5C%5CK_%7B324%5EoC%7D%3D61.29M%5E%7B-1%7Ds%5E%7B-1%7D)
Hence, the rate constant at 324°C is
Answer:
It has to have a problem base and a realistic explanation.
Explanation:
It needs to have enough information for you to be able to come up with an answer and realistic explanation.
Hope I helped :)
Answer:
A. 6N
B. 4H, 2O
C. 4H, 4N, 12O
D. 2Ca, 4O, 4H
E. 3Ba, 6Cl, 18O
F. 5Fe, 10N, 30O
G. 12Mg, 8P, 32O
H. 4N, 16H, 2S, 8O
I. 12Al, 18Se, 72O
J. 12C, 32H
I am 90% sure this is correct
Answer:
[Cr(NH3)6.]C13
Explanation:
Alfred Werner's coordination theory (1893) recognized two kinds of valency;
Primary valency which are nondirectional and secondary valency which are directional.
Hence, the number of counter ions precipitated from a complex depends on the primary valency of the central metal ion in the complex.
We must note that it is only these counter ions that occur outside the coordination sphere that can be precipitated by AgNO3.
If we consider the options carefully, only [Cr(NH3)6.]C13 possess counter ions outside the coordination sphere which can be precipitated when treated with aqueous AgNO3.
