Answer:
On the basis of your knowledge of the reaction of halogens with alkanes, decide which product you would not expect to be formed in even small quantities in the bromination of ethane?
A) BrCH2CH2Br
B) CH3CH2CH2Br
C) CH3CHBr2
D) CH3CH2CH2CH3
E) BrCH2CH2CH2CH2Br
Explanation:
The reaction of ethane with bromine in presence of UV light forms mono substituted ethane at all primary and secondary carbons.
This is an example of free radical substitution.
The structure of ethane and its bromination is shown below:
Among the given options that which is not possible to form is option B) that is CH3CH2CH2Br(propyl bromide).
Remaining all other products are possisble to form on free radical substitution of ethane.
The greatest concentration of atomic mass is the atom's nucleus. This is because the nucleus is made up out of protons and neutrons while the electrons surrounding the nucleus have a very small mass.
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
10) Arsenic is a chemical agent which was used in preserving dead bodies in the past. It is mainly a formaldehyde mixture with coloring agents to give a dead body the look of life. Sylvester was a third mummy that was embalmed with arsenic in the late 1800s and is now on exhibit at Ye Olde Curiosity Shop in Seattle, Washington. His mummified body showed an extremely unusual since the remains weighed approximately 80 lbs when a body is composed of 70-80% water, after dehydration, the weight should be about 20-30% the premortem weight. The solution was also used during Civil War to preserve dead bodies <span>as the dead were being shipped home from the battlefield.
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9) R<span>esearch can yield dates when certain plants/animals became domesticated and entered the standard diets of people, meaning you would not see masses eating beef before the cow became domesticated. To further this you can look at when a specific food was introduced into the diets of people in the geographic area of the person you are studying. </span>
Okay, to do this you have to work with the relative molecular mass (RMM). You can get this from looking at the periodic table.
The RMM for the whole molecule is:
58.933 + (2 x 35.453) + (12 x 1.008) + (6 x 15.9994) = 237.96
Then you work out the RMM for water:
(2 x 1.008) + 15.9994 = 18.0154
As there are 6 moles of water in this molecule then multiply the RMM of H2O by 6 = 108.0924
Finally, divide the total H2O RMM by the total molecule RMM and multiply by 100 to get a percentage:
(108.0925 / 237.96) x 100 = 45.42%
