Answer:
The answer to your question is: letter E
Explanation:
Normally, the correct order of boiling points is:
Alcohols > Ketones > Ether > Alkane
Then
A. n-butane < 1-butanol < diethyl ether < 2-butanone
B. n-butane < 2-butanone < diethyl ether < 1-butanol
C. 2-butanone < n-butane < diethyl ether < 1-butanol
D. n-butane < diethyl ether < 1-butanol < 2-butanone
E. n-butane < diethyl ether < 2-butanone < 1-butanol
(- 1°C) < 34.6°C < 79.64°C < 117.7°C
Answer:
canmann
Explanation:
because it is setting for that only
Answer:
it is a membrane-linked glycoproteins that connects with other cells which are also called transmembrane
Hope This Helped
Answer:
Explanation:
![\Delta H_{rxn}=\sum [n_{i}\times \Delta H_{f}^{0}(product)_{i}]-\sum [n_{j}\times \Delta H_{f}^{0}(reactant_{j})]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5Csum%20%5Bn_%7Bi%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28product%29_%7Bi%7D%5D-%5Csum%20%5Bn_%7Bj%7D%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28reactant_%7Bj%7D%29%5D)
Where 
and 
are number of moles of product and reactant respectively (equal to their stoichiometric coefficient).
is standard heat of formation.
So, ![\Delta H_{rxn}=[2mol\times \Delta H_{f}^{0}(CO_{2})_{g}]+[3mol\times \Delta H_{f}^{0}(H_{2}O)_{g}]-[1mol\times \Delta H_{f}^{0}(C_{2}H_{5}OH)_{l}]-[3mol\times \Delta H_{f}^{0}(O_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B2mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28CO_%7B2%7D%29_%7Bg%7D%5D%2B%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28H_%7B2%7DO%29_%7Bg%7D%5D-%5B1mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28C_%7B2%7DH_%7B5%7DOH%29_%7Bl%7D%5D-%5B3mol%5Ctimes%20%5CDelta%20H_%7Bf%7D%5E%7B0%7D%28O_%7B2%7D%29_%7Bg%7D%5D)
or, ![\Delta H_{rxn}=[2mol\times -393.509kJ/mol]+[3mol\times -241.818kJ/mol]-[1mol\times -277.69kJ/mol]-[3mol\times 0kJ/mol]](https://tex.z-dn.net/?f=%5CDelta%20H_%7Brxn%7D%3D%5B2mol%5Ctimes%20-393.509kJ%2Fmol%5D%2B%5B3mol%5Ctimes%20-241.818kJ%2Fmol%5D-%5B1mol%5Ctimes%20-277.69kJ%2Fmol%5D-%5B3mol%5Ctimes%200kJ%2Fmol%5D)
or,
