Answer:
Pentan_1,5_di-al
Explanation:
OHC-CH₂-CH₂-CH₂-CHO
This is Pentan_1,5_di-al
If we break this compound, we will observe that there is presence aldehyde group and hence the functional group "al". This aldehyde is bonded to carbon 1 and carbon 5 respectively.
Also the pentan is due to presence of 5 carbon atoms.
Therefore, the IUPAC name of this compound (OHC-CH₂-CH₂-CH₂-CHO) is Pentan_1,5_di-al
Answer:
Check the electronic configuration of elements.
Explanation:
▪Valence electrons are the elwctrons present in the outermost shell of any element.
For example,
Electronic Configuration of Sodium = 2,8,1
Here , Sodium has 1 valence electrons.
▪Valency of an element is the total no. of electrons to be gained/losed in order to achieve duplet/octate state.
For example,
Electronic configuration of Sodium = 2,8,1
Sodium can achieve octate state either by losing 1 electron or gaining 7 electrons. But losing 1 electron is eay than gaining 7 electrons. So Valency of Sodium = +1
☆Metals have 1 or 2 or 3 valence electrons.
☆Non metals have 4 or 5 or 6 or 7 valence electrons.
☆Noble gases tend to stay in duplet/octate state i.e they have 2 or 8 valence electrons.
Explanation:
The given data is as follows.
= 100 mm Hg or 
= 0.13157 atm
= 
= (1080 + 273) K = 1357 K
= 
= (1220 + 273) K = 1493 K
= 600 mm Hg or 
= 0.7895 atm
R = 8.314 J/K mol
According to Clasius-Clapeyron equation,
![log(\frac{0.7895}{0.13157}) = \frac{\Delta H_{vap}}{2.303 \times 8.314 J/mol K}[\frac{1}{1357 K} - \frac{1}{1493 K}]](https://tex.z-dn.net/?f=log%28%5Cfrac%7B0.7895%7D%7B0.13157%7D%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B2.303%20%5Ctimes%208.314%20J%2Fmol%20K%7D%5B%5Cfrac%7B1%7D%7B1357%20K%7D%20-%20%5Cfrac%7B1%7D%7B1493%20K%7D%5D)
![log (6) = \frac{\Delta H_{vap}}{19.147}[\frac{(1493 - 1357) K}{1493 K \times 1357 K}]](https://tex.z-dn.net/?f=log%20%286%29%20%3D%20%5Cfrac%7B%5CDelta%20H_%7Bvap%7D%7D%7B19.147%7D%5B%5Cfrac%7B%281493%20-%201357%29%20K%7D%7B1493%20K%20%5Ctimes%201357%20K%7D%5D)
0.77815 = 
= 
J/mol
= 
= 221.9 kJ/mol
Thus, we can conclude that molar heat of vaporization of substance X is 221.9 kJ/mol.
(Can I have Brainlist)?
Answer:
2C4H10 + 13O2 —> 8CO2 + 10H2O. Oxidation reaction
8 (4 moles CO2 per mole butane)
Explanation:
could be written C4H10 + 6 1/2 O2 —> 4CO2 + 5H2O
