The molecular structure of 1-nitrobutane is 
. The structure of 1-nitrobutane is shown below.
An atom's formal charge would be determined by the covalent model of chemical bonding, which assumes that almost all chemical bonds include equal sharing of electrons among all atoms, regardless their relative electronegativity.
The structure for 1-nitrobutane, making sure to add all non-zero formal charges
There are four kind of molecule present in 1-nitrobutane and they are carbon, hydrogen , nitrogen and oxygen. Nitrogen is bonded with two oxygen atom out of them one oxygen atom is attached with single bond and second oxygen atom is bonded with double bond. Nitrogen has positive charge whereas oxygen has negative charge.
It is a kind of alkane in with nitro group is attached with alkane group.
To know more about 1-nitrobutane
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Answer:
It is difficult, if not impossible, to heat a solid above its melting point because the heat that ... in a solid are packed in a regular structure that is characteristic of that particular substance.
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Answer:
H2O> NH3> CH4
Explanation:
According to valence shell electron pair repulsion theory (VSEPR), bond angles and repulsion of electron pairs depends on the nature of electron pairs on the central atom of the molecule. Lone pairs cause more repulsion (and distortion of bond angles) than bond pairs). Lone pair- lone pair repulsion is greater than lone pair bond pair repulsion.
Water contains two lone pairs on oxygen hence it experiences the greatest repulsion. Ammonia has only one lone pair on nitrogen hence there is lesser repulsion between lone pairs and bond pairs. Methane possess only bond pairs of electrons hence it has the least repulsion.
According to the PH formula:
PH= Pka +㏒ [strong base/weak acid]
when we have PH at the first equivalence =3.35 and the Pka1 = 1.4
So, by substitution, we can get the value of ㏒[strong base / weak acid]
3.35 = 1.4 + ㏒[strong base/ weak acid]
∴㏒[strong base/weak acid] = 3.35-1.4 = 1.95
to get the Pka2 we will substitute with the value of ㏒[strong base/ weak acid] and the value of PH of the second equivalence point
∴Pk2 = PH2 - ㏒[strong base/ weak acid]
= 7.55 - 1.95 = 5.6
The standard atomic weight of a C is 12, and the standard atomic weight of a H is 1. So to find molar ratio of C and H in the compound: 60.0/12=5, 5.05/1=5. This means the molar ratio of C and H is 5:5, thus 1:1. Assuming the molecular formula is CnHn, to find molar mass: 12n + 1n = 78.12. n=78.12/(12+1) = 6. So the compound's molecular formula is C6H6, benzene.
