Answer:
267.57 kPa
Explanation:
Ideal gas law:
PV = n RT R = 8.314462 L-kPa/K-mol
P (16.5) = 1.5 (8.314462)(354) P = 267.57 kPa
Rutherford theorized that atoms have their charge concentrated in a very small nucleus.
This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).
Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.
That is different from Plum Pudding model, because it shows that most of the atom is empty space.
According to Rutherford model of the atom:
1) Atoms have their charge concentrated in a very small nucleus.
2) Major space in an atom is empty.
3) Atoms nucleus is surrounded by negatively charged particles called electrons.
4) An atom is electrically neutral.
CH3CH2CH2CH3 < CH3CH2CHO < CH3CHOHCH3
Explanation:
Boiling point trend of Butane, Propan-1-ol and Propanal.
Butane is a member of the CnH2n+2 homologous series is an alkane. Alkanes have C-H and C-C bonds which have Van der waals dispersion forces which are temporary dipole-dipole forces (forces caused by the electron movement in a corner of the atom). This bond is weak but increases as the carbon chain/molecule increases.
In Propan-1-ol(Primaryalcohol), there is a hydrogen bond present in the -OH group. Hydrogen bond is caused by the attraction of hydrogen to a highly electronegative element like Cl-, O- etc. This bond is stronger than dispersion forces because of the relative energy required to break the hydrogen bond. Alcohols (CnH2n+1OH) also experience van der waals dispersion forces on its C-C chain and C-H so as the Carbon chain increases the boiling point increases in the homologous series.
Propanal which is an Aldehyde (Alkanal) with the general formula CnH2n+1CHO. This molecule has a C-O, C-C and C-H bonds only. If you notice, the Oxygen is not bonded to the Hydrogen so there is no hydrogen bond but the C-O bond has a permanent dipole-dipole force caused by the electronegativity of oxygen which is bonded to carbon. It also has van der waals dispersion forces caused by the C-C and C-H as the carbon chain increases down the homologous series. The permanent dipole-dipole forces are not as easy to break as van der waals forces.
In conclusion, the hydrogen bonds present in alcohols are stronger than the permanent dipole-dipole bonds in the aldehyde and the van der waals forces in alkanes (irrespective of the carbon chain in Butane). So Butane < Propanal < Propan-1-ol
