Answer: remains constant.
Justification:
1) The phase changes are:
i) Boiling: pass from liquid to gas (absorbs heat energy)
ii) Condensation: pass from gas to liquid (release heat energy)
iii) Melting: pass from solid to liquid (absorb heat energy)
iv) Freezing: pass from liquid to solid (release heat energy)
v) Sublimation: pass from solid to gas (absorbs heat energy)
vii) Deposition: pass from gas to solid (release heat energy)
2) When a phase change occurs, whichever it is, the heat energy related with the process, either absortion or release, is used, to overcome the intermolecular forces (in the case of heat energy absortion) or to create stronger intermolecular forces (in the case of heat energy release).
Because of that, the heat energy exchange does not change the temperature of the substance.
20.7 Degree celsius is the temperature of Zippy's room.
Explanation:
The initial conditions of the kookie cola will be depicted by P1,V1 and T1.
P1= 36.5 psi or 2.4816 atm
V1= 355ml or 0.35 L
T1= 275.5K
The final conditions are given P2,V2 and T2
P2= 38.9 psi or 2.6469 atm
V2 = 355 ml or 0.35 L
also the solubility of the carbon dioxide does not change with temperature.
Applying the formula
P1VI/T1=P2V2/T2
since volume does not change and remains constant.
It can be written as:
P1/T1=P2/T2
= 2.4816/275.5=2.6469/T2
T2= 293.85 K
So you have a balloon rising through the atmosphere. use the formula p1/v1=p2/v2 and add the variables into the equation, leaving 295/52.5=252/x. multiply 252 by 52.5 and divide that number by 295.
52.5*252=13230. divide by 295 =44.9 L
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
