Answer:
individual isotopes
Re 185 = 139.094
Re 187 = 232.815
Explanation:Correcting the typo error for natural abundance as 37 40 as 37.40 and 62 60 as 62.60
Answer is found as follows
For Re 185 natural abundance 37.40%
371.9087 x 37.40/100 = 139.0938538
six significant figures after rounding off = 139.094
For Re 187 natural abundance 62.60
371.9087 x 62.60/100 = 232.8148462
six significant figures after rounding off = 232.815
Answer:
The net charge on each lysine molecule would be -1.
Explanation:
- <u>When the pH is above 2.2</u> the deprotonated form of the carboxylic acid is more present, while the amino group and side chain (which is also amino) remain protonated (with a positive charge):
R-COOH ↔ R-COO⁻
R-NH₃⁺
R'-NH₃⁺
Net charge = +1
- <u>When pH is above 9.0</u>, the carboxyl group remains deprotonated, while the amino group is deprotonated and the side chain is protonated:
R-COOH ↔ R-COO⁻
R-NH₂
R'-NH₃⁺
Net charge = 0
- <u>When pH is above 10.5</u>, the carboxyl group remains deprotonated, while both the amino group and the side chain are deprotonated:
R-COOH ↔ R-COO⁻
R-NH₂
R'-NH₂
Net charge = -1
So at pH=13 (which is above 10.5) the net charge is -1.
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
Answer:
all are forms of asexual reproduction
T K = ºC + 273
T = 18 + 273
T = 291 K
hope this helps!
