Diphosphorus tetraiodide is a covalent compound.
It has low melting point as compared to ionic compounds
It is a rare compound where the oxidation state of Phosphorous is +2.
It is also termed as subhalide of phosphorous.
Answer:
4Al203 + 9Fe -> 3Fe3O4 + 8Al
Amount of Al on reactant side: 8
Amount of Al on product side: 8
Amount of O on reactant side: 12
Amount of O on product side: 12
Amount of Fe on reactant side: 9
Amount of Fe on product side: 9
Answer:
Explanation:
a) for 1.000 g X: 0.1621 g Y
ratio of mass of element Y = 2.100g : 0.1621g
= 1 : 0.07
b) 1.000 g X: 0.7391 g Y
ratio of mass of element Y = 2.100g : 0.7391g
= 1: 0.35 , = 20:7
c) 1.000 g X: 0.2579 g Y
ratio of mass of element Y = 2.100g : 0.2579g
= 1 : 0.12
d) 1.000 g X: 0.2376 g Y
ratio of mass of element Y = 2.100g : 0.2376g
= 1: 0.11
e) 1.000 g X: 0.2733 g Y
ratio of mass of element Y = 2.100g : 0.2733g
= 1 : 0.13
From the values obtained , the closest that is in compliance with the law f multiple proportions is option B
<h2><u>Full Question:</u></h2>
In hemoglobin, a single amino acid change at position 6 from Glu to Val has major consequences on hemoglobin structure that makes the molecule defective leading to sickle cell anemia. Predict whether the following hypothetical change would or would not have a major effect at position 6. Briefly explain (1-2 sentences). Glu to Leu Hint: Look at the structures of the R groups and consider their chemical properties
<h2><u>Answer:</u></h2>
The structure of the haemoglobin, hence the RBC won't be same as normal.
<h3><u>Explanation:</u></h3>
Both the leucine and glutamic acid are alpha amino acids which have an alpha carboxylic acid group and an alpha amino group. The variable in case of glutamic acid is propyl acid while the variable in case of leucine is isobutyl.
The glutamic acid is the normal amino acid of the 6th position of Beta chain of hemoglobin. Its an acid group, so can form bonds with another base inside the haemoglobin, or can form other hydrogen bonds. But the isobutyl group is an alkyl group. So it doesn't have that much effect in the recovering the structure, and sickle cell anemia prevails.
Answer:
Initially the function is symmetric with respect to the axis of the one dimensional box. In the final state it is also symmetrical, however you can envision a snapshot of the system as the light field is interacting with the wave-function wherein a node begins to develop as is shown in the middle and the wave function is evolving from the initial to final state. Now consider that the electron density during process is the square of the wave function:
Electron density during transition
As can be seen in the initial and final states the electron density is symmetrically distributed with respect to the axis of the box. However with the field on, the electron density is not symmetrically distributed and a transitory dipole moment can be present. To relate back to real molecules think of each of those orbitals as a linear combination of atomic orbitals. One important factor is the symmetry. But there may be one other factor that will be just as important as symmetry. If you treat orbital 1 as a linear combination over n orbitals and orbital 2 as a linear combinations of orbitals as well, there will be a spatial over lap between the orbital in the ground state and the orbital in the excited state. If there is no spatial overlap between the ground state and excited state orbitals there will be no transition dipole moment. However, if the electrons are in the same place spatially, a large transition dipole moment will result.
Explanation:
