Two optical isomers can form:
An equimolar mixture of two optical isomers is called a racemic mixture.
What is a Racemic mixture?
An equimolar mixture of two enantiomers that is optically inactive is known as a racemic mixture (or racemate) (i.e. does not rotate plane-polarized light).
The racemic mixture can be created by:
- Combining enantiomers in equal amounts, or (more frequently)
- Reactions that create one or more new chiral centers without the influence of chirality (i.e. chiral reagent, catalyst, etc.)
The light that is plane-polarized does not spin in an optically inactive mixture.
Since each enantiomer rotates plane-polarized light to an equal and opposite extent, an equimolar mixture of two enantiomers will typically result in a racemic mixture. There won't be any net rotation if they are both present in equal proportions.
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#1
<span>The rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.
#2
shape, molecular weight. I don't know for sure though
#3
</span>O2, N2, Ar, H2O vapor
<h2>Plasma is most likely inflence by - Option C</h2>
This plasma is most likely influenced by magnetic and electric fields. The massive positively charged ions which are influenced the lighter electrons than replace the ions to keep charge impartiality.
Therefore electric field influence the plasma. The movement of charged particles identical to a magnetic field line is not influenced. Plasma contains charged particles which are positive nuclei and negative electrons which can be molded and restricted by magnetic forces.
Alike iron filings in the closeness of a magnet, bits in the plasma will ensue magnetic field lines.
Answer:
What make Boron Boron is that it has 5 protons, and will therefor have 5 electrons in the unionized state. While I was looking this up I learned that there are two stable isotopes, Boron 10 with five neutrons, and boron 11 with six. The more common is boron 11, which is 80.1% of naturally occuring boron.
To answer this item, we must take note that the ligand that binds the tightest is the one with the lowest dissociation constant, Kd. Kd's for both A and B are already given so, we only need to solve Kds for C and D.
Kd of C
0.3 = (1x10⁻⁶)/(1x10⁻⁶ + Kd) ; Kd = 2.3x10⁻⁶
Kd of D
0.8 = (1x10⁻⁹)/(1x10⁻⁹ + Kd) ; Kd = 2.5x10⁻10
Since Ligand D has the least value of dissociation constant then, it can be concluded that it binds the tightest.
