Answer:
D. pUC-chloramphenicol(minus)
Explanation:
It contains chloramphenicol resistance gen, the PMB1 posses origin of replication (ori), beta-galactosidase coding gen Laz. It also has pUC18 with many cloning site in the Lac Z gene which makes the recombinant clones to be verified via culture plates which is made up of IPTG and X- Gal.
Answer:
We can Interprete it as 1mole of Sodium Chloride and 1mole of Silver Nitrate React to Produce
1Mole of Silver Chloride and 1Mole of Sodium Nitrate
Answer:
8.33mol/L
Explanation:
First, let us calculate the molar mass of of formaldehyde (CH2O). This is illustrated below:
Molar Mass of CH2O = 12 + (2x1) + 16 = 12 + 2 + 16 = 30g/mol
Mass of CH2O from the question = 0.25g
Number of mole CH2O =?
Number of mole = Mass /Molar Mass
Number of mole of CH2O = 0.25/30 = 8.33x10^-3mole
Now we can calculate the molarity of formaldehyde (CH2O) as follow:
Number of mole of CH2O = 8.33x10^-3mole
Volume = 1mL
Converting 1mL to L, we have:
1000mL = 1L
Therefore 1mL = 1/1000 = 1x10^-3L
Molarity =?
Molarity = mole /Volume
Molarity = 8.33x10^-3mole/1x10^-3L
Molarity = 8.33mol/L
Therefore, the molarity of formaldehyde (CH2O) is 8.33mol/L
the compounds in which phosphorous posses the highest possible oxidation have to mention here.
The species in which phosphorous have the highest oxidation state are: H₃PO₄, P₂O₅, PCl₅
The possible oxidation state of phosphorous is III and V. The highest oxidation state is V. There are several compounds in which phosphorous posses the +5 oxidation state. Like- Phosphoric acid (H₃PO₄), phosphorous pentoxide (P₂O₅), Phosphorous chloride (PCl₅) etc.
The oxidation state of an element depends upon the valence electron the valence shell of phosphorous is 3s² 3p³. Thus there are 5 electrons, as it has vacant 3d orbital thus it can easily form compound having +5 oxidation state.
B. White Dwarf.
<h3>Explanation</h3>
The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.
As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:
Chandrasekhar Limit: 
.
A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 
, which is much smaller than the Chandrasekhar Limit.
As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.
