D = m/v
m = Dv = 10.5 * 100 = 1050 grams
1050/Ag's amu = 1050/47 = around 22.5
22.5 moles is your answer
Answer: capsule and fimbriae
Explanation:
Many prokaryotes have sticky outermost layer called the capsule.It is made up of polysaccharides and it helps them clinging to each other or to other surfaces around them.It also helps prevent the cell from drying out.
Fimbriae,also known as attachment pilus,used by bacteria to adhere to one and other,animal cells or to any inanimate objects.
Answer:
A and D takes much slower
Explanation:
Here, we want to select, out of the four given reactions, the one that is slower than the other two
The answers in these case are reactions 1 and 4 ( A and D)
The two reactions show what is called rust (as directly seen in reaction 4)
When we speak of rust, we simply mean a reaction that occurs over time
For example, non coated roofings of houses doesn’t get to change color at an instant
The color degradation that occurs is something that takes some time from the initial time they were used to roof the house
Hence, from these analogy, we can see that these reactions need an an external support to thrive or to come into existence
These external supports are natural forces and they contributing efforts occur over time and cannot be seen immediately
These reactions are thus ones that take much slower time than conventional laboratory reactions in the case of the formation of the precipitate or a reaction that requires a low flash point temperature such as that of black powder to produce such explosive effects
So in conclusion, what we are saying is that the two selected reactions are subjected to the availability of some conditions and may take time to manifest and these absolutely differentiates them from reactions that are spontaneous such as the one having an explosive effect or the other one leading to the formation of a precipitate which takes far less times
Answer:
% recovery
MP range of product
mass of product
Explanation:
Liquid–liquid extraction (LLE) is a process of transferring one (or more) solute(s) which are present in a feed solution to another immiscible liquid (solvent). The other solvent that becomes enriched in the target solute(s) is called extract. The original feed solution that is depleted in solute(s) is subsequently referred to as the raffinate.
This method is used to purify compounds and separate mixtures of compounds. This is very important when we want to isolate a product from a reaction mixture.
The percent recovery is the amount of solute that is transferred to the extract. This is the most important data to be recorded in an LLE experiment.
The melting point range necessarily helps us to identify the product and the mass of solid tells us the quantity of the solid obtained after extraction.
Answer:
18.01528
Explanation:
find this by adding the molecular masses of two hydrogen atoms and oxygen atom
