You have given no demonstration based on your microscopic investigation so I cant tell you the answer to the question. I will try to help you by elaborating how to decipher..
Three terms hypotonic, hypertonic and isotonic are used when referring to two solutions separated by a selectively permeable membrane.
The hypertonic solution has a great concentration of OAS than the solution on the other side of the membrane. It is described, therefore, as having a great osmolarity. The hypotonic solution has a lower concentration of OAS, or osmolarity, than the solution on the other side of the membrane. When the two solutions are at an equilibrium, the concentration of OAS being equal on both sides of the membrane, the osmolarities are equal and are said to be isotonic.
The net flow of water is from the hypotonic to the hypertonic solution. When the solutions are isotonic, there is no net flow of water across the membrane.
If red blood cells are placed in a solution with a lower solute concentration than is found in the cells, water moves into the cells by osmosis, causing the cells to swell; such a solution is hypotonic to the cells.
So, look at the information and data you have on your microscopic investigation and use these guidelines to tell you which is which.
1.) D.
2.) B.
3.) D.
4.) A.
5.) B
Oops I accidentally hit the two stars :( these are right
Water is know to be biotic
There are no various sorts of bacterial provinces in Anna's example. Just Serratia marcescens was discovered in light of the fact that the microorganisms in the specimen all had a similar shape and size. Microbiologists need to detach bacterial provinces from an example to have the capacity to take a gander at what sorts of microscopic organisms are in the example.
Answer:
0.05 mg/mL ( B )
Explanation:
Given data:
20 mg/ml starch
2% solution = 2g of solute is in 100g of solvent
<u>Determine the new concentration in mg/ml </u>
Dilution equation = C1V1 = C2V2
new concentration ; applying the dilution factor
dilution factor = 1 : 400 ; ( 2 /400 )g = 0.005 g of solute is present in every 100 mL
∴ new concentration = 0.00005 g / 1 mL * ( 1000 mg / 1g ) = 0.05 mg/mL
