no. but, .1 is the same as 10 percent. .1 "percent" is different than just .1
1 percent of a percent is 1/10000 or 0.0001; this is the same as saying 1 percent percent. Example: 5 percent percent is 0.0005, so 5 percent percent of 100 is 0.0005 x 100 or 0.05.
1/100 as a percent = 1%
NO. 1 percent is less than 5 percent.
1 percent of 3200 is 32.
The glucose was able to go through the sac. The glucose went from high concentration to low concentration. The glucose is permeable.
1:10 ratio.
The three types of hemolysis bacteria can exhibit are alpha-hemolysis, beta-hemolysis, and gamma-hemolysis. Alpha-hemolysis causes partial destruction of red blood cells, resulting in a greenish discoloration around the bacterial colonies. Beta-hemolysis causes complete lysis of red blood cells, leading to a clear zone around the bacterial colonies. Gamma-hemolysis is when there is no hemolysis of red blood cells.
Another way to express the concentration of a glucose solution that is 0.01 percent by weight is as 100 parts per million (ppm). This means there are 100 grams of glucose in 1 million grams of solution.
Another way to express the concentration of a 0.01 percent by weight glucose solution is to say it is a 100 parts per million (ppm) solution. This means there are 100 grams of glucose in every 1 million grams of solution.
To prepare a 10% glucose solution, you would mix 10 grams of glucose with enough water to make a total solution volume of 100 ml. This means the final solution would contain 10 grams of glucose and 90 ml of water.
1% (presumably m/v) glucose solution would contain 1 g of glucose per 100 ml of solution. Therefore the conversion of 1 g / 100 ml to units of mol/L requires that we divide by the molar mass and multiply by the conversion factor of ml to L. Therefore: (1 g / 100 ml) * (1 mol / 180.16 g) * (1000 ml / 1 L) = 0.0555 M = 0.06 M.
If you added more glucose that the Kliger Iron Agar test than is called for you risk getting false results. The bacteria need to exhaust the glucose to turn the solution red or yellow. If too much glucose is present you may not be able to determine if the bacteria can ferment animal proteins.
The cleavage of glycogen phosphorylase releases glucose-1-phosphate by breaking the glycosidic bond within glycogen. This glucose-1-phosphate can then be further processed to yield free glucose for energy production.
In this scenario, the glucose in sac 1 will diffuse out of the sac into the distilled water due to the concentration gradient. However, since distilled water is hypotonic compared to the 40% glucose solution, water molecules will also move into the sac via osmosis to try to balance the concentration inside and outside the sac. This will cause the sac to swell as water moves in, reaching an equilibrium point where the movement of glucose and water is balanced.
In glycolysis, one glucose molecule produces a net yield of two ATP molecules at the end of the process.
To make a 1 molar solution of glucose (C6H12O6) in 1 liter of water, you would need to dissolve 180 grams of glucose. This is because the molar mass of glucose is 180 g/mol, so 1 mole of glucose weighs 180 grams.