Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
= 
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.
Answer:
Bohr model A
Explanation:
It has more valance electrons therefore has more interaction between the atoms and has more electronegativity.
Answer:
Explanation:
From the given information:
TO start with the molarity of the solution:
= 0.601 mol/kg
= 0.601 m
At the freezing point, the depression of the solution is 
Using the depression in freezing point, the molar depression constant of the solvent 
The freezing point of the solution 
The molality of the solution is:
Molar depression constant of solvent X, 
Hence, using the elevation in boiling point;
the Vant'Hoff factor 
the melting point of water is 32 degrees Fahrenheit , 0 degrees Celsius. <span />
The purpose of the uninoculated control tubes used in this test is that two uninoculated control tubes are needed to show the results of the medium in both aerobic and anaerobic environments. It is used to show it is sterile and also as a color comparison, used also to show that the medium remains green under both conditions.
