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:
b) 5.87 E23 molecules
Explanation:
∴ mm SO3 = 80.066 g/mol
⇒ molecules SO3 = (78.0 g)(mol/80.066 g)(6.022 E23 molec/mol)
⇒ molec SO3 = 5.866 E23 molecules SO3
Answer:
a) T
b) T
c) F
d) F
e) T
f) T
g) T
h) F
I) F
j) F
k) F
l) F
Explanation:
The w/v concentration is obtained from, mass/volume. Hence;
%w/v= 50/1000= 5%
In the %w/w we have;
25g/100 g = 25% w/w
In combustion reaction, energy is given out hence it is exothermic.
Neutralization reaction yields a salt and water
% by mass of carbon is obtained from;
8× 12/114 × 100 = 84.1%
All the ionic substances mentioned have very low solubility in water.
One mole of a substance contains the Avogadro's number of each atom in the compound.
There are two iron atoms so one mole contains 2× 55.85 g of iron.
Some sulphates such as BaSO4 are insoluble in water.
Halides are soluble in water hence NaI is soluble in water.
The equation does not balance with the given coefficients because the number of atoms of each element on both sides differ.
The equation represents a decomposition of calcium carbonate as written.
HEY mate here your answer
CoBr2 . 6 h2o = Cobalt(II) Bromide Hexahydrate
