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2Na + Cl2 → 2NaCl
46 g. ➡️ 117 g
x. ➡️ 29.3 g
46 g. ➡️ 117 g
x. ➡️ 29.3 g
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.
An element is a substance that can't be broken down further. Lithium (Li) and calcium (Ca) are examples of elements.
Carbon monoxide (CO) and potassium bromide (KBr) are examples of compounds, which are made up of more than one element put together.