Aflexible plastic container contains 0.860g of helium has in a volume of 19.2l if 0.205g of helium is removed at contact pressure and temperature what will be the new volume

Given:

Initial volume of He, V1 = 19.2 L

Initial mass of He, m1 = 0.0860 g

Mass of He removed = 0.205 g

To determine:

The new volume of He i.e V2

Explanation:

Based on Avogadro's law:

Volume of a gas is directly proportional to the # moles of the gas

Volume (V) α moles (n) (1)

Atomic mass of He = 4 g/mol

Initial moles of He, n1 = 0.860 g/4 g.mol-1 = 0.215 moles

Final moles of He, n2 = (0.860-0.205)g/4 g.mol-1 = 0.164 moles

Based on eq(1) we have:

V1/V2 = n1/n2

V2 = V1 n2/n1 = 19.2 L * 0.164 moles/0,215 moles = 14.6 L

Ans: New volume is 14.6 L

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answer : one can design an experiment using a liquid and an additive that can be added to the liquid and observe the chemical changes in the reaction species.

one would need a beaker which is filled with a known amount of the liquid in which thermometer is placed to measure the temperature changes; after addition of the additive to the liquid solution one can note down the temperature change and find whether the reaction is endothermic or exothermic in nature. by subtracting the initial temperature value with the final temperature one can find the temperature changes.

various combinations of liquid and additive can be used. liquid in this case can be a known quantity of water and additives can be epsom salt or without epsom salt the experiment can be conducted to notice the changes. water and ice, hydrogen peroxide and dry yeast, vinegar and baking soda,etc.