Answer:
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Answer:
Anhydrous sodium carbonate is stable to heat and does not decompose even when it is heated to redness. This is because sodium carbonate salt on heating with acids react to release carbon dioxide.
Answer:
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Explanation:
As given
60 min = 50 gm (1)
then we know half-life mean half amount decay time
so we can write as the half of 200 is 100 gm hence
T 1/2 = 100 (2)
solving these two equation by cross multiplication we will get
T 1/2 = 120 min
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