The half life of Radon-222 is 3.8 days. If you have a 90.0g sample of Radon-222 in the lab, how much will be left after 19 days?
1 answer:
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Answer:
Explanation:
<em>Ferrous Sulphate</em><em> is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide </em>
<em>, Sulphur Dioxide </em>
<em> as well as Ferric Oxide </em>
<em>.</em>
<em>We can hence, frame a skeletal equation of this reaction and try to balance it.</em>
<em>Hence,</em>
<em>Now,</em>
<em>a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of </em><em>steps</em><em>:</em>
<em>1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside </em>.
<em>2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:</em>
<em> Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :</em>
<em> </em>
<em>b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds </em><em>decompose </em><em>into smaller elements and compounds. Here, as Ferrous Sulphate </em><em>decomposes </em><em>into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is </em><em>Decomposition Reaction.</em>
