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>
Answer: B) 2 (as indicated by electron distribution shown), but taking into account the real properties of this element, 4,7,8 also occur (see below).
Explanation:
This is the electron complement/atomic number of ruthenium, which actually has the structure [Kr] 5s1 4d7
Nevertheless, Ru does not form Ru(I) compounds and few Ru(II) compounds (RuCl2, RuBr2, RuI2). It also forms Ru(III)Cl3 and a larger number of Ru(IV) compounds, e.g. RuO2, RuS2. It also forms RuO4
With the info given i would have to say their is no kinetic energy, it's all potential energy.
Empirical formula is the simplest ratio of whole numbers of components in a compound.
Assuming for 100 g of the compound
Cu As S
mass 48.41 g 19.02 g 32.57 g
number of moles 48.41 / 63.5 g/mol 19.02 / 75 g/mol 32.57 / 32 g/mol
= 0.762 mol = 0.2536 mol = 1.018 mol
divide by the least number of moles
0.762 / 0.2536 0.2536 / 0.2536 1.018 / 0.2536
= 3.00 = 1.00 = 4.01
once they are rounded off
Cu - 3
As - 1
S - 4
therefore empirical formula is Cu₃AsS₄
