Fe 3+ + SCN- --> FeSCN 2+
<span>.......Fe 3+ .......SCN-.........FeSCN 2+ </span>
<span>I.......0.04..........0.001.............. </span>
<span>C........-x...............-x............. </span>
<span>E.....0.04-x.....0.001-x...........x </span>
<span>Keq = 203.4 = x / (0.04-x)(0.001-x) </span>
<span>203.4 = x / (x^2 - 0.041x + 4x10^-5) </span>
<span>203.4x^2 - 8.34x + 0.00094 = x </span>
<span>203.4x^2 - 9.34x + 0.00094 = 0 </span>
<span>x = -0.0001M or 0.0458M </span>
<span>so, using your Keq, there would be no SCN- or Fe 3+ left.....all would be in the form of FeSCN 2+</span>
The correct answer is Shale
Answer:
10 molecules of NH₃.
Explanation:
N₂ + 3H₂ --> 2NH₃
As the N₂ supply is unlimited, what we need to do to solve this problem is <u>convert molecules of H₂ into molecules of NH₃</u>. To do so we use the <em>stoichiometric coefficients</em> of the balanced reaction:
- 15 molecules H₂ *
= 10 molecules NH₃
10 NH₃ molecules could be prepared from 15 molecules of H₂ and unlimited N₂.
Answer:
2 half-lives=0.8
6 half-lives= 0.05
Explanation
Half-lives are constant and always decrease by half, implying that the concentration decreases by half at a consistent rate.
3.2/2= 1.6/2= 0.8 is two half-lives
3.2/2= 1.6/2= 0.8/2= 0.4/2= 0.2/2= 0.1/2=0.05 is six half-lives
