Answer:
Explanation:
By definition one <em>half-life</em> is the time to reduce the initial concentration to half.
For a <em>second order reaction </em>the rate law equations are:
![\dfrac{d[B]}{dt}=-k[B]^2](https://tex.z-dn.net/?f=%5Cdfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D-k%5BB%5D%5E2)
![\dfrac{1}{[B]}=\dfrac{1}{[B]_0}+kt](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B%5BB%5D%7D%3D%5Cdfrac%7B1%7D%7B%5BB%5D_0%7D%2Bkt)
The <em>half-life</em> equation is:
![t_{1/2}=\dfrac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cdfrac%7B1%7D%7Bk%5BA%5D_0%7D)
Thus, substitute the<em> rate constant</em> 
and the <em>half-life </em>time <em>224s</em> to find [A]₀:
![224s=\dfrac{1}{1.30\times10^{-3}M^{-1}\cdot s^{-1}[A]_0}](https://tex.z-dn.net/?f=224s%3D%5Cdfrac%7B1%7D%7B1.30%5Ctimes10%5E%7B-3%7DM%5E%7B-1%7D%5Ccdot%20s%5E%7B-1%7D%5BA%5D_0%7D)
![[A]_o=0.291M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.291M)
The Iron (III) ion is
The sulfate ion is
Thus when forming the compound iron (iii) sulfate you would need three sulfate ions bonded to two iron ions in order to fill the valency requirement
∴ the formula would be Fe₂(SO₄)₃
Whenever you are adding a number to a radical you must bracket it.
To answer this, we look at the polarities of these molecules given. A molecule is said to be polar when there is an unequal shraing of electrons the opposite is called nonpolar.
CH4 = nonpolar
CH3OH = polar
CH3 CH3 = nonpolar
CH3 CH2 CH2 OH = polar
<span>CH3 CH2 CH2 CH2 CH2 CH3 = nonpolar</span>
The percent composition of what? It should ask to find the percent composition of one of the parts like copper, sulfer, or water.
