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)
1) Balanced chemical equation
H2SO4 + 2NaOH ---> Na2 SO4 + 2H2O
=> 1 mol H2SO4 : 2 moles NaOH
2) Convert 89.3 g of H2SO4 and 96.0 g of NaOH to moles
Molar mass of H2SO4 = 98.1 g/mol
Molar mass of NaOH = 40.0 g/mol
moles = mass in grams / molar mass
moles H2SO4 = 89.3 g / 98.1 g/mol = 0.910 mol
moles NaOH = 96.0 g / 40.0 g/mol = 2.40 mol
3) Theoretical molar ratio = 2 moles NaOH / 1 mol H2SO4
So, all the 0.91 mol of H2SO4 will be consumed along with 1.820 (2*0.91) moles of NaOH, and 0.580 moles (2.40 - 1.82) of NaOH will be left over by the chemical reaction.
4) Convert 0.580 moles NaOH to mass
0.580 moles * 40.0 g/mol = 23.2 g of NaOH will be left over
Answer:
0.880 (0.88039867109 to be exact)
Explanation:
To convert from molecules to moles, simply divide by Avogadro's number which is 6.02 x 10^23
So, 5.30x10^23/6.02x10^23 = 0.880
