Answer:
emf generated by cell is 2.32 V
Explanation:
Oxidation: ![2Al-6e^{-}\rightarrow 2Al^{3+}](https://tex.z-dn.net/?f=2Al-6e%5E%7B-%7D%5Crightarrow%202Al%5E%7B3%2B%7D)
Reduction: ![3I_{2}+6e^{-}\rightarrow 6I^{-}](https://tex.z-dn.net/?f=3I_%7B2%7D%2B6e%5E%7B-%7D%5Crightarrow%206I%5E%7B-%7D)
---------------------------------------------------------------------------------
Overall: ![2Al+3I_{2}\rightarrow 2Al^{3+}+6I^{-}](https://tex.z-dn.net/?f=2Al%2B3I_%7B2%7D%5Crightarrow%202Al%5E%7B3%2B%7D%2B6I%5E%7B-%7D)
Nernst equation for this cell reaction at
-
![E_{cell}=E_{cell}^{0}-\frac{0.059}{n}log{[Al^{3+}]^{2}[I^{-}]^{6}}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE_%7Bcell%7D%5E%7B0%7D-%5Cfrac%7B0.059%7D%7Bn%7Dlog%7B%5BAl%5E%7B3%2B%7D%5D%5E%7B2%7D%5BI%5E%7B-%7D%5D%5E%7B6%7D%7D)
where n is number of electrons exchanged during cell reaction,
is standard cell emf ,
is cell emf ,
is concentration of
and
is concentration of ![Cl^{-}](https://tex.z-dn.net/?f=Cl%5E%7B-%7D)
Plug in all the given values in the above equation -
![E_{cell}=2.20-\frac{0.059}{6}log[(4.5\times 10^{-3})^{2}\times (0.15)^{6}]V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.20-%5Cfrac%7B0.059%7D%7B6%7Dlog%5B%284.5%5Ctimes%2010%5E%7B-3%7D%29%5E%7B2%7D%5Ctimes%20%280.15%29%5E%7B6%7D%5DV)
So, ![E_{cell}=2.32V](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.32V)
Answer:
![\large\boxed{\large\boxed{0.529M}}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Clarge%5Cboxed%7B0.529M%7D%7D)
Explanation:
Since the <em>rate constant</em> has units of <em>s⁻¹</em>, you can tell that the order of the reaction is 1.
Hence, the rate law is:
![r=d[A]/dt=-k[A]](https://tex.z-dn.net/?f=r%3Dd%5BA%5D%2Fdt%3D-k%5BA%5D)
Solving that differential equation yields to the well known equation for the rates of a first order chemical reaction:
![[A]=[A]_0e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_0e%5E%7B-kt%7D)
You know [A]₀, k, and t, thus you can calculate [A].
![[A]=0.548M\times e^{-3.6\cdot 10^{-4}/s\times99.2s}](https://tex.z-dn.net/?f=%5BA%5D%3D0.548M%5Ctimes%20e%5E%7B-3.6%5Ccdot%2010%5E%7B-4%7D%2Fs%5Ctimes99.2s%7D)
![[A]=0.529M](https://tex.z-dn.net/?f=%5BA%5D%3D0.529M)
Basically this is used in calculating the nuclear binding energy by converting the mass defect (calculated first) to energy and if we recall, Einstein's equation E=mc2 is the perfection equation to use because E=mc2 in which E represents units of energy, m represents units of mass, and c 2 is the speed of light squared.
The molar concentration of the original HF solution : 0.342 M
Further explanation
Given
31.2 ml of 0.200 M NaOH
18.2 ml of HF
Required
The molar concentration of HF
Solution
Titration formula
M₁V₁n₁=M₂V₂n₂
n=acid/base valence (amount of H⁺/OH⁻, for NaOH and HF n =1)
Titrant = NaOH(1)
Titrate = HF(2)
Input the value :
![\tt 0.2\times 31.2\times 1=M_2\times 18.25\times 1\\\\M_2=0.342](https://tex.z-dn.net/?f=%5Ctt%200.2%5Ctimes%2031.2%5Ctimes%201%3DM_2%5Ctimes%2018.25%5Ctimes%201%5C%5C%5C%5CM_2%3D0.342)
(46 sec) x √ ((253.80894 g I2/mol) / (44.0128 g N2O/mol)) = 110 sec