Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, 
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
![[Cl^{-}{diluted}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bdiluted%7D%5D)
= 0.0222 M
![[Cl^{-}{concentrated}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D)
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V
Answer:
How can you distinguish a physical change from a chemical change?
Explanation:
Answer:
Explanation:
q = (mass) (temp change) (specific heat)
q = (10000 g) (40 °C) (0.385 J/g⋅°C) = 154000 J = 154 kJ
154 kJ / 2220 kJ/mol = 0.069369369 mol
0.069369369 mol times 44.0962 g/mol = 3.06 g (to three sig figs)
answer choice 4
The moles of oxygen that are produced when 26.5 mol of Al2O3 decomposes is 39.8 mol
<u>calculation</u>
<u> </u> 2Al2O3 + 4Al +3 O2
- use the mole ratio of Al2O3 to O2 to determine the moles of O2.
- that is from the equation above the mole ratio of Al2O3 : O2 is 2:3
- the moles of O2 is therefore=n 26.5 mol x3/2= 39.8 moles
Amines are derivatives of
Ammonia (NH₃) in which atleast one hydrogen atom is replaced by an alkyl group. Amines are further classifies as;
Primary Amines: In primary amines the nitrogen atom is attached to two hydrogen atoms and one alkyl group.
Secondary Amines: In secondary amines the nitrogen atom is attached to two alkyl groups and one hydrogen atom.
Tertiary Amines: In tertiary amines the nitrogen atom is attached to three alkyl groups, hence it has no hydrogen atom.
Below are three isomers of tertiary amines with molecular formula
C₅H₁₃N.
