Answer : The correct option is, (b) +0.799 V
Solution :
The values of standard reduction electrode potential of the cell are:
![E^0_{[H^{+}/H_2]}=+0.00V](https://tex.z-dn.net/?f=E%5E0_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D%3D%2B0.00V)
![E^0_{[Ag^{+}/Ag]}=+0.799V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.799V)
From the cell representation we conclude that, the hydrogen (H) undergoes oxidation by loss of electrons and thus act as anode. Silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.
The half reaction will be:
Reaction at anode (oxidation) : 
Reaction at cathode (reduction) : 
The balanced cell reaction will be,
Now we have to calculate the standard electrode potential of the cell.
![E^o_{cell}=E^o_{[Ag^{+}/Ag]}-E^o_{[H^{+}/H_2]}](https://tex.z-dn.net/?f=E%5Eo_%7Bcell%7D%3DE%5Eo_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D-E%5Eo_%7B%5BH%5E%7B%2B%7D%2FH_2%5D%7D)
Therefore, the standard cell potential will be +0.799 V
Answer:
44.2 L
Explanation:
Use Charles Law:
We have all the values except for V₂; this is what we're solving for. Input the values:
- make sure that your temperature is in Kelvin
From here, we need to get V₂ by itself. To do this, multiply by 273 on both sides:
Therefore, V₂ = 44.2 L
It's also helpful to know that temperature and volume are linearly related. So, when temperature drops, so will volume and vice versa.
Given :
A 250 ml beaker weighs 13.473 g .
The same beaker plus 2.2 ml of water weighs 15.346 g.
To Find :
How much does the 2.2 ml of water, alone, weigh .
Solution :
Now, mass of water is given by :
Therefore , mass of 2.2 ml of water alone is 1.873 g .
Hence , this is the required solution .
Answer:
1. V2.
2. 299K.
3. 451K
4. 0.25 x 451 = V2 x 299
Explanation:
1. The data obtained from the question include:
Initial volume (V1) = 0.25mL
Initial temperature (T1) = 26°C
Final temperature (T2) = 178°C
Final volume (V2) =.?
2. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Initial temperature (T1) = 26°C
Initial temperature (T1) = 26°C + 273 = 299K
3. Conversion from celsius to Kelvin temperature.
T(K) = T (°C) + 273
Final temperature (T2) = 178°C
Final temperature (T1) = 178°C + 273 = 451K
4. Initial volume (V1) = 0.25mL
Initial temperature (T1) = 299K
Final temperature (T2) = 451K
Final volume (V2) =.?
V1 x T2 = V2 x T1
0.25 x 451 = V2 x 299
1.6456 x 10^3 (ten to the third power)
