Answer:
12.15, 2.23
Explanation:
Using the dilution formula
C₁V₁ = C₂ V₂
where C₁ = 1.46 M
V₁ = 1.48 ml = 0.00148 L
C₂ = unknown
V₂ = 150 ml + 1.48ml = 151.48 ml = 0.15148L
1.46 M × 0.00148 L = C₂×0.15148L
C₂ = 0.0143 M
pOH = - log ( OH⁻) = - log ( 0.0143) = 1.85
pH + pOH = 14
pH = 14 - 1.85 = 12.15
b) using the same formula
C₁V₁ = C₂ V₂
C₁ = 0.99 M
V₁ = 1.49 ml = 0.00149 L
C₂ = unknown
V₂ = 250 ml + 1.49 ml = 251.49 ml = 0.25149 L
0.99 M × 0.00149 L = C₂ × 0.25149 L
C₂ = 0.00587
pH = - log ( H⁺) = - ( -2.23) = 2.23
Answer:
T = 4.062V
Explanation:
from PV = nRT => T = PV/RT
P = 1 atm
V = Final Volume
n = 3 moles
R = 0.08206 L·atm/mol·K
T = ?
T = 1 atm · V(Liters)/(3 moles)(0.08206L·atm/mol·K) = 4.062·V(final) Kelvin
Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Explanation :
Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.
In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.
We are taking the value of standard reduction potential form the standard table.
![E^0_{[Ag^{+}/Ag]}=+0.80V](https://tex.z-dn.net/?f=E%5E0_%7B%5BAg%5E%7B%2B%7D%2FAg%5D%7D%3D%2B0.80V)
![E^0_{[Cu^{2+}/Cu]}=+0.34V](https://tex.z-dn.net/?f=E%5E0_%7B%5BCu%5E%7B2%2B%7D%2FCu%5D%7D%3D%2B0.34V)
In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.
The balanced two-half reactions will be,
Oxidation half reaction (Anode) : ![Cu(s)\rightarrow Cu^{2+}(aq)+2e^-](https://tex.z-dn.net/?f=Cu%28s%29%5Crightarrow%20Cu%5E%7B2%2B%7D%28aq%29%2B2e%5E-)
Reduction half reaction (Cathode) : ![Ag^{+}(aq)+e^-\rightarrow Ag(s)](https://tex.z-dn.net/?f=Ag%5E%7B%2B%7D%28aq%29%2Be%5E-%5Crightarrow%20Ag%28s%29)
Thus the overall reaction will be,
![Cu(s)+2Ag^{+}(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)](https://tex.z-dn.net/?f=Cu%28s%29%2B2Ag%5E%7B%2B%7D%28aq%29%5Crightarrow%20Cu%5E%7B2%2B%7D%28aq%29%2B2Ag%28s%29)
From this we conclude that, 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Hence, the 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.
Answer:
B Temperature is clear cause of carbon dioxide levels
Answer:The hybridisation of C1 and C2 are sp3 and sp2 respectively.
Explanation:
Here the first carbon (C-1) formed four straight single bonds.
Linking to the electron theory (a fig.1 shown below) one electron from 2s orbital moves to the empty 2pz orbital.
The 2s and the three 2p orbitals hybridise and each orbital will be completed by one electron shared mutually with and from N, H, H, and the other C.
C-2 the second carbon has an sp2 because it forms a double bond with oxygen.
In the sp2 hybridisation the 2s orbital hybridises with only two 2p orbitals leaving the other 2p orbital unhybridised. The orbitals in sp2 is to be completed with one more electron each by sharing covalently with the C, O and the other O.
Note: The unhybridised p orbital is completed with the other shared electron of the oxygen. This forms the π bond in the carbon-oxygen double bond. (Fig 2. Shows the electronic arrangements.)