Liquid silver is less dense than solid silver, so the solid silver would sink
Answer:
6.22 × 10⁻⁵
Explanation:
Step 1: Write the dissociation reaction
HC₆H₅COO ⇄ C₆H₅COO⁻ + H⁺
Step 2: Calculate the concentration of H⁺
The pH of the solution is 2.78.
pH = -log [H⁺]
[H⁺] = antilog -pH = antilog -2.78 = 1.66 × 10⁻³ M
Step 3: Calculate the molar concentration of the benzoic acid
We will use the following expression.
Ca = mass HC₆H₅COO/molar mass HC₆H₅COO × liters of solution
Ca = 0.541 g/(122.12 g/mol) × 0.100 L = 0.0443 M
Step 4: Calculate the acid dissociation constant (Ka) for benzoic acid
We will use the following expression.
Ka = [H⁺]²/Ca
Ka = (1.66 × 10⁻³)²/0.0443 = 6.22 × 10⁻⁵
Answer:
Subtract them.
Explanation:
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I was actually learned about this in school just found an source.
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: An oxygen atom in heavy water has an extra neutron. A hydrogen atom in heavy water has an extra proton.
Explanation:
