Answer:
Explanation:
At the cathode
In case of molten AgI
Silver will be collected
In case of molten LiI
lithium will be collected
in case of aqueous LiI,
hydrogen gas will be collected as reduction potential of H⁺ is more than Li⁺
in case of aqueous AgI,
Silver will be obtained at cathode because reduction potential of silver is more than H⁺
At the Anode
In case of molten NaBr
Bromine will be collected
In case of molten NaF
Fluorine will be collected
in case of aqueous NaBr ,
Bromine will be collected as reduction potential of Br⁻ is less than O⁻²
in case of aqueous NaF ,
oxygen will be obtained because reduction potential of F⁻ is more than O⁻² .
Answer:
The molarity of a sugar solution is 2 M.
Explanation:
Molarity is a concentration measure that expresses the moles of solute per liter of solution. In this case it is calculated with the simple rule of three:
4 L of solution--------8 moles of sugar
1 L of solution ------x= (1 L of solution x 8 moles of sugar)/4 L of solution
x=2 moles of sugar---> <em>The solution is 2M</em>
Answer:
The over all reaction :
The standard cell potential of the reaction is 0,.897 Volts.
Explanation:
Reduction at cathode :
..[1]
Reduction potential of 
to 
Oxidation at anode:
.[2]
Reduction potential of 
to 
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
The over all reaction : 2 × [1] + [2]
The standard cell potential of the reaction is 0,.897 Volts.
Answer:
Moles of BCl₃ needed = 0.089 mol
Explanation:
Given data:
Moles of BCl₃ needed = ?
Mass of HCl produced = 10.0 g
Solution:
Chemical equation:
BCl₃ + 3H₂O → 3HCl + B(OH)₃
Number of moles of HCl:
Number of moles = mass/molar mass
Number of moles = 10.0 g/ 36.46 g/mol
Number of moles = 0.27 mol
Now we will compare the moles of HCl with BCl₃.
HCl : BCl₃
3 : 1
0.27 : 1/3×0.27 = 0.089 mol
