Answer:
E cell = +1.95 V
Explanation:
At Anode : Oxidation reaction takes place
At Cathode : Reduction reaction takes place
The reaction with lower value of reduction potential will undergo Oxidation
E = -1.18 V
This equation undergo oxidation reaction and become:
Anode(Oxidation-Half) :
E = +1.18 V
Cathode(Reduction-Half) :
E =+0.77 V
To balance the reaction multiply reduction-Half with 2.We get :
Note that E is intensive property , do not multiply E of oxidation-half with 2
Ecell = 0.77 -(-1.18)
E = +1.95 V
Answer:
V1 / T1 = V2 / T2
V₁ = 420 mL
T₁ = 210 K
V₂ = ?
T₂ = 250 K
420 / 210 = V2 / 250
V2 = (420 / 210) x 250 = 500 mL
Explanation:
Answer: The empirical formula for the given compound is 
Explanation : Given,
Percentage of C = 84.4 %
Percentage of H = 15.6 %
Let the mass of compound be 100 g. So, percentages given are taken as mass.
Mass of C = 84.4 g
Mass of H = 15.6 g
To formulate the empirical formula, we need to follow some steps:
Step 1: Converting the given masses into moles.
Moles of Carbon =
Moles of Hydrogen = 
Step 2: Calculating the mole ratio of the given elements.
For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 7.03 moles.
For Carbon = 
For Hydrogen = 
Step 3: Taking the mole ratio as their subscripts.
The ratio of C : H = 1 : 2
Hence, the empirical formula for the given compound is 
Answer:
5.4 moles CaO 1 mole CaCl₂ 5.4 mole CaCl₂
1 mole CaO
Explanation:
