Explanation:
this is the correct answer
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Answer:
for combustion of naphthalene is -5164 kJ/mol
Explanation:
where, C refers heat capacity and 
refers change in temperature.
Here, 
So, 
is generally expressed in terms of per mole unit of reactant. Also, should be negative as it is an exothermic reaction (temperature increases).
Molar mass of naphthalene is 128.17 g/mol
So, 1.025 g of naphthalene = 
of naphthalene
= 0.007997 moles of naphthalene
The balanced equation
6CO₂ + 6H₂O = C₆H₁₂O₁₆ + O₂
<h3>Further explanation</h3>
Given
Reaction
CO2 + _H2O = _C6H12O16 + _O2
Required
Balanced equation
Solution
Give a coefficient
aCO₂ + bH₂O = C₆H₁₂O₁₆ + cO₂
Make an equation
C, left = a, right = 6⇒a=6
H, left = 2b, right = 12⇒2b=12⇒b=6
O, left=2a+b, right = 16+2c⇒2a+b=16+2c⇒2.6+6=16+2c⇒2c=2⇒c=1
The equation becomes :
<em>6CO₂ + 6H₂O = C₆H₁₂O₁₆ + O₂</em>
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
