CH=benzene
Why?
benzene is represented by the empirical formula CH, which indicates that a typical sample of the compound contains one atom of carbon (C) to one atom of hydrogen (H).
----(<em>Is</em><em> </em><em>this</em><em> </em><em>what</em><em> </em><em>you</em><em> </em><em>meant</em><em>?</em><em>?</em><em>?</em><em>)</em>
Answer : The fuel value and the fuel density of pentane is, 49.09 kJ/g and 
respectively.
Explanation :
Fuel value : It is defined as the amount of energy released from the combustion of hydrocarbon fuels. The fuel value always in positive and in kilojoule per gram (kJ/g).
As we are given that:
Fuel value = 
Molar mass of pentane = 72 g/mol
Fuel value = 
Fuel value = 49.09 kJ/g
Now we have to calculate the fuel density of pentane.
Fuel density = Fuel value × Density
Fuel density = (49.09 kJ/g) × (0.626g/mL)
Fuel density = 30.73 kJ/mL =
Thus, the fuel density of pentane is
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
Answer:
#1 Exposition
#2 Background information
#3 Complication
this is right unless you're speaking of theme plot conflict climax falling action or conclusion
Answer:
54.99% yield
Explanation:
percent yield is just the amount you obtained over the amount expected times 100%.
(experimental value/theoretical value) x 100%
= (107.9 g/196.2 g) x 100%
=54.99% yield
