Answer:
1.18 V
Explanation:
The given cell is:
Half reactions for the given cell follows:
Oxidation half reaction: 
Reduction half reaction: 
Multiply Oxidation half reaction by 2 and Reduction half reaction by 3
Net reaction: 
Oxidation reaction occurs at anode and reduction reaction occurs at cathode.
To calculate the 
of the reaction, we use the equation:
Putting values in above equation, we get:
To calculate the EMF of the cell, we use the Nernst equation, which is:
![E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Al^{3+}]^2}{[Fe^{2+}]^3}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.059%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAl%5E%7B3%2B%7D%5D%5E2%7D%7B%5BFe%5E%7B2%2B%7D%5D%5E3%7D)
where,
= electrode potential of the cell = ?V
= standard electrode potential of the cell = +1.21 V
n = number of electrons exchanged = 6
Putting values in above equation, we get:
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The balanced equation for the above reaction is
HBr + KOH ---> KBr + H₂O
stoichiometry of HBr to KOH is 1:1
HBr is a strong acid and KOH is a strong base and they both completely dissociate.
The number of HBr moles present - 0.25 M / 1000 mL/L x 52.0 mL = 0.013 mol
The number of KOH moles added - 0.50 M / 1000 mL/L x 26.0 mL = 0.013 mol
the number of H⁺ ions = number of OH⁻ ions
therefore complete neutralisation occurs.
Therefore solution is neutral. At 25 °C, when the solution is neutral, pH = 7.
Then pH of solution is 7
In this question we need to find the new volume of the gas. Since we have been given the pressure and temperature change, we can used to combined gas law equation.
the parameters for 1st instance are given on the left side and parameters for the second instance are given on the right side of the equation
(319 mmHg x 0.558 L)/ 115 K = (215 mmHg x V)/387 K
V = 2.79 L
During this phase Change heat energy is being absorbed by the molecules, and as a result the molecules possess a greater ability to move around and possess higher kinetic energy because of this. The molecules also possess a higher potential energy.
