The correct Lewis structure for Fluorine is A.
<span>Answer:
M in this equation is molar mass. If A is He and B is O2 then MA = 4 g/mol and MB = 32 g/mol
rate A = 1.5L/24hr rate B = 1.5L/?hr and rateA/rateB = ?/24</span>
The k is the proportionality constant of the reaction. Graphically, this is the slope of the graph. Since the graph is linear, then there is only 1 value of k. To calculate this, choose two random points in the line. Suppose we use (0.15,10) and (0.30,20), calculate for the slope.
Slope = k = (10 - 20)/(0.15 - 0.30) = 66.67 mL CO₂/g CaCO₃
<u>Answer:</u> The concentration of other (diluted) 
solution is 0.0026 M
<u>Explanation:</u>
The half reactions for the cell is:
<u>Oxidation half reaction (anode):</u> 
<u>Reduction half reaction (cathode):</u> 
In this case, the cathode and anode both are same. So, 
will be equal to zero.
To calculate cell potential of the cell, we use the equation given by Nernst, which is:
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}_{(diluted)}]}{[Fe^{2+}_{(concentrated)}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D%7D%7B%5BFe%5E%7B2%2B%7D_%7B%28concentrated%29%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 2
= 0.047 V
![[Fe^{2+}_{(diluted)}]](https://tex.z-dn.net/?f=%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D)
= ? M
![[Fe^{2+}_{(concentrated)}]](https://tex.z-dn.net/?f=%5BFe%5E%7B2%2B%7D_%7B%28concentrated%29%7D%5D)
= 0.10 M
Putting values in above equation, we get:
![0.047=0-\frac{0.0592}{2}\log \frac{[Fe^{2+}_{(diluted)}]}{0.10M}](https://tex.z-dn.net/?f=0.047%3D0-%5Cfrac%7B0.0592%7D%7B2%7D%5Clog%20%5Cfrac%7B%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D%7D%7B0.10M%7D)
![[Fe^{2+}_{(diluted)}]=0.0026M](https://tex.z-dn.net/?f=%5BFe%5E%7B2%2B%7D_%7B%28diluted%29%7D%5D%3D0.0026M)
Hence, the concentration of other (diluted) solution is 0.0026 M
The series of metals shown above is what we call the activity series. This is a guide for us to know what products are formed during a metal displacement reaction. A metal must be above the other metallic compound for displacement to occur.
1. Ag is way below the series compared to Na. No reaction occurs.
2. Al is above Fe, no reaction occurs.
3. Na is above Ni, no reaction will occur.
4. Fe is a more active metal than copper. A displacement reaction will occur.
