Answer: CuI₂ + Br₂
Explanation:
1) The activity series F > Cl > Br > I means that F is the most active and I is the least active of those four elements (the halogens, group 17 in the periodic table).
The activity is a measure of how eager is an element to react compared to other elements in the series in a single replacement reaction.
2) Choice 1: CuI₂ + Br₂
Since the activity of Br is higher than that of I, Br will react with CuI₂, displacing I, which will be left alone, as per this chemical equation:
CuI₂ + Br₂ → CuBr₂ + I₂
Being I less active than Br, it cannot displace Br in CuBr₂.
3) Choice 2: Cl₂ + AlF₃
Being Cl less active than F, the former will not displace the latter, and the reaction will not proceed.
4) Choice 3: Br₂ + NaCl
Again, being Br less active than Cl, the former will not displace the latter, and the reaction will not proceed.
5) Choice 4: CuF₂ + I₂
Once more, being I less active than F, the former will not displace the latter, and the reaction will not proceed.
Answer:
2.12×10²³ atoms.
Explanation:
From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ atoms. This simply means that 1 mole of zirconium also 6.02×10²³ atoms.
Thus, we can obtain the number of atoms present in 0.3521 mole of zirconium as follow:
1 mole of zirconium also 6.02×10²³ atoms.
Therefore, 0.3521 mole of zirconium will contain = 0.3521 × 6.02×10²³ = 2.12×10²³ atoms.
Therefore, 0.3521 mole of zirconium contains 2.12×10²³ atoms.
Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, 
is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BCl%5E%7B-%7D%7Bdiluted%7D%5D%7D%7B%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D%7D)
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
![[Cl^{-}{diluted}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bdiluted%7D%5D)
= 0.0222 M
![[Cl^{-}{concentrated}]](https://tex.z-dn.net/?f=%5BCl%5E%7B-%7D%7Bconcentrated%7D%5D)
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V
Molybdenum Arsenide
I think that’s right but not %100 sure
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