Answer:
Cobalt chloride
Explanation:
It is a molecule formed of one Cobalt atom, and two Chlorine atoms.
The d-block elements are found in groups 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 of the periodic table.
So your answer is A) Transition metals
<h3><u>Answer;</u></h3>
Nitrogen (N)
<h3><u>Explanation;</u></h3>
- Atoms may lose or gain electrons to achieve stability. Metal atoms lose electrons to attain stable configuration, and as a result they form positively charged ions called cations.
- <u>Non-metals on the other hand gain electrons to attain stable configuration, and as a result form negatively charged ions called anions.</u>
- <u>Among the elements given above nitrogen (N) gains electrons to form a negatively charged ion called anion. </u>
First step is to balance the reaction equation. Hence we get
P4 + 5 O2 => 2 P2O5
Second, we calculate the amounts we start with
P4: 112 g = 112 g/ 124 g/mol – 0.903 mol
O2: 112 g = 112 g / 32 g/mol = 3.5 mol
Lastly, we calculate the amount of P2O5 produced.
2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4
mol of P2O5.
This is 1.4 * (31*2 + 16*5) = 198.8 g
<u>Answer:</u> Copper (I) iodide will precipitate first.
<u>Explanation:</u>
We are given:
of CuCl = 
of CuI = 
Concentration of 
Concentration of 
Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.
![K_{sp}=[Cu^+][Cl^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BCl%5E-%5D)
Putting values in above equation, we get:
![1.0\times 10^{-6}=[Cu^+]\times 0.021](https://tex.z-dn.net/?f=1.0%5Ctimes%2010%5E%7B-6%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.021)
![[Cu^+]=\frac{1.0\times 10^{-6}}{0.021}=4.76\times 10^{-5}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B1.0%5Ctimes%2010%5E%7B-6%7D%7D%7B0.021%7D%3D4.76%5Ctimes%2010%5E%7B-5%7DM)
Concentration of copper (I) ion = 
![K_{sp}=[Cu^+][I^-]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCu%5E%2B%5D%5BI%5E-%5D)
Putting values in above equation, we get:
![5.1\times 10^{-12}=[Cu^+]\times 0.017](https://tex.z-dn.net/?f=5.1%5Ctimes%2010%5E%7B-12%7D%3D%5BCu%5E%2B%5D%5Ctimes%200.017)
![[Cu^+]=\frac{5.1\times 10^{-12}}{0.017}=3.00\times 10^{-10}M](https://tex.z-dn.net/?f=%5BCu%5E%2B%5D%3D%5Cfrac%7B5.1%5Ctimes%2010%5E%7B-12%7D%7D%7B0.017%7D%3D3.00%5Ctimes%2010%5E%7B-10%7DM)
Concentration of copper (I) ion = 
For the precipitation of copper (I) ions, we need less concentration of copper (I) ions. So, copper (I) iodide will precipitate first.
