Refer to the table below. Credits to https://terpconnect.umd.edu/~wbreslyn/chemistry/naming/IonicCharge2.jpg
Cations with (+ ) charges lose electrons in order to obtain an octet (8 valence electrons) when they ionically bond with another ion. We're looking for the ions that loses electrons here. So, from the table:
Al 3+ , S 2- , O 2-, Ag + , Ne ( noble gas, no charge)
Since Al and Ag has (+) charges, they are going to lose electrons to form ionic bonds with other atoms.
Answer:
A. One unpaired electron
B. 5 unpaired electrons
Explanation:
In A ,Fe is in +3 oxidation state and Electronic configuration- [Ar]3d5
And NO2 is a strong field ligand hence it causes pairing in t2g orbitals and results one unpaired electron in dZX orbital.
In B, also Fe is in +3 oxidation state but F is weak field ligand hence causes no pairing of Electrons hence it results 5 unpaired electrons with electronic configuration t2g^3 eg^2
Answer:
0.01 moles of SrCO₃
Explanation:
In this excersise we need to propose the reaction:
K₂CO₃ + Sr(NO₃)₂ → 2KNO₃ + SrCO₃
As we only have data about the potassium carbonate we assume the strontium nitrite as the excess reactant.
1 mol of K₂CO₃ react to 1 mol of Sr(NO₃)₂ in order to produce 2 moles of potassium nitrite and 1 mol of strontium carbonate.
Ratio is 1:1. In conclussion,
0.01 mol of K₂CO₃ must produce 0.01 moles of SrCO₃