Explanation:
Molecular formula of the compound =
The carbon chain in the molecule will look like :
1. 
2. 
3. 
4. 
No branching is present, so that means the valency of the carbon will be fulfilled by 10 hydrogen atoms and 1 oxygen atom
Answer:
At end point there will a transition from pink to colorless.
Explanation:
As the student put the vinegar in the titrator and NaOH in the beaker, it means that he has poured phenolphthalein in the NaOH solution.
The pH range of phenolphthalein is 8.3-10 (approx), it means it will show pink color in basic medium.
So on addition of phenolphthalein in NaOH the solution will become pink in color.
When we start pouring vinegar from titrator neutralization of NaOH will begin.
On complete neutralization , on addition of single drop of vinegar the solution will become acidic and there will be complete disappearance of pink color solution in the beaker.
Answer:
CN^- is a strong field ligand
Explanation:
The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).
Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.
Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.
Answer:
Explanation:
Hello,
Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:
Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:
Best regards.
Answer:
Scandium(III) fluoride, ScF3, is an ionic compound. It is slightly soluble in water but dissolves in the presence of excess fluoride to form the ScF63− anion.
hope it will help you......
