Answer:
The empirical formula for this compound is thus CH2.
Explanation:
Answer:
Zn2+ is colourless
Explanation:
We know that transition metal salts are usually coloured due to the possibility of d-d transition.
This d-d transition can only occur when there are vacant d-orbitals. The electronic configuration [Ar] 4s23d8 suggests the presence of vacant d-orbitals and the possibility of the compounds of Zn2+ being coloured.
However, the absence of colours in Zn2+ compounds shows that there is no d-d transition(electronic) spectra observed for Zn2+ because the d orbitals are completely filled. This means that the correct electronic configuration of the ion is [Ar] 3d10.
Shiny
Somewhat reactive
Standard temp. and pressure
The empirical formula of the compound is C. NiF₂.
<em>Step 1</em>. Calculate the <em>moles of each element</em>
The empirical formula is the simplest whole-number ratio of atoms in a compound.
The ratio of atoms is the same as the ratio of moles.
So, our job is to calculate the molar ratio of Ni to F.
Moles of Ni = 9.11 g Ni × (1 mol Ni /(58.69 g Ni) = 0.1552 mol Ni
Moles of F = 5.89 g F × (1 mol F/19.00 g F) = 0.3100 mol F
<em>Step 2</em>. Calculate the <em>molar ratio</em> of the elements
Divide each number by the smallest number of moles
Ni:F = 0.1552:0.3100 = 1:1.997 ≈ 1:2
<em>Step 3</em>: Write the <em>empirical formula</em>
EF = NiF₂
