Calculate the enthalpy of atomization per unit volume for Rb. Express your answer in kJ/m3.

Determine the volume (in mL) occupied at room temperature by 3.5e22 mercury (Hg) atoms in the liquid state.

WTF???

Nearly every one of your questions has been an EDX exam question!
Please follow the honour code and preserve the integrity of the course. Don't spoil it for the rest of us.
Thank you.

Please don't spam.

You have one hell of a cheek, schoolgirl.

How DARE you criticise someone else when you are cheating on an exam?

Now stop being so pathetic as to ask other people for answers, and either work it out yourself or get off the course

Please stop spamming or I have to report you.

What's more what I do is none of your concern.
Thank you.

Ya. i got 1st 1. Rb=82.0/55.79*1*10^-6= 1.47*10^6

To calculate the enthalpy of atomization per unit volume for Rb, you need the enthalpy of atomization and the molar volume of Rb.

1) Enthalpy of atomization (ΔH):
The enthalpy of atomization represents the energy required to convert one mole of a substance into its individual gaseous atoms. The enthalpy of atomization for Rb can be found in a chemistry reference book or online database. Let's assume it is ΔH = -77 kJ/mol.

2) Molar volume (V):
The molar volume is the volume occupied by one mole of a substance. For Rb, the molar volume can also be found in reference sources. Let's assume it is V = 55.76 cm³/mol.

Now, to calculate the enthalpy of atomization per unit volume for Rb (Hv):

Hv = ΔH / V

Hv = (-77 kJ/mol) / (55.76 cm³/mol)

To express the answer in kJ/m³, we need to convert cm³ to m³:

1 m³ = 10^6 cm³

So, 1 cm³ = 10^-6 m³

Hv = (-77 kJ/mol) / (55.76 cm³/mol) * (10^6 cm³/m³)

Hv ≈ -1380.31 kJ/m³

Therefore, the enthalpy of atomization per unit volume for Rb is approximately -1380.31 kJ/m³.

Moving on to the second question:

To determine the volume occupied at room temperature by 3.5e22 mercury (Hg) atoms in the liquid state, we need to use the concept of molar volume and Avogadro's number.

1) Avogadro's number (NA):
Avogadro's number represents the number of atoms or molecules in one mole of any substance. It is approximately 6.022 × 10²³ atoms/mol.

2) Number of moles:
Given 3.5e22 Hg atoms, we need to convert this to moles:

Number of moles = (Number of atoms) / (Avogadro's number)

Number of moles = (3.5e22 atoms) / (6.022 × 10²³ atoms/mol)

Number of moles ≈ 0.580 mol

3) Molar volume (V):
The molar volume for mercury can be found in reference sources. Let's assume it is V = 14.01 cm³/mol.

4) Conversion from cm³ to mL:
1 cm³ is equal to 1 mL.

Now, to calculate the volume occupied by 3.5e22 Hg atoms:

Volume = (Number of moles) * (Molar volume)

Volume = (0.580 mol) * (14.01 cm³/mol)

Volume ≈ 8.114 mL

Therefore, at room temperature, 3.5e22 mercury (Hg) atoms in the liquid state would occupy approximately 8.114 mL of volume.