Answer:
Substance is Aluminium
Explanation:
We are given;
Mass; m = 13.5 grams
Volume; V = 5 cm³
Formula for density is;
density = m/V
Density = 13.5/5
Density = 2.7 g/cm³
From the table attached, we can see that the element with a corresponding density of 2.7 is Aluminium
Answer:
Molarity = 0.5 M
Osmolarity = 0.5 x 2 = 1 Osmpl.
Molecules of Cl2 = 6.02 x / 4= 1.505 x
no. of molecules
Explanation:
If we add half mole in 1L volume than molarity will obviously be 0.5 M.
The osmolarity is molarity multiplies by number of dissociates of solute that for CaCl2 are 2. So, 2 x 0.5 = 1
Half will be molecules of Ca and half will be of Cl2 for 0.5M.
Answer:
The correct answer is -1085 KJ/mol
Explanation:
To calculate the formation enthalphy of a compound by knowing its lattice energy, you have to draw the Born-Haber cycle step by step until you obtain each element in its gaseous ions. Find attached the correspondent Born-Haber cycle.
In the cycle, Mg(s) is sublimated (ΔHsub= 150 KJ/mol) to Mg(g) and then atoms are ionizated twice (first ionization: ΔH1PI= 735 KJ/mol, second ionization= 1445 KJ/mol) to give the magnesium ions in gaseous state.
By other hand, the covalent bonds in F₂(g) are broken into 2 F(g) (Edis= 154 KJ/mol) and then they are ionizated to give the fluor ions in gaseous state 2 F⁻(g) (2 x ΔHafinity=-328 KJ/mol). The ions together form the solid by lattice energy (ΔElat=-2913 KJ/mol).
The formation enthalphy of MgF₂ is:
ΔHºf= ΔHsub + Edis + ΔH1PI + ΔH2PI + (2 x ΔHaffinity) + ΔElat
ΔHºf= 150 KJ/mol + 154 KJ/mol + 735 KJ/mol + 1445 KJ/mol + (2 x (-328 KJ/mol) + (-2913 KJ/mol).
ΔHºf= -1085 KJ/mol
