Answer:
This question appears incomplete
Explanation:
However, it should be noted that addition of soluble salts generally lowers the freezing point of water hence after the addition, water will no longer freeze at 0°C but lower.
Soluble salts tend to form more ions in water, it is these ions that are responsible for interfering with the hydrogen bonds hence lowering the freezing. Thus, (since each bag are of the same weight) <u>the bag that contains the salt that ionizes more in water will lower the freezing point by the greatest amount</u>.
NOTE: Different weight of the salts could lead to more ions been formed in the water by some salts as against the other.
The grams of glucose are needed to prepare 400g of a 2.00%(m/m) glucose solution g is calculated as below
=% m/m =mass of the solute/mass of the solution x100
let mass of solute be represented by y
mass of solution = 400 g
% (m/m) = 2% = 2/100
grams of glucose is therefore =2/100 = y/400
by cross multiplication
100y = 800
divide both side by 100
y= 8.0 grams
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