Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
V<span>an't
Hoff factor (i) for NaCl solution is approximately 2.
</span>Van't Hoff factor (i) for glucose solution is 1.<span>
Change in freezing point from pure solvent to
solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per
kilogram of solvent.
</span>Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.
Answer:
1859.4 g of ZnCrO₄ in 10.25 moles
Explanation:
First of all, we determine the molecular formula of the compound:
Zinc → Zn²⁺ (cation)
Chromate → CrO₄⁻² (anion)
Zinc chromate → ZnCrO₄
Molar mass for the compound is:
Molar mass of Zn + Molar mass of Cr + (Molar mass of O) . 4 = 181.41 g/mol
65.41 g/mol + 52 g/mol + 16 g/mol . 4 = 181.41 g/mol
Let's apply this conversion factor: 10.25 mol . 181.41 g/mol = 1859.4 g
Answer:
See Explanation
Explanation:
Let us find the number of moles of each element
For iron = 2.24g/56 = 0.04 moles
For oxygen = 0.96 g/16 = 0.06 moles
Then we divide by the lowest ratio;
0.04/0.04 and 0.06/0.04
1 and 2
We Thus obtain 1 : 1.5
Multiplying through by 2 we have; 2 : 3
The formula of the oxide formed is Fe2O3
The balanced equation is;
4Fe + 3O2 ------> 2Fe2O3
Answer:
Le Chatelier's principle predicts that equilibrium will shift to decrease the concentration of reactants. Increasing the rate of the forward reaction will mean a decrease in reactants. ... When the concentration of reactants is increased, the equilibrium shifts to the right and there will be more product than before.
Answer:
Saturated
Explanation:
If the sugar sits undissolved at the bottom even after vigorous stirring, the lemonade has dissolved all the sugar it can hold. The lemonade is saturated.
If the lemonade were <em>unsaturated</em>, it could hold more sugar and that at the bottom would continue to dissolve.
The lemonade cannot be <em>supersaturated </em>because, if it were, the solid at the bottom would serve as nuclei on which the excess sugar in the solution could form more crystals.
