Answer : The molal freezing point depression constant of X is 
Explanation : Given,
Mass of urea (solute) = 5.90 g
Mass of X liquid (solvent) = 450.0 g
Molar mass of urea = 60 g/mole
Formula used :
where,
= change in freezing point
= freezing point of solution = 
= freezing point of liquid X= 
i = Van't Hoff factor = 1 (for non-electrolyte)
= molal freezing point depression constant of X = ?
m = molality
Now put all the given values in this formula, we get
![[0.4-(-0.5)]^oC=1\times k_f\times \frac{5.90g\times 1000}{60g/mol\times 450.0g}](https://tex.z-dn.net/?f=%5B0.4-%28-0.5%29%5D%5EoC%3D1%5Ctimes%20k_f%5Ctimes%20%5Cfrac%7B5.90g%5Ctimes%201000%7D%7B60g%2Fmol%5Ctimes%20450.0g%7D)
Therefore, the molal freezing point depression constant of X is
From the equation above the reacting ratio of KClO3 to O2 is 2:3 therefore the number of moles of oxygen produced is ( 4 x3)/2 = 6 moles since four moles of KClO3 was consumed
mass=relative formula mass x number of moles
That is 32g/mol x 6 moles =192grams
Answer:
8 cm3
Explanation:
The volume of this irregular solid will calculated as the difference between the final volume and the initial volume;
The final volume of the water and the solid is 25 ml
The initial volume of the water alone was 17 ml
The volume of the irregular solid is thus approximately;
25 - 17 = 8 ml
We then use the conversion;
1 cm3 = 1 mL
Thus the volume of the solid is 8 cm3
Peter is taller, convert cm to in.
