Freezing point depression depends of the number of particles of the solute in the solution.
1)Pure water have highest freezing point. All other solutions with given solutes will have lower temperatures.
2) The more particles of the solute in the solution the lower freezing point is going to be.
<span>b. 1.0 m NaCl ( dissociates and give 2 mol ions (1 mol Na⁺ and 1 mol Cl⁻))
c. 1.0 m K3PO4 (</span>dissociates and give 4 mol ions (3 mol K⁺ and 1 mol PO4³⁻)<span>
d. 1.0 m CaCl2 (</span>dissociates and give 3 mol ions (1 mol Ca²⁺ and 2 mol Cl⁻))<span>
e. 1.0 m glucose (c6h12o6) (glucose does not dissociate, and solution have
1 mole of particles of the solute(glucose))
The largest number of particles has </span>1.0 m K3PO4 solution, and it is has lowest freezing point . Answer is C.
Answer:
The average atomic mass of copper is 63.55 amu.
Explanation:
hope this help
Answer:
[CaCl₂·2H₂O] = 1.43 m
Explanation:
Molality is mol of solute / kg of solvent.
Mass of solvent = 40 g
Let's convert g to kg → 40 g / 1000 = 0.04 kg
Let's determine the moles of solute (mass / molar mass)
8.43 g / 146.98 g/mol = 0.057 mol
Molality = 0.057 mol / 0.04 kg → 1.43
Answer:
Mass = 15.20 g of KCl
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
2 KClO₃ = 2 KCl + 3 O₂
Step 1: Calculate moles of KClO₃ as;
Moles = Mass / M/Mass
Moles = 25.0 g / 122.55 g/mol
Moles = 0.204 moles
Step 2: Find moles of KCl as;
According to equation,
2 moles of KClO₃ produces = 2 moles of KCl
So,
0.204 moles of KClO₃ will produce = X moles of KCl
Solving for X,
X = 2 mol × 0.204 mol / 2 mol
X = 0.204 mol of KCl
Step 3: Calculate mass of KCl as,
Mass = Moles × M.Mass
Mass = 0.204 mol × 74.55 g/mol
Mass = 15.20 g of KCl
Answer: B. The element called tin is a metal with the chemical symbol Sn.
Explanation:
A is incorrect because Ni is Nickel, not Niobium.
C is incorrect because Carbon is a nonmetal and its symbol is C, not Cr.
D is incorrect because Copper's symbol is Cu, not Ce.
E is incorrect because As is the symbol for Arsenic, not Astatine.
