The freezing point depression is a colligative property, which means that it depends on the number of particles of solute disolved in the solution.
When you have solutes that are ionic compounds they dissociate in water into ions, then the compound that dissociates more ions will produce more particles and will decrease the freezing point the most.
Given theses aqueous solutions Na2 CO3, Co Cl3, and Li NO3 you can predict the order of the freezing points.
First, write the dissociation equations>
Na2CO3 -> 2Na(+) + CO3 (2-) These are 3 ions: two of Na(+) and one of CO3(2-)
The number inside parenthesis are number of charge not number of molecules.
Co Cl3 -> Co(3+) + 3 Cl (1-) Those are 4 ions: one of Co (+) and three of Cl (-)
Li NO3 -> Li (+) + NO3 (-) those are two ions: one of Li (+) and one of NO3(-)
Then the ionic compound that dissociates into more ions give the solution with lower freezing point, and these is the rank from higher to lower freezing point:
Li NO3 > Na2 CO3 > Co Cl3.
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Answer is: volume of oxygen is 4.63 liters.
Balanced chemical reaction: 2C + O₂ → 2CO.
m(C) = 4.50 g.
n(C) = m(C) ÷ M(C).
n(C) = 4.50 g ÷ 12 g/mol.
n(C) = 0.375 mol.
From chemical reaction: n(C) : n(O₂) = 2 : 1.
n(O₂) = 0.1875 mol.
T = 48°C = 321.15 K.
p = 810 mmHg ÷ 760 mmHg/atm= 1.066 atm.
<span>R = 0.08206
L·atm/mol·K.
Ideal gas law: p·V = n·R·T.</span>
V(O₂) =
n·R·T / p.<span>
V(O₂) =
0.1875 mol · 0.08206 L·atm/mol·K · 321.15 K / 1.066 atm.</span><span>
V(O₂<span>) =
4.63 L.</span></span>
Na 2Co3
1*23=23. 2*12=24. 6*16=96
23+24+96=143
(23*100)/143. (24*100)/143. (96*100)/143
=16%. =16.7%. =67.1%
Answer:
There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.
Explanation:
