The increase in the boiling point of a solvent is a colligative property.
That means that the increase in the boling point will be related to the number of particles (molecules or ions) present in the solution.
The higher the number of particles (molecules or ions) the higher the increase in the boiling point.
All the aqueous solutions presented are electrolytes, i.e. the solutes are ionic compounds.
Then, you have to compare the number of ions that you have in each solution.
A) 1.0 M KCl ---> 1.0 M K+ + 1.0 MCl- = 2 moles of particles / liter
B) 1.0 M CaCl2 --> 1.0M Ca(2+) + 1.0M * 2 Cl (-) = 3 moles of particle / liter
C) 2.0M KCl ---> 2.0 M K+ + 2.0 M Cl- = 4 moles of particle / liter
D) 2.0 M CaCl2 ----> 2.0 M Ca (2+) + 2.0M * 2 Cl (-) = 6 moles of particle / liter.
Then, the solution 2.0M CaCl2(aq) has the highest increase in the boiling point.
Answer: option D) 2.0 M Ca Cl2(aq)
Answer:
= 29.64 g NaNO3
Explanation:
Molarity is given by the formula;
Molarity = Moles/Volume in liters
Therefore;
Number of moles = Molarity × Volume in liters
= 1.55 M × 0.225 L
= 0.34875 moles NaNO3
Thus; 0.34875 moles of NaNO3 is needed equivalent to;
= 0.34875 moles × 84.99 g/mol
= 29.64 g
For the following question(s), consider a 4% starch solution and a 10% starch solution separated by a semipermeable membrane.
Which of the following also occurs in this system?
There is a net flow of water from the 4% starch solution into the 10% starch solution
Answer:
nano3 is a solid so its not that one it has to be NH3
Explanation:
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Answer:
12.9
Explanation:
From the question given, the concentration of the hydroxide ion is given:
[OH^-] = 1.4 x 10^ -13M
pOH =?
pOH = —Log [OH^-]
pOH = —Log 1.4 x 10^ -13
pOH = 12.9
