4) would be your correct answer
Answer:
Manganese decreases from 4+ to 2+ (reduced and oxidizing agent) and nitrogen increases from 2+ to 5+ (oxidized and reducing agent).
Explanation:
Hello there!
In this case, according to the given redox reaction, we rewrite it as a convenient first step:
Next, we assign the oxidation numbers as follows:
Thus, we can see that both manganese and nitrogen undergo a change in their oxidation number, the former decreases from 4+ to 2+ (reduced and oxidizing agent) and the latter increases from 2+ to 5+ (oxidized and reducing agent).
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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:
The molarity of the formed CaBr2 solution is 0.48 M
Explanation:
Step 1: Data given
Number of moles CaBr2 = 0.72 moles
Volume of water = 1.50 L
Step 2: Calculate the molarity of the solution
Molarity of CaBr2 solution = moles CaBr2 / volume water
Molarity of CaBr2 solution = 0.72 moles / 1.50 L
Molarity of CaBr2 solution = 0.48 mol / = 0.48 M
The molarity of the formed CaBr2 solution is 0.48 M
