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IgorC [24]
4 years ago
10

What aqueous solution has the highest boiling point at standard pressure? A) 1.0 M KCl(aq) B) 1.0 M CaCl2(aq) C) 2.0 M KCl(aq) D

) 2.0 M CaCl2(aq)
Chemistry
1 answer:
Lapatulllka [165]4 years ago
5 0
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)
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Answer:

329.7%

Explanation:

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3 years ago
Suppose 50.0g of silver nitrate is reacted with 50g of hydrochloric acid producing silver chloride and a mixture of other produc
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Answer:

53.6 grams of silver chloride was produced.

Explanation:

AgNO_3+HCl+\rightarrow AgCl+HNO_3

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

Mass of silver nitrate = 50.0 g

Mass of hydrogen chloride = 50.0 g

Mass of silver chloride = x

Mass of  nitric acid = 46.4 g

Mass of silver nitrate + Mass of hydrogen chloride =

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[te]50.0 g+50.0 g=x+46.4 g[/tex]

x=50.0 g+50.0 g - 46.4 g = 53.6 g

53.6 grams of silver chloride was produced.

8 0
3 years ago
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
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Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

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