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Stels [109]
3 years ago
11

3. Match each of the following descriptions with one of the beakers in Model 1. In each case, assume the change in volume as the

solid(s) are added is minimal. a. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of water in beaker . b. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.500 M sodium hydroxide solution in beaker . c. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M sodium hydroxide solution in beaker . d. A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M calcium nitrate solution in beaker . 4. Based on the solubility product constant, Ksp, for calcium hydroxide given in Model 1, do you expect most of the 1.00 mole sample of solid to dissolve in any of the four beakers
Chemistry
1 answer:
vampirchik [111]3 years ago
6 0

Answer:

Explanation:

When calcium hydroxide is dissolved in water , it ionizes as follows .

Ca( OH)₂ = Ca⁺² + 2 OH ⁻

When it is dissolved in water which contains minimal OH⁻ , so there is almost no common ion effect . Hence calcium hydroxide is fully dissolved in pure water solvent .

When 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.500 M sodium hydroxide solution in beaker , it is not fully dissolved due to common ion of hydroxide ion ( OH⁻ )

NaOH = Na⁺ + OH⁻

OH⁻ ion from NaOH , suppresses the dissolution of calcium hydroxide .

Similarly

When A 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M sodium hydroxide solution in beaker , it is not fully dissolved due to common ion of hydroxide ion ( OH⁻ )  

NaOH = Na⁺ + OH⁻

OH⁻ ion from NaOH , suppresses the dissolution of calcium hydroxide

When 1.00 mole sample of solid calcium hydroxide is added to 500.0 mL of 0.200 M calcium nitrate solution , it is not fully dissolved due to common ion of calcium  ion ( Ca⁺² )

Ca( NO₃)₂ = Ca⁺² + 2NO₃⁻

Ca⁺²  ion from Ca( NO₃)₂ , suppresses the dissolution of calcium hydroxide .

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For the following reaction,
jeka94

Answer:

The answer to your question is:

1.- CO

2.- 0.414 moles of CO2

Explanation:

Data

                               2CO   + O2      ⇒    2CO2

CO = 0.414 moles

O2 = 0.418

 

Process

theoretical ratio   CO/O2 = 2/1 = 1

experimental ratio  CO/O2 = 0.414/0.418 = 0.99

Then the limiting reactant is CO

2.-

                    2 moles of CO ---------------  2 moles of CO2

                    0.414 moles of CO ---------  x

                   x = (0.414 x 2) / 2

                   x = 0.414 moles of CO2

                   

5 0
3 years ago
Nitrogen forms a surprising number of compounds with oxygen. A number of these, often given the collective symbol NOx (for "nitr
kvv77 [185]

Answer:

9.2

Explanation:

Let's do an equilibrium chart of this reaction:

2NO(g) + O₂(g) ⇄ 2NO₂(g)

4.9 atm    5.1 atm    0       Initial

-2x             -x           +2x    Reacts (stoichiometry is 2:1:2)

4.9-2x      5.1-x        2x      Equilibrium

The mole fraction of NO₂ (y) can be calculated by the Raoult's law, that states that the mole fraction is the partial pressure divided by the total pressure:

y = 2x/(4.9 - 2x + 5.1 -x + 2x)

0.52 = 2x/(10 - x)

2x = 5.2 -0.52x

2.52x = 5.2

x = 2.06 atm

Thus, the partial pressure at equilibrium are:

pNO = 4.9 -2*2.06 = 0.78 atm

pO₂ = 5.1 - 2.06 = 3.04 atm

pNO₂ = 2*2.06 = 4.12 atm

Thus, the pressure equilibrium constant Kp is:

Kp = [(pNO₂)²]/[(pNO)²*(pO₂)]

Kp = [(4.12)²]/[(0.78)²*3.04]

Kp = [16.9744]/[1.849536]

Kp = 9.2

4 0
3 years ago
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