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gogolik [260]
2 years ago
15

Potassium chloride (KCl), sodium sulfate (Na2SO4), glucose (C6H12O6), and sodium phosphate (Na3PO4) are soluble in water. Rank t

he solution from the lowest impact on phase change temperatures to highest based on colligartve properties if they all have the same molarity concentration.
a. KCl
b. Na2SO4
c. C6H12O6
d. CO2
e. (NH4)3PO4
Chemistry
1 answer:
Nimfa-mama [501]2 years ago
8 0

Answer:

(NH4)3PO4 > Na2SO4 > KCl > C6H12O6 > CO2

Explanation:

Colligative properties are those properties that depend on the amount of solute present in the system.

Usually, the more the number of particles in the system, the greater the impact of colligative properties on phase change temperatures if all the solutions have the same molarity concentration.

Hence, the order of impact on phase change temperatures is as follows; (NH4)3PO4 > Na2SO4 > KCl > C6H12O6 > CO2

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For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?
Sonja [21]

Question:

<em>For an exothermic reaction at equilibrium, how will increasing the temperature affect Keq?</em>

Answer:

<em>The reaction will proceed towards the liquid phase. Heat is on the reactant side of the equation. Lowering temperature will shift equilibrium left, creating more liquid water. A reaction that is exothermic releases heat, while an endothermic reaction absorbs heat.</em>

<em>If you increase the temperature, the position of equilibrium will move in such a way as to reduce the temperature again. It will do that by favouring the reaction which absorbs heat. In the equilibrium, that will be the back reaction because the forward reaction is exothermic.</em>

Hope this helps, have a good day. c;


8 0
3 years ago
A student uses red litmus paper to test the pH value of different solutions. Which would result in a color change of the litmus
OverLord2011 [107]
The answer is all of the above but fertilizer 
5 0
3 years ago
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What is the solute in the mixture of muddy water
yarga [219]

Answer:

A solute is a solid which dissolves in a solvent. Muddy water contains water and soil, soil does not dissolve in water

6 0
3 years ago
If the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10, how much does the pH change:
andreev551 [17]

Answer:

The pH changes by 2.0 if the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10.

Explanation:

To solve this problem we use the<em> Henderson-Hasselbach equation</em>:

  • pH = pKa + log [A⁻]/[HA]

Let's say we have a weak acid whose pKa is 7.0:

  • pH = 7.0 + log [A⁻]/[HA]

If the [A⁻]/[HA] ratio is 10/1, we're left with:

  • pH = 7.0 + log (10/1)
  • pH = 7.0 + 1
  • pH = 8.0

Now if the ratio is 1/10:

  • pH = 7.0 + log (1/10)
  • pH = 7.0 - 1
  • pH = 6.0

The difference in pH from one case to the other is (8.0-6.0) 2.0.

<em>So the pH changes by 2.0</em> if the [A-]/[HA] ratio of a base/weak acid mixture changes from 10/1 to 1/10.

<u>Keep in mind that no matter the value of pKa, the answer to this question will be the same.</u>

3 0
3 years ago
How many grams of copper (II) hydroxide can be prepared from 2.4 grams of copper (II) nitrate (Cu(NO3)2 ) and excess sodium hydr
andreyandreev [35.5K]

Answer:

percentage yield = 67%

Explanation:

Mass of Cu(NO₃)₂  = 15.25 g

Mass of NaOH   = 12.75 g

Percentage yield = ?

Solution:

Cu(NO₃)₂ + 2NaOH   →  Cu(OH)₂ + 2NaNO₃

Moles of Cu(NO₃)₂:

Number of moles = mass/ molar mass

Number of moles = 15.25 g /187.56 g/mol

Number of moles = 0.08 mol

Moles of NaOH :

Number of moles = mass/ molar mass

Number of moles = 12.75 g / 40 g/mol

Number of moles = 0.32 mol

Now we will compare the moles of Cu(OH)₂ with NaOH and Cu(NO₃)₂.      NaOH             :      Cu(OH)₂

                               2                   :          1

                               0.32              :           1/2×0.32 = 0.16 mol

                            Cu(NO₃)₂         :           Cu(OH)₂

                                  1                  :               1

                             0.08                :              0.08

The number of moles produced by  Cu(NO₃)₂  are less so it will limiting reactant.

Mass of Cu(OH)₂:

Mass = number of moles × molar mass

Mass = 0.08 mol × 97.6 g/mol

Mass = 7.808 g

Theoretical yield = 7.808 g

Percent yield:

percentage yield = Actual yield/ theoretical yield ×  100

percentage yield = 5.23 g/  7.808 g ×  100

percentage yield = 0.67 ×  100

percentage yield = 67%

5 0
2 years ago
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