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Otrada [13]
3 years ago
5

Absalon adds 1 g of salt to 1 L of room temperature water (25 °C). Then, he starts a timer and observes what happens. He notices

that it takes 1 minute for the salt to dissolve. He decides to repeat his experiment, and he adds 1 g of salt to another 1 L of room temperature water (25 °C). After he adds the salt, he starts a timer. But, instead of watching the salt dissolve, he stirs the salt and water with a spoon until it dissolves. He notices that it only takes 30 seconds for the salt to dissolve in his second experiment.
Why does the salt dissolve faster in Absalon's second experiment?


Stirring the salt and water increases the polarity of the water molecules, which causes the ionic bonds of the salt to break.

Stirring the salt and water increases particle motion, which causes more collisions to occur between the water and salt.

Stirring the salt and water increases the surface area of the water, which causes more collisions to occur between the water and salt.

Stirring the salt and water increases the pressure on the solution, which causes the ionic bonds of the salt to break.
Chemistry
2 answers:
igor_vitrenko [27]3 years ago
7 0

Answer: Option (b) is the correct answer.

Explanation:

When Absalon stirred the solution which is containing salt in water then there will occur disturbance into solution due to stirring.

As a result, particles will come in motion and hence, they tend to gain more kinetic energy. Due to which there will be more number of collisions between the solute (salt) and solvent (water) particles.

Hence, rate of reaction will increase and because of this salt will readily dissolve into water.

Thus, we can conclude that the salt dissolve faster in Absalon's second experiment because stirring the salt and water increases particle motion, which causes more collisions to occur between the water and salt.

lana [24]3 years ago
6 0

Answer:

B

Explanation:

I looked it up and found the answer lol

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Which of the compounds above are strong enough acids to react almost completely with a hydroxide ion (pka of h2o = 15.74) or wit
luda_lava [24]

The compounds can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Further explanation </em></h3>

In an acid-base reaction, it can be determined whether or not a reaction occurs by knowing the value of pKa or Ka from acid and conjugate acid (acid from the reaction)

Acids and bases according to Bronsted-Lowry

Acid = donor (donor) proton (H + ion)

Base = proton (receiver) acceptor (H + ion)

If the acid gives (H +), then the remaining acid is a conjugate base because it accepts protons. Conversely, if a base receives (H +), then the base formed can release protons and is called the conjugate acid from the original base.

From this, it can be seen whether the acid in the product can give its proton to a base (or acid which has a lower Ka value) so that the reaction can go to the right to produce the product.

The step that needs to be done is to know the pKa value of the two acids (one on the left side and one on the right side of the arrow), then just determine the value of the equilibrium constant

Can be formulated:

K acid-base reaction = Ka acid on the left : K acid on the right.

or:

pK = acid pKa on the left - pKa acid on the right

K = equilibrium constant for acid-base reactions

pK = -log K;

K~=~10^{-pK}

K value> 1 indicates the reaction can take place, or the position of equilibrium to the right.

There is some data that we need to complete from the problem above, which is the pKa value of some compounds that will react, namely:

pyridinium pKa = 5.25

acetone pKa = 19.3

butan-2-one pKa = 19

Let's look at the K value of each possible reaction:

pka H₂O = 15.74, pka of H₂CO₃ = 6.37)

  • 1. C₅H₆N pyridinium

* with OH⁻

C₅H₆N + OH- ---> C₅H₅N- + H₂O

pK = pKa pyridinium - pKa H₂O

pK = 5.25 - 15.74

pK = -10.49

K~=~10^{4.9}

K values> 1 indicate the reaction can take place

* with HCO3⁻

C₅H₆N + HCO₃⁻-- ---> C₅H₅N⁻ + H₂CO₃

pK = 5.25 - 6.37

pK = -1.12

K`=~10^{1.12]

Reaction can take place

  • 2. Acetone C₃H₆O

* with OH-

C₃H₆O + OH⁻ ---> C₃H₅O- + H₂O

pK = 19.3 - 15.74

pK = 3.56

K~=~10^{ -3.56}

Reaction does not happen

* with HCO₃-

C₃H₆O + HCO₃⁻ ----> C₃H₅O⁻ + H₂CO₃

pK = 19.3 - 6.37

pK = 12.93

K`=~10 ^{-12.93}

Reaction does not happen

  • 3. butan-2-one C₄H₇O

* with OH-

C₄H₇O + OH- ---> C₄H₆O- + H₂O

pK = 19 - 15.74

pK = 3.26

K~=~10^{-3.26}

Reaction does not happen

* with HCO₃⁻

C₄H₇O + HCO₃⁻ ---> C₄H₆O⁻ + H₂CO₃

pK = 19 - 6.37

pK = 12.63

K~=~ 10^{-12.63}

Reaction does not happen

So that can react with OH⁻ and HCO₃⁻ only C₅H₆N pyridinium

<h3><em>Learn more </em></h3>

the lowest ph

brainly.com/question/9875355

the concentrations at equilibrium.

brainly.com/question/8918040

the ph of a solution

brainly.com/question/9560687

Keywords : acid base reaction, the equilibrium constant

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Answer:

Explanation:

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Molecular weight / 2 = equivalent weight .

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Answer: Different plant structure can increase chances of reproduction by the animals and other living organisms in the ecosystem.

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Explanation:

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Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

This equation tells that ; when  1 mole of Zn react with 1 mole of sulfuric acid, it produces 1 mole of zinc sulfate and 1 mole of hydrogen.

Since sulfuric acid is in excess so Zinc is the limiting reagent

(Limiting reagent : Substance which get consumed when the reaction completes, limiting reagent helps in predicting the amount of products formed)

Limiting reagent (Zn) will decide the amount of Hydrogen produced

Zn(s) + H_{2}SO_{4}(aq) \rightarrow ZnSO_{4}(aq) + H_{2}(g)

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5\ mole\ zinc\rightarrow 5\ mole\ H_{2}

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