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s344n2d4d5 [400]
3 years ago
15

RATE LAW QUESTION !

Chemistry
1 answer:
vivado [14]3 years ago
6 0
In general, we have this rate law express.:

\mathrm{Rate} = k \cdot [A]^x [B]^y
we need to find x and y

ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).

then we go to compare two experiments in which only one concentration is changed

compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4)  by the smaller [B] (experiment 1) and call it Δ[B]

Δ[B]= 0.3 / 0.1 = 3

now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:

ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...

solve for y in the equation \Delta \mathrm{Rate} = \Delta [B]^y

3.09 = (3)^y \implies y \approx 1

To this point, \mathrm{Rate} = k \cdot [A]^x [B]^1

do the same to find x.
choose two experiments in which only the concentration of B is unchanged:

Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4

solve for x for \Delta \mathrm{Rate} = \Delta [A]^x

4=  (2)^x \implies x = 2

the rate law is

Rate = k·[A]²[B]
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Answer:

1.634 molL-1

Explanation:

The mol ration between NH3 and HCl is 1 : 1

Using Ca Va / Cb Vb = Na / Nb   where a = acid and b = base

Na = 1

Nb = 1

Ca = 0.208 molL-1

Cb = ?

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Cb = Ca Va / Vb

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Cb = 0.1634 molL-1 (Concentration of diluted ammonia solution)

Using the dilution equation;

C1V1 = C2V2

Initial Concentration, C1 = ?

Initial Volume, V1 = 25.00 mL

Final Volume, V2 =  250 mL

Final Concentration, C2 = 0.1634 molL-1

Solving for C1;

C1 = C2 * V2 / V1

C1 = 0.1634 * 250 / 25.00

C1 = 1.634 molL-1

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3 years ago
(6 points) Calculate the maximum number of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g
Schach [20]

Answer:107.1 g, 124.1 g

Explanation:

The equation of the reaction is;

Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)

Hence;

For Al2S3

Number of moles= reacting mass/molar mass

Number of moles = 158g/150gmol-1 =1.05 moles

If 1 mole of Al2S3 yields 3 moles of H2S

1.05 moles of Al2S will yield

1.05 × 3/1 = 3.15 moles

Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g

For water

Number of moles of water = 131g/18gmol-1= 7.3 moles

6 moles of water yields 3 moles of H2S

7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S

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3 years ago
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Read 2 more answers
In a common laboratory experiment in general chemistry students are asked to determine:_____.
Alex787 [66]
<h3>The <u>complete question</u> would be:</h3>

In a common laboratory experiment in general chemistry, students are asked to determine the relative amounts of benzoic acid and charcoal in a solid mixture. Benzoic acid is relatively soluble in hot water, but charcoal is not. Devise a method for separating the two components in the mixture.

<h3>The <u>solution</u> for that would be:</h3>

First you put the mixture in hot water and allow the benzoic acid to separate. Next funnel it and then see what's on the funnel paper. Lastly, heat the leftover water and see what's left after it evaporates.

<h3><u>What are laboratory experiments ?</u></h3>
  • Using controlled surroundings to test theories, researchers may develop laboratory experiments.
  • Laboratories are rooms or specially constructed facilities within buildings that are often found on college and university campuses and are utilized for academic research.
  • It's crucial to differentiate between experiments and other kinds of research investigations carried out in lab settings.
  • Even if a research study takes place in a lab, it is not always an experiment.
  • In order to evaluate causal conclusions regarding the links between independent and dependent variables, laboratory studies, like all genuine experimental designs, use procedures of random assignment of participants and control groups.

To view more about experiments, refer to:

brainly.com/question/17143317

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1 year ago
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masya89 [10]

Answer:

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Explanation:

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