Chemistry High School

1) Imran did an experiment to see how changing the surface area of limestone affected the rate with which it reacts with hydrochloric acid. He timed how long it took for the reaction to produce 50 cm³ of carbon dioxide gas from which he could calculate the rate in cm³ of carbon dioxide produced per second. size time (s) rate (cm³/s) large chips 150 0.33 a) What was the independent variable? b) What was the dependent variable? c) 1) List four variables that must be controlled. ii) Explain why they must be controlled. small chips 110 powder 15 d) Complete the table by calculating and filling in the missing rates. e) Which had the biggest surface area, large chips, small chips or powder? f) What is the relationship between the surface area and the time the reaction takes? . g) What is the relationship between the surface area and the reaction rate? h) Explain why changing the surface area affects the reaction rate in this way. ​

1) Imran Did An Experiment To See How Changing The Surface Area Of Limestone Affected The Rate With Which

Answers

Answer 1

a) The independent variable was the surface area of the limestone (large chips, small chips, and powder).

b) The dependent variable was the rate of the reaction, measured in cm³ of carbon dioxide produced per second.

c) i) Four variables that must be controlled are:

Concentration of hydrochloric acidMass of limestoneTemperaturePressure

ii) These variables must be controlled to ensure that the only variable affecting the reaction rate is the surface area of the limestone.

What is the relationship between surface area and the rate of a reaction?

The relationship between surface area and the rate of a reaction is that the rate of a reaction increases as the surface area of the reactants increases.

The missing rates can be calculated using the formula:

Rate = Volume of carbon dioxide produced ÷ Time taken to produce it

Size of limestone Time taken (s) Volume of carbon dioxide produced (cm³) Rate (cm³/s)

Large chips 150 50 0.33

Small chips 110 50 0.45

Powder 15 50 3.33

e) The powder had the biggest surface area.

f) The time taken for the reaction to occur decreases as the surface area of the limestone increases.

g) The reaction rate increases as the surface area of the limestone increases.

h) Changing the surface area of the limestone increases the frequency of collisions between the reactant particles (limestone and hydrochloric acid). This increase in the frequency of collisions results in an increase in the reaction rate.

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Related Questions

which of the following solutions is a buffer? check all that apply. which of the following solutions is a buffer?check all that apply. a solution made by mixing 100 ml of 0.100 m hcook and 50 ml of 0.100 m kcl a solution made by mixing 100 ml of 0.100 m hcooh and 50 ml of 0.100 m hcl a solution made by mixing 100 ml of 0.100 m hcooh and 50 ml of 0.100 m naoh a solution made by mixing 100 ml of 0.100 m hcooh and 500 ml of 0.100 m naoh

Answers

The following solutions are buffers:

A solution made by mixing 100 ml of 0.100 M HCOOH and 50 ml of 0.100 M KCl, and a solution made by mixing 100 ml of 0.100 M HCOOH and 500 ml of 0.100 M NaOH.

Let's learn more about buffer solutions:

1. Buffer solutions can withstand or resist pH changes, even when adding a strong acid or base. A buffer solution comprises a weak acid and its conjugate base or a weak base and its conjugate acid. The pH change is insignificant when an acid or a base is added to a buffer solution.

2. The chemical equation for the ionization of HCOOH in water is as follows:

HCOOH + H2O ⇌ H3O+ + HCOO-

The salt made from the acid and the base in the buffer solution must be able to act as a source of ions to neutralize either the excess H3O+ ions or OH- ions introduced into the solution. For example, KCl and NaOH are used in the buffer solutions listed above, as they produce K+ and Na+ ions, which can interact with HCOO- ions in the first buffer and HCOO- ions and HCOOH molecules in the second buffer.

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Ag2S forms a black precipitate and CdS forms a dark yellow precipitate. A solution of CdS is prepared by dissolving 0. 25 moles of solid CdS in distilled water. When equilibrium is established, a small amount of dark yellow solid remains on the bottom. A 0. 25 mol if solid AgNO3 is then stirred into the beaker, and the dark yellow precipitate is replaced with a black precipitate.


I. Did the following equilibrium shift after the AgNO3 was added? Explain using 2-3 sentences.


CdS (s) equilibrium arrows Cd 2+ (aq) + S 2- (aq)


II. Which compound has the larger Ksp value? Justify your answer

Answers

Yes, the equilibrium shifted after the AgNO₃ was added and Ag₂S having a larger Ksp value as compared to CdS.

The addition of AgNO₃ provides Ag⁺ ions, which react with S₂⁻ ions in the solution to form insoluble Ag₂S, causing the equilibrium to shift towards the products. This results in the formation of a black precipitate, indicating the presence of Ag₂S.

Ag₂S has a larger Ksp value compared to CdS. This is because Ag₂S has a lower solubility product constant (Ksp) than CdS, which means it is less soluble in water. The Ksp value for Ag₂S is 8.5 x 10⁻⁵¹, while the Ksp value for CdS is 4.0 x 10⁻²⁷. Therefore, Ag₂S has a larger Ksp value and is less soluble in water than CdS.

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