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Plaster of paris hardens by (1) combining with water (2) giving out water (3) giving of \(\mathrm{CO}_{2}\) (4) changing into \(\mathrm{CaCO}_{3}\)

Short Answer

Expert verified
Plaster of Paris hardens by combining with water (Option 1).

Step by step solution

01

Understanding Plaster of Paris

Plaster of Paris (POP) is a white powder that is scientifically known as calcium sulfate hemihydrate \((\text{CaSO}_4 \frac{1}{2} \text{H}_2 \text{O})\). It is commonly used for casting and molding.
02

Reaction with Water

When Plaster of Paris is mixed with water, it undergoes an exothermic reaction. This means it interacts with water and releases heat. POP reverts to its original form, gypsum \((\text{CaSO}_4 \text{2H}_2 \text{O})\), which is a hardened material.
03

Eliminating Incorrect Options

Option (2) can be eliminated because POP does not give out water but absorbs it. Option (3) is incorrect as \(\text{CO}_2\) is not involved in the process. Option (4) is incorrect because POP does not change into \( \text{CaCO}_3 \) during hardening.
04

Correct Option

Based on the above steps, the correct way Plaster of Paris hardens is by combining with water.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Calcium Sulfate Hemihydrate
Calcium sulfate hemihydrate is the chemical name for Plaster of Paris. Its chemical formula is \((\text{CaSO}_4 \frac{1}{2} \text{H}_2 \text{O})\).
This compound is known for its ability to form a paste when mixed with water, and subsequently harden into a solid form.
The term 'hemihydrate' indicates that for every molecule of calcium sulfate, there is half a molecule of water.
The white powder of Plaster of Paris is easy to work with, making it popular for creating casts, molds, and sculptures. It serves a variety of industrial needs as well.
Gypsum Formation
When Plaster of Paris is combined with water, it undergoes a chemical reaction.
The compound absorbs water and transforms back into its original form, known as gypsum. Gypsum is chemically termed \((\text{CaSO}_4 2\text{H}_2 \text{O})\).
This absorption of water and change to gypsum causes the mixture to harden.
Understanding this transformation helps explain why plaster becomes solid after mixing with water. The reaction can be represented as:
\( \text{CaSO}_4 \frac{1}{2} \text{H}_2 \text{O} + 1.5\text{H}_2 \text{O} \rightarrow \text{CaSO}_4 \text{2H}_2 \text{O} \)
Exothermic Reaction
The hardening of Plaster of Paris is an exothermic reaction. This means that during the process, heat is released.
As Plaster of Paris absorbs water and changes into gypsum, energy is given off in the form of heat.
This is why you can feel warmth coming from the plaster as it sets.
Exothermic reactions are characterized by:
  • Release of heat.
  • Temperature increase of the surroundings.
  • Common in many everyday processes like rusting iron and burning wood.
Recognizing that the hardening of Plaster of Paris is exothermic helps understand why it releases heat while setting.

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Most popular questions from this chapter

The correct order regarding the solubility of alkaline carth metal halide in water is (1) \(\mathrm{BeCl}_{2}>\mathrm{MgCl}_{2}>\mathrm{CaCl}_{2}>\mathrm{SrCl}_{2}>\mathrm{BaC} \mathrm{l}_{2}\) (2) \(\mathrm{MgCl}_{2}>\mathrm{CaCl}_{2}>\mathrm{BeCl}_{2}>\mathrm{BaCl}_{2}>\mathrm{SrCl}_{2}\) (3) \(\mathrm{BaCl}_{2}>\mathrm{MgCl}_{2}>\mathrm{CaCl}_{2}>\mathrm{BeCl}_{2}>\mathrm{SrCl}_{2}\) (4) \(\mathrm{CaCl}_{2}>\mathrm{MgCl}_{2}>\mathrm{SrCl}_{2}>\mathrm{BaCl}_{2}>\mathrm{BeCl}_{2}\)

Which of the following represents marble? (1) \(\mathrm{CaSO}_{4}\) (2) \(\mathrm{CaCO}_{3}\) (3) \(\mathrm{CaCl}_{2}\) (4) \(\mathrm{MgSO}_{4}\)

Epsom salt on strong heating gives (1) anhydrous salt (2) \(\mathrm{MgO}+\mathrm{SO}_{2}+\mathrm{O}_{2}\) (3) \(\mathrm{MgS}+\mathrm{O}_{2}\) (4) \(\mathrm{Mg} \mathrm{SO}_{3}+\mathrm{O}_{2}\)

A certain metal is present in the soil, plants, bones, egg shells, sea shells and coral. It is also used to remove oxygen from molten steel and its hydroxide is used to detect \(\mathrm{CO}_{2}\). The metal is (1) \(\mathrm{Mg}\) (2) \(\Lambda 1\) (3) \(\mathrm{Ca}\) (4) \(\mathrm{Na}\)

Which of the following statements is false? (1) \(\Lambda\) lkaline carth metals are weaker reducing agents than alkali metals becausc of their comparatively high ionization cncrgics. (2) The reducing nature of alkaline carth metals follows the incrcasing order as \(\mathrm{Be}<\mathrm{Mg}<\mathrm{Ca}<\) \(\mathrm{Sr}<\mathrm{Ba}\) (3) Compounds of II group clements are colourless and diamagnctic duc to the abscnce of umpaircd clcctrons. (4) \Lambdalkaline carth metals have low clectrical and thermal conductivitics than alkali metals.

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