Plaster of Paris should be stored in a moisture-proof container. Explain why ?
Plaster of Paris should be stored in a moisture-proof container. Explain why ?
Plaster of Paris $(POP)$ should be stored in a moisture-proof container because Plaster of Paris, a powdery mass, absorbs water (moisture) to form a hard solid known as gypsum.
$\underset{(POP)}{\mathop{CaS{{O}_{4}}.\frac{1}{2}{{H}_{2}}O}}\,\,+1\,\frac{1}{2}{{H}_{2}}O\,\to \,\underset{Gypsum\,\,(Hard\,\,solid)}{\mathop{CaS{{O}_{4}}.2{{H}_{2}}O}}\,$
Similar Questions
Why should curd and sour substances not be kept in brass and copper vessels ?
Why should curd and sour substances not be kept in brass and copper vessels ?
What will happen if a solution of sodium hydrocarbonate is heated ? Give the equation of the reaction involved.
What will happen if a solution of sodium hydrocarbonate is heated ? Give the equation of the reaction involved.
Do basic solutions also have $H^+_{(aq)}$ ions ? If yes, then why are these basic ?
Do basic solutions also have $H^+_{(aq)}$ ions ? If yes, then why are these basic ?
Metal compound $A$ reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
Metal compound $A$ reacts with dilute hydrochloric acid to produce effervescence. The gas evolved extinguishes a burning candle. Write a balanced chemical equation for the reaction if one of the compounds formed is calcium chloride.
$10 \,mL$ of a solution of $NaOH$ is found to be completely neutralised by $8 \,mL$ of a given solution of $HCl$. If we take $20 \,mL$ of the same solution of $NaOH$, the amount of $HCl$ solution (the same solution as before) required to neutralise it will be ...... $mL$
$10 \,mL$ of a solution of $NaOH$ is found to be completely neutralised by $8 \,mL$ of a given solution of $HCl$. If we take $20 \,mL$ of the same solution of $NaOH$, the amount of $HCl$ solution (the same solution as before) required to neutralise it will be ...... $mL$