Questions from the first batch - Chemistry📌✅
This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
ScanSolve AI Answer
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1. The oxides that are only soluble in basic oxide are acidic oxides.
2. Soda lime is used instead of caustic soda because it is less deliquescent, less corrosive, and has a higher capacity for absorbing carbon dioxide and moisture.
3. Hydrogen reacts with chlorine to form hydrogen chloride under the condition of light (especially UV light) or heat.
4. Ethylethanoate reacts with ammonia to form ethanamide (acetamide) and ethanol.
$$\text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NH}_3 \to \text{CH}_3\text{CONH}_2 + \text{CH}_3\text{CH}_2\text{OH}$$
5. Water that does not lather easily is hard water.
6. A strong acid is an acid that completely dissociates (ionizes) in water to produce hydrogen ions ($\text{H}^+$).
7. Electrolysis of brine (concentrated aqueous sodium chloride) gives chlorine gas ($\text{Cl}_2$), hydrogen gas ($\text{H}_2$), and sodium hydroxide ($\text{NaOH}$).
8. The electronegativity series describes the tendency of an atom to attract electrons in a chemical bond. Electronegativity generally increases across a period and decreases down a group in the periodic table.
9. Faraday's laws of electrolysis are used to calculate the amount of substance produced or consumed at an electrode during electrolysis based on the quantity of electricity passed.
10. The extraction of sulfur is done by the Frasch process.
11. A neutralization reaction is a chemical reaction between an acid and a base that produces a salt and water.
12. The term used for the combination of two or more elements to form a compound is synthesis.
13. Calcium carbonate ($\text{CaCO}_3$) is largely insoluble in water and does not react significantly with water under normal conditions.
14. Yes, sodium bicarbonate ($\text{NaHCO}_3$) will react with a base to give a salt and water. For example:
$$\text{NaHCO}_3(\text{aq}) + \text{NaOH}(\text{aq}) \to \text{Na}_2\text{CO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l})$$
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