Chemistry, asked by diyakhrz12109, 3 months ago

ғʀᴀᴍᴇ ᴀɴʏ 5 ᴄʜʀᴍɪᴄᴀʟ ᴇǫᴜᴀᴛɪᴏɴ ᴡɪᴛʜ 5 ᴇxᴀᴍᴘʟᴇs​

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Answered by Anonymous
4

Answer:

(i) Change in colour: Reaction between lead nitrate solution and potassium iodide solution. Pb(NO3)2(aq)+2KI → PbI2(s)+2KNO3(aq) In this reaction, colour changes from colourless to yellow. (ii)Change in temperature: Action of dilute sulphuric acid on zinc. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2 In this reaction, heat is evolved (iii) Formation of precipitate: Action of barium chloride on sodium sulphate. BaCl2(aq) +Na2SO4(aq) → BaSO4(s) +2NaCl(aq) BaSO4(s) -

Explanation:

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Answered by Anonymous
2

Answer:

(i) Change in colour: Reaction between lead nitrate solution and potassium iodide solution. Pb(NO3)2(aq)+2KI → PbI2(s)+2KNO3(aq) In this reaction, colour changes from colourless to yellow. (ii)Change in temperature: Action of dilute sulphuric acid on zinc. Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2 In this reaction, heat is evolved (iii) Formation of precipitate: Action of barium chloride on sodium sulphate. BaCl2(aq) +Na2SO4(aq) → BaSO4(s) +2NaCl(aq) BaSO4(s)

Explanation:

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