Question

State your observation.

i. When hydrogen sulphide gas is passed through lead acetate solution.

ii. When sulphur dioxide is passed over red rose petals.

iii. When sulphur dioxide passed through acidified potassium dichromate.

iv. When bottle of ferric chloride kept open.

v. When acetic acid is added to methyl orange.​

Answer 1

Answer:

When hydrogen sulphide gas is passed through lead acetate solution.

is correct answer please mark as best answer and

Answer 2

Answer 1 :-

1) When hydrogen sulphide gas passed through lead acetate .

H2S + Pb( CH3COOH )^2 = PbS + 2CH3COOH

Here, You can see that it forms a black colour precipitate of Lead sulphide ( PbS) .

Here, Lead is more reactive than hydrogen so it displaces hydrogen and displacement of ions also take place here.

So it is double displacement and precipitate reaction.

Answer 2 :-

When sulphur di oxide passed over the rose petals leave then it decolourise the colour of rose petals and it turns white in colour but it only happens in the absence of moisture.

When SO2 got moisture then it regains it colours and again turn red.

Answer 3 :-

When sulphur dioxide passed through acidified potassium dichromate then it forms chromium sulphate, potassium sulphate and water

3SO2 + K2Cr2O7 + H2SO4 = Cr2( SO4)3 + K2SO4 + H2O

It changes the colour of acidified potassium dichromate into orange colour.

Answer 4 :-

When bottle of ferric chloride kept open then it is changes into solid to liquid form. As it react with moisture and form ferric chloride and gain some water molecules

FeCl3 + nH2O = FeCl3 . nH2O

Answer 5 :-

Methyl orange is a synthetic indicator which is used to check the nature of substances that it is acidic or basic. So When acidic acid is added to methyl orange it changes it colour from orange to red colour .$.$