When a strip of red-brown metal X is placed in a colourless salt solution $Y N O_{3}$ then metal Y is set free and a blue coloured salt solution $X (N O_{3})_{2}$ is formed. The liberated metal Y forms a shining white deposit on the strip of metal X.

(a) What do you think metal X is ?

(b) Name the salt $Y N O_{3}$ .

(c) What could be metal Y ?

(d) Name the salt $X (N O_{3})_{2}$ .

(e) what type of reaction takes place between metal X and salt soultion $Y N O_{3}$ ?

Open in App

Solution

(a) Copper.
(b) Silver nitrate.
(c) Silver.
(d) Copper nitrate.
(e) Displacement reaction.

Suggest Corrections

Similar questions

Q. A strip of metal X is dipped in a blue coloured salt solution YSO₄. After some time, a layer of metal Y from the salt solution is formed on the surface of metal strip X. Metal X is used in galvanisation whereas metal Y is used in making electric wires. Metal X and metal Y together form an alloy Z.
(a) What could metal X be?
(b) What could metal Y be?
(c) Name the metal salt YSO₄.
(d) What type of chemical reaction takes place when metal X reacts with salt solution YSO₄? Write the equation of the chemical reaction involved.
(e) Name the alloy Z.

Q. A metal salt MX when exposed to light splits up to form metal M and a gas X₂. Metal M is used in making ornaments whereas gas X₂ is used in making bleaching powder. The salt MX is itself used in black and white photography.
(a) What do you think metal M is?
(b) What could be gas X₂?
(c) Name the metal salt MX.
(d) Name any two salt solutions which on mixing together can produce a precipitate of salt MX.
(e) What type of chemical reaction takes place when salt MX is exposed to light? Write the equation of the reaction?

Two metals X and Y form the salts $X S O_{4}$ and $Y_{2} S O_{4}$ , respectively. The solution of salt $X S O_{4}$ is blue in colour whereas that of $Y_{2} S O_{4}$ is colourless. When barium chloride solution is added to $X S O_{4}$ solution, then a white precipitate Z is formed along with a salt which turns the solution green. And when barium chloride solution is added to $Y_{2} S O_{4}$ solution, then the same white precipitate Z is formed along with colourless common salt solution.

(a) What could the metal X and Y be ?

(b) Write the name and formula of salt $X S O_{4}$ .

(d) What is the name and formla of white precipitate Z ?

(e) Write the name and formula of the salt which turns the solution green in the first case.

Q. Two metals X and Y form the salts XSO₄ and Y₂SO₄, respectively. The solution of salt XSO₄ is blue in colour whereas that of Y₂SO₄ is colourless. When barium chloride solution is added to XSO₄ solution, then a white precipitate Z is formed alongwith a salt which turns the solution green. And when barium chloride solution is added to Y₂SO₄ solution, then the same white precipitate Z is formed alongwith colourless common salt solution.
(a) What could the metals X and Y be?
(b) Write the name and formula of salt XSO₄.
(c) Write the name and formula of salt Y₂SO₄.
(d) What is the name and formula of white precipitate Z?
(e) Write the name and formula of the salt which turns the solution green in the first case.

Join BYJU'S Learning Program

(a) Copper. (b) Silver nitrate. (c) Silver. (d) Copper nitrate. (e) Displacement reaction.

(a) Copper.
(b) Silver nitrate.
(c) Silver.
(d) Copper nitrate.
(e) Displacement reaction.