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In lab H(2)O(2) is prepared by...

In lab `H_(2)O_(2)` is prepared by

A

Cold `H_(2)SO_(4) + BaO_(2)`

B

HCl + `BaO_(2)`

C

Conc `H_(2)SO_(4) + Na_(2)O_(2)`

D

`H_(2) + O_(2)`

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The correct Answer is:
To solve the question regarding the preparation of hydrogen peroxide (H₂O₂) in the laboratory, we will analyze the given options step by step. ### Step 1: Analyze Option 1 - **Option 1:** Cold H₂SO₄ + Barium Peroxide - **Reaction:** When barium peroxide (BaO₂) reacts with cold dilute sulfuric acid (H₂SO₄), it produces barium sulfate (BaSO₄) and hydrogen peroxide (H₂O₂). - **Equation:** \[ BaO_2 + H_2SO_4 \rightarrow BaSO_4 + H_2O_2 \] - **Conclusion:** This option is correct as it successfully produces hydrogen peroxide. ### Step 2: Analyze Option 2 - **Option 2:** HCl + Barium Peroxide - **Reaction:** Barium peroxide does not react with hydrochloric acid (HCl) to produce hydrogen peroxide. It primarily reacts with sulfuric acid, phosphoric acid, or carbonic acid. - **Conclusion:** This option is incorrect. ### Step 3: Analyze Option 3 - **Option 3:** Concentrated H₂SO₄ + Sodium Peroxide - **Reaction:** Sodium peroxide (Na₂O₂) reacts with cold dilute sulfuric acid to produce hydrogen peroxide and sodium sulfate (Na₂SO₄). However, it does not react with concentrated sulfuric acid. - **Equation:** \[ Na_2O_2 + H_2SO_4 \rightarrow H_2O_2 + Na_2SO_4 \] - **Conclusion:** This option is incorrect as concentrated sulfuric acid does not yield hydrogen peroxide. ### Step 4: Analyze Option 4 - **Option 4:** H₂ + O₂ - **Reaction:** The reaction between hydrogen (H₂) and oxygen (O₂) produces water (H₂O) and not hydrogen peroxide. - **Equation:** \[ 2H_2 + O_2 \rightarrow 2H_2O \] - **Conclusion:** This option is incorrect as it does not produce hydrogen peroxide. ### Final Conclusion After analyzing all the options, the correct method for preparing hydrogen peroxide in the laboratory is: - **Correct Option:** Cold H₂SO₄ + Barium Peroxide (Option 1). ---
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The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " "of" H_(2)O_(2) " solution " =2xx " molarity of" H_(2)O_(2) solution What is the molarity of "11.2 V" H_(2)O_(2) ?

The strength of H_(2)O_(2) is expressed in several ways like molarity, normality,% (w/V), volume strength, etc. The strength of "10 V" means 1 volume of H_(2)O_(2) on decomposition gives 10 volumes of oxygen at 1 atm and 273 K or 1 litre of H_(2)O_(2) gives 10 litre of O_(2) at 1 atm and 273 K The decomposition of H_(2)O_(2) is shown as under : H_(2)O_(2)(aq) to H_(2)O(l)+(1)/(2)O_(2)(g) H_(2)O_(2) can acts as oxidising as well as reducing agent. As oxidizing agent H_(2)O_(2) is converted into H_(2)O and as reducing agent H_(2)O_(2) is converted into O_(2) . For both cases its n-factor is 2. :. "Normality " " of " H_(2)O_(2) solution =2xx "molarity of" H_(2)O_(2) solution What is thepercentage strength (%w/V) of "11.2 V" H_(2)O_(2)

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