Hydrogen practice · Medium

Option A ($CO$ and $H_2$) is correct because synthesis gas, also known as syngas, is a mixture of carbon monoxide (CO) and hydrogen (H₂).

Option B is correct because when H2O2 acts as an oxidizing agent, it is reduced to H2O.

Option A is correct because only $C{u^{2 + }}$ precipitates as CuS when HCl is added followed by H2S in the given aqueous solution.

Option D (All of the above) is correct because all the listed methods can be used to obtain pure hydrogen gas.

Option D ($\mathrm{H}_2 \mathrm{O}_2$) is correct because hydrogen peroxide liberates oxygen upon heating, turns acidic KI solution brown due to iodine formation, and reduces acidified $\mathrm{KMnO}_4$ solution.

Option D is correct because fluoride ions convert hydroxyapatite to fluorapatite, which is represented by the formula $\left[ {3\{C{a_3}{{\left( {P{O_4}} \right)}_2}\} \cdot Ca{F_2}} \right]$.

Option D (${97^0}$) is correct because the O-O-H bond angle in H2O2 is approximately 97 degrees due to the non-planar structure and repulsion between lone pairs.

The correct answer is A ($2{H}_2{O}+2 {Na} \rightarrow 2{NaOH}+{H}_2$) because in this reaction, water is reduced to hydrogen gas.

Option A (Both S-I and S-II are correct) is correct because hydrogen does resemble both alkali metals and halogens in its properties, and it exists in atomic form only at high temperatures.

Option B (Permutit process) is correct because the Permutit process uses zeolites to exchange ions and remove permanent hardness.

Option A (75% Ortho Hydrogen + 25% para Hydrogen) is correct because at room temperature, ordinary hydrogen exists as a mixture of ortho and para hydrogen in a 3:1 ratio.

Option A (Weak acid) is correct because hydrogen peroxide ($H_2O_2$) can donate protons, behaving as a weak acid.

Option A (${H_2}O,\,\,N{H_3},\,\,\,C{O_2}$) is correct because molecular sieves are designed to trap small molecules like water, ammonia, and carbon dioxide within their pores.

Option D (Oxygen, ozone) is correct because fluorine reacts with water to produce both oxygen and ozone as products.

Option B is correct. In reaction (A), H₂O₂ acts as an oxidizing agent, and in reaction (B), it acts as a reducing agent.

Option A (Passing $\mathrm{CO_2}$ into $\mathrm{BaO_2}$) is correct because hydrogen peroxide can be prepared in the laboratory by reacting barium peroxide with carbon dioxide and water.

Option D ($\mathrm{H}_2$) is correct because hydrogen is the component present in the greatest proportion in water gas.

Option D ($250 \mathrm{~mL}$) is correct because $20 \mathrm{~vol} \mathrm{~H}_2 \mathrm{O}_2$ means 1 L produces 20 L of $\mathrm{O}_2$ at STP, thus 0.25 L produces 5 L.

Option C (Reaction of zinc with aqueous alkali) is correct because it represents a common laboratory method for producing dihydrogen gas.

Option C ($H_2$) is correct because atomic hydrogen produced by the oxy-hydrogen flame can reach temperatures of 4000 K, suitable for welding metals with high melting points.

Option A (Basic solution) is correct because ionic hydrides react with water to form a metal hydroxide and hydrogen gas, resulting in a basic solution.

Option D (If Statement I is false but Statement II is correct) is correct because molarity can be less than, equal to, or greater than molality, and the ratio depends on the solvent's density.

Option C (It cannot be placed in first group of periodic table) is correct because Hydrogen can be placed in the first group due to its resemblance with alkali metals.

Option B is correct because potassium superoxide (KO2) reacts with water to produce potassium hydroxide, hydrogen peroxide, and oxygen gas.

Option D (Passing silen electric discharge through Hydrogen at low pressure) is correct because atomic hydrogen is produced by passing hydrogen gas through a silent electric discharge at low pressure, breaking the H-H bond.

Option C is correct because $H_2O_2$ does not decompose $Na_2CO_3$ to give $CO_2$ directly, and it is not a strong acid.

Option C (Both (a) and (b)) is correct because a 5.6 volume H₂O₂ solution is both 1.7% weight by volume and 1 N.

Option B is correct because water's high dielectric constant allows it to effectively reduce the electrostatic forces between ions, making it a good solvent for ionic compounds and therefore a universal solvent.

Option C (Carbon monoxide) is correct because in the water-gas shift reaction, carbon monoxide reacts with steam to produce hydrogen gas and carbon dioxide.

Option B (Na₂O₂) is correct because sodium peroxide reacts with water to produce hydrogen peroxide and sodium hydroxide.