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Which one of the following must be added to dilute hydrochloric acid to produce hydrogen?

A. Iron

B. Iron sulfide

C. Copper chloride

D. Sulfur

Answer Explanation:

A. Iron (Correct Answer): Iron can react with hydrochloric acid to produce hydrogen gas according to the following reaction:

2HCl + Fe => FeCl2 + H2

So, adding iron to dilute hydrochloric acid would produce hydrogen gas.

B. Iron sulfide: Iron sulfide may react with hydrochloric acid, but it does not directly produce hydrogen gas. The reaction would likely produce hydrogen sulfide gas instead.

C. Copper chloride: Copper chloride does not react with hydrochloric acid to produce hydrogen gas. The reaction between copper chloride and hydrochloric acid would likely produce copper chloride and hydrogen chloride gas.

D. Sulfur: Sulfur does not react with hydrochloric acid to produce hydrogen gas. The reaction between sulfur and hydrochloric acid would likely produce hydrogen sulfide gas.

Therefore, the Correct Answer is A.

More Questions on TEAS 7 Science

  • Q #1: Why is KOH a strong alkali?

    A. Because it reacts vigorously with acids

    B. Because it forms a basic solution when dissolved in water

    C. Because in solution it fully dissociates into K+ and OH-

    D. Because it forms insoluble precipitates with metal cations

    Answer Explanation

    A) Because it reacts vigorously with acids: While KOH does react with acids to form salts and water, the strength of an alkali is not solely determined by its reactivity with acids.

    B) Because it forms a basic solution when dissolved in water: This statement is true, but it does not fully explain why KOH is considered a strong alkali. Many compounds can form basic solutions when dissolved in water.

    C) Because in solution it fully dissociates into K+ and OH- (Correct Answer): KOH is considered a strong alkali because it fully dissociates in aqueous solution into potassium ions (K+) and hydroxide ions (OH-). This dissociation leads to a high concentration of hydroxide ions in solution, making it strongly alkaline.

    D) Because it forms insoluble precipitates with metal cations: This statement describes the formation of insoluble hydroxide precipitates when alkali solutions are added to solutions of metal salts. However, it does not fully explain why KOH itself is considered a strong alkali.

  • Q #2: What color do acids make litmus paper turn?

    A. Red

    B. Yellow

    C. Blue

    D. Black

    Answer Explanation

    A) Red (Correct Answer): Acids turn litmus paper red. Litmus paper is a common indicator that changes color in response to changes in acidity or alkalinity. Acids change the blue litmus paper to red.

    B) Yellow: Acids do not typically turn litmus paper yellow. Litmus paper is generally unaffected by acids in a way that would cause it to turn yellow.

    C) Blue: Acids do not turn litmus paper blue. Instead, they change blue litmus paper to red.

    D) Black: Acids do not turn litmus paper black. Litmus paper is not expected to change to black in the presence of acids.

  • Q #3: How can solid lead iodide be separated from solution?

    A. Filtration

    B. Distillation

    C. Decantation

    D. Chromatography

    Answer Explanation

    A) Filtration: Solid lead iodide can be separated from the solution using filtration. Filtration involves passing the mixture through a filter paper or porous material, allowing the solid particles to be retained while the liquid passes through.

    B) Distillation: Distillation is a process used to separate components of a mixture based on differences in their boiling points. It is not typically used to separate solid lead iodide from a solution.

    C) Decantation: Decantation involves carefully pouring off the liquid portion of a mixture, leaving the solid behind. While it can be used to separate solid and liquid phases, it may not be as effective as filtration for separating fine solid particles like lead iodide.

    D) Chromatography: Chromatography is a technique used to separate and analyze mixtures based on differences in the components' distribution between two phases: a stationary phase and a mobile phase. It is not typically used for separating solid lead iodide from a solution.