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xxTIMURxx [149]
3 years ago
13

Potassium hydroxide is classified as an Arrhenius base because KOH contains(1) OH- ions (2) O2- ions (3) K+ ions (4) H+ ions

Chemistry
2 answers:
Readme [11.4K]3 years ago
6 0

Answer : The correct option is, (1) OH^- ions.

Explanation :

According to the Arrhenius concept, an acid is a type of substance which ionizes in water to gives H^+ ions (hydronium ions or hydrogen ions) and a base is a type of substance which ionizes in water to gives OH^- ions (hydroxide ions).

As we know that KOH is a base which ionizes in water to gives OH^- ions (hydroxide ions) and K^+ ions.

Hence, potassium hydroxide is classified as an Arrhenius base because KOH contains OH^- ions.

Sloan [31]3 years ago
3 0
The answer is (1)OH- ions. The base has more OH- ions while the acid has more H+ ions. And if you dissolve them into water, the pH of the solution is also different. Base has pH>7 and acid has pH<7 at room temperature.
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A student placed 18.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
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Answer:

1.30464 grams of glucose was present in 100.0 mL of final solution.

Explanation:

Molarity=\frac{moles}{\text{Volume of solution(L)}}

Moles of glucose = \frac{18.5 g}{180 g/mol}=0.1028 mol

Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)

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A 30.0 mL sample of above glucose solution was diluted to 0.500 L:

Molarity of the solution before dilution = M_1=1.208 mol

Volume of the solution taken = V_1=30.0 mL

Molarity of the solution after dilution = M_2

Volume of the solution after dilution= V_2=0.500L = 500 mL

M_1V_1=M_2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{1.208 mol/L\times 30.0 mL}{500 mL}

M_2=0.07248 mol/L

Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:

Volume of solution = 100.0 mL = 0.1 L

0.07248 mol/L=\frac{\text{moles of glucose}}{0.1 L}

Moles of glucose = 0.07248 mol/L\times 0.1 L=0.007248 mol

Mass of 0.007248 moles of glucose :

0.007248 mol × 180 g/mol = 1.30464 grams

1.30464 grams of glucose was present in 100.0 mL of final solution.

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