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8090 [49]
3 years ago
14

The purpose of an indicator is to A. test and determine the pH values of unknown solutions. B. help acids and bases dissociate w

hen placed in water. C. make a solution more neutral. D. conduct electricity.

Chemistry
2 answers:
ad-work [718]3 years ago
7 0

Answer is: A. test and determine the pH values of unknown solutions.

Acid-base indicators are usually weak acids or bases and they are chemical detectors for hydrogen or hydronium cations (pH or pOH values).

Example for acid-base indicator is phenolphthalein (molecular formula C₂₀H₁₄O₄). Phenolphthalein is colorless in acidic solutions and pink in basic solutions.

Another example is methyl orange. It is red colour in acidic solutions and yellow colour in basic solutions.

IRISSAK [1]3 years ago
3 0

Answer: The correct answer is option A.

Explanation:

An indicator is a substance which changes color to show any chemical change happening in the reaction. It gives the visible sign of any change.

Hence, main role of indicators from the given options is to determine the pH values of unknown solutions.

pH is the scale which shows the acidity and basicity of the solutions. It is the negative logarithmic value of hydrogen ion concentration of the solution. More the hydrogen ion concentration, less will be the pH and more acidic will be the solution.

pH=-\log[H^+]

The scale ranges from 0 to 14.

The value ranging from 0 to 6.9 are acidic solutions.

The value ranging from 7.1 to 14 are basic solutions.

The value having pH = 7 is neutral solution.

Indicator used to determine the pH of the solution is pH-paper which is shown in the image.

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Answer:

the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

Explanation:

Given: radius of disk, R = 2.0 cm = 2 × 10⁻² cm, surface charge density,σ = 6.3 μC/m² = 6.3 × 10⁻⁶ C/m², distance on central axis, z = 12 cm = 12 × 10⁻² cm.

The electric field, E at a point on the central axis of a charged disk is given by E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  })

Substituting the values into the equation, it becomes

E = σ/ε₀(1 - \frac{z}{\sqrt{z^{2} + R^{2} }  }) = 6.3 × 10⁻⁶/8.854 × 10⁻¹²(1 - \frac{0.12}{\sqrt{0.12^{2} + 0.02^{2} } }) = 7.12 × 10⁵(1 - \frac{0.12}{0.1216}) = 7.12 × 10⁵(1 - 0.9864) = 7.12 × 10⁵ × 0.0136 = 0.0968 × 10⁵ = 9.68 × 10³ N/C = 9.68 kN/C

Therefore, the electric field at Z = 12 cm is E =   9.68 × 10³ N/C = 9.68 kN/C

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